SOME PROBLEMS YOU SHOULD BE ABLE TO DO

SOME PROBLEMS YOU SHOULD BE ABLE TO DO I ve attempted to make a list of the main calculations you should be ready for on the exam, and included a handful of the more important formulas. There are no examples here: for that, check my lecture worksheets accompanying the corresponding sections. Please let me know if you notice any mistakes or omissions! I don t recommend studying exclusively from this list it is inevitable that I have left something out. Be sure to look at old exams, problem sets, etc. as well. A few topics we covered only briefly are in italics. It couldn t hurt to look, but don t worry about these too much. Chapter. and.: Vectors in the plane and vectors in three dimensions. Compute basic vector operations: addition/subtraction, scalar multiplication, magnitude. Know how to interpret these both algebraically and geometrically. Find vector from point P to point Q. Find a unit vector in the direction of a given vector (or a vector with some other specified length) Midpoint of a line segment. Equation of a sphere. Statics problems (balancing forces)..3: Dot products. Compute the dot product of two vectors algebraically (a b + a b + a 3 b 3 ) and geometrically ( a b cos θ) Compute the angle between two vectors Compute and interpret proj v u and scal v (u): proj v u = ( u v v v Work done by a constant force. Parallel and normal components of a force..4: Cross products. ) v, scal v u = u cos θ = u v v. Compute the cross product of two vectors. Find the area of a triangle with given vertices. Find the area of a parallelogram with given vertices. version: 989c3b / 7-4-8 :46:48-5

SOME PROBLEMS YOU SHOULD BE ABLE TO DO.5: Lines and curves in space. Parametrize a straight line You always need to know two things: a point r that the line goes through, and a vector v in the direction of the line (often you will need to do some work to find v, depending on what information is given to you!). Then use the formula r(t) = r +tv (and think about the bounds) between two given points through a point and perpendicular to a given plane through a point and perpendicular to two given lines tangent to a curve r(t) at t = a given as the intersection of two planes Parametrize other simple curves (circles) Check whether lines intersect Take a limit (by taking the limit of each component).6: Calculus of vector-valued functions. Find the derivative r (t) = dr dt Find tangent vector to a curve at time t Find unit tangent vector to a curve Integrate a vector-valued function (by integrating each component).7: Motion in space. Find velocity, acceleration, and speed. Find position from velocity and velocity from acceleration, given an initial condition v() or a(). Movement in a gravitational field..7: Length of curves. The arc length of r(t) = f(t), g(t), h(t) is ˆ b a ˆ b f (t) + g (t) + h (t) dt = r (t) dt. This is the distance traveled by a particle moving along the curve from t = a to t = b. Find arc length in polar. Check whether a path is parametrized by arc length. a.: Planes and surfaces. Chapter Find the equation for a plane with normal vector a, b, c passing through (x, y, z ). Find the equation for a plane through three given points Find the equation for a line given as the intersection of two planes. Check whether two planes are orthogonal. Equations for cylinders

.: Quadric surfaces. SOME PROBLEMS YOU SHOULD BE ABLE TO DO 3 Sketch the graph of a quadric surface by drawing the xy-, xz-, and yz-traces, or some other traces parallel to the coordinate planes. Find the intersection of a line with a quadric surface..3: Limits of functions of several variables. Compute the limit of a function of two variables. Use the two-path test to show that a limit does not exist..4 &.5: Partial derivatives. Compute the partial derivative of a function f(x, y) Compute the four second-order partial derivatives of a function Use the chain rule to compute partial derivatives of functions of two or three variables. Apply implicit differentiation to an expression F (x, y) =.6: Directional derivatives and the gradient. Compute (and interpret) the directional derivative D u f Find the gradient f. Find direction of fastest ascent/descent and the rate of fastest ascent/descent for a function f(x, y). Find directions of increase. Sketch level curves of a function, and tangent directions to level curves..7: tangent planes and linear approximation. Find tangent plane to implicit surface F (x, y, z) = at a point (x, y, z ). Find tangent plane to explicit surface z = f(x, y) at a point (x, y, z ). Find the linear approximation to f(x, y) near a point (a, b) and use this to approximate values of f. Work with differentials..8: Maxima and minima. Find the critical points of a function. Use the second derivative test to classify the critical points as max/min/saddle. Find the global max/min of a function f(x, y) on a region R: () Find critical points inside R () Find relative max/min on the boundary (3) Make a list of all the interesting points and compute values of f there to find the true max and min..9: Lagrange multipliers. Use Lagrange multipliers to find the maxima and minima of f(x, y) subject to the constraint g(x, y) = Key formula: f(x, y) = λ g(x, y). Translate this into three equations in the three variables x, y, and λ and solve Translate a word problem into a constrained optimization problem.

4 SOME PROBLEMS YOU SHOULD BE ABLE TO DO 3.: Double integrals. Chapter 3 Set up and compute the double integral of a function f(x, y) on a rectangular region a x b, c y d. Find the average value of f(x, y) on such a region. Fubini s theorem: you can change the order of the integral. 3.: Double integrals over other regions. Evaluate a double integral where the inner bounds depend on the outer variable. Sketch the region of integration for a double integral, given the bounds. Give the bounds on a double integral, given a description of the region. Change the order of integration in a double integral over a non-rectangular region. 3.3: Double integrals in polar coordinates. Sketch the region of integration for a double integral in polar Evaluate a double integral in polar. Convert a double integral in rectangular coordinates into polar: () Write the bounds in polar () Write the function in polar (substitute r cos θ for x, r sin θ for Y (3) Write r dr dθ instead of dx dy. 3.4: Triple integrals. Evaluate a triple integral. Set up the bounds for a triple integral in rectangular coordinates (important shapes: rectangular regions, tetrahedral regions) Change the order of integration in a triple integral. 3.5: Cylindrical spherical coordinates. Find the rectangular coordinates for a point given (r, θ, z) x = r cos θ, y = r sin θ, z = z. Find the cylindrical coordinates for a point given (x, y, z): Set up the bounds for a triple integral in cylindrical coordinates (Important shapes: cylinder, paraboloid.) Convert a rectangular integral into cylindrical coordinates and evaluate (usual three steps: convert the bounds, convert the function, use r dr dθ dz). Find the rectangular coordinates for a point given (ρ, θ, φ): x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos φ. Set up the bounds for a triple integral into spherical coordinates (Important shapes: spheres, hemispheres, ice cream cones.) Convert a rectangular integral into spherical coordinates and evaluate (usual three steps: convert the bounds, convert the function, use ρ sin φ dρ dφ dθ.

SOME PROBLEMS YOU SHOULD BE ABLE TO DO 5 3.7: Change of variables in multiple integrals. Given a change of coordinates x = g(u, v), y = h(u, v), compute the Jacobian J(u, v): x x J(u, v) = u v y y = x y u v x y (x, y) = v u (u, v). u v Change an integral in terms of x and y to an integral in terms of u and v: f(x, y) da = f(g(u, v), h(u, v)) J(u, v) da. R S Find the uv bounds for an integral from the xy bounds (often helpful; convert the equation for each edge of the region into an equation involving uv, and sketch the corresponding region in the uv-plane). 4.: Vector fields. Chapter 4 Sketch a vector field from the equation. Write a formula for a vector field given a description. Compute the gradient field F = f(x, y) for a function f(x, y) 4.: Line integrals. ˆ Compute the line integral C f(x, y) ds of a scalar function along a path. Method: parametrize the path C by r(t) = x(t), y(t). Use bounds as parametrization as bounds on integral, plug in x and y to f, and use r (T ) dt for the ds: ˆ ˆ b f ds = f(x(t), y(t)) r (t) dt. Compute the line integral of a C a ˆ C F dr of a vector function along a path (i.e. a circulation integral). Parametrize C by r(t), then plug in x and y to F, and dot that with r (t) to get a function of t. Compute the flux of a vector field across a path C Same method, but use n = y(t), x(t) instead of r (t) in the above. 4.3: Conservative vector fields. Check whether a D vector field F = f, g is conservative. Check whether a 3D vector field F = f, g, h is conservative. Find a potential function φ(x, y) for a conservative vector field. Use the fundamental theorem for line integrals to compute the line integral of a conservative field along a path C: ˆ F dr = φ(end) φ(begin). Note: integral is indepedent of path. C

6 SOME PROBLEMS YOU SHOULD BE ABLE TO DO 4.4: Green s theorem. Compute the divergence and curl of a D vector field (both are scalar functions, not vetor fields). Be able to apply both versions of Green s theorem to double integrals over a region R: Circulation form: F dr = curl F da Flux form: C C R F n ds = R div F da Use Green s theorem to turn a line integral over a complicated path from A to B into a line integral over a simpler path from A to B and a double integral over the region between the two paths. 4.5: Divergence and curl. Compute the divergence of a 3D vector field F (this is a scalar function). Compute the curl of a 3D vector field: i j k F = x y z f g h ( h = y g ) i + z NB: This is another vector field. ( f z h x ) ( g j + x f ) k y 4.6: Surface integrals. Parametrize a surface in 3D () Cylinder of radius a; x = a cos u, y = a sin u, z = v () Sphere of radius a: x = a sin u cos v, y = a sin u sin v, z = a cos u. (3) Graph z = f(x, y) (for example, z = (x + y )): x = u, y = v, z = f(u, v). Next, know how to compute the tangent vectors t u, t v, and the normal vector t u t v. Compute surface integrals of scalar functions: f(x, y, z) ds. Steps to convert to a double integral in u, v: () The bounds are the bounds for your parametrization () Rewrite the function in terms of u and v (plug in your parametrization for x, y, and z) (3) Use t u t v du dv for ds. Compute flux integrals of vector fields across a surface: F n ds. Steps to convert to a double integral in u, v: () The bounds are the bounds for your parametrization () Rewrite the function in terms of u and v (plug in your parametrization for x, y, and z) (3) Plug in x, y, z from parametrization to F (4) Dot that with t u t v and integrate the result du dv. S S

SOME PROBLEMS YOU SHOULD BE ABLE TO DO 7 4.7: Stokes theorem. Be able to apply Stokes theorem, either to turn a computation of the flux of a curl into a circulation integral, or to turn a circulation integral into the flux of a curl. ( F) n ds = F dr. S Know how to compute both sides! Know which way to go around the curve C to make this work (right-hand rule). Apply Stokes theorem on regions with multiple boundaries. 4.8: Divergence theorem. Be able to apply divergence theorem, in either direction. F n ds = F dv. Know how to compute both sides! S D C

Math (Lesieutre).: Vectors in the plane January 9, 7 Problem. Let u =, 3 and v =,. a) Compute u + v, both geometrically and algebraically. Do your answers match? We have u + v =, 3 +, =, 3 +, 4 =,. b) Compute u v, both geometrically and algebraically. For this one, u v =, 3, =, 5. c) Compute 3u, both geometrically and algebraically. Multiplying both components by 3, we obtain 3u = 3, 9. d) What is the magnitude of v (i.e. v?) Does the formula match the picture? For this one, the formula gives v = + ( ) = 4 =. This obviously matches the length we get if we draw the vector, which points straight towards the bottom of the page and has length. Problem. a) What is the vector pointing from (, ) to (4, 3)? Again, try to do this one by drawing the picture. We want the vector v = 4, 3, = 3, 4. b) Find a unit vector parallel to the vector in your answer from (a). We need u = v v = 3, 4 3, 4 3 = = 3 + ( 4) 5 5, 4. 5 c) Find a vector with length 7 parallel to the vector in your answer from (a). We want 7 times our answer from (b), which is w = 5, 8. 5 Problem 3. Relative to the air, an airplane is flying 3 degrees west of north, with speed 5 MPH. The wind is traveling due north at MPH. What is the velocity vector of the airplane relative to the ground?

First we need to translate to find the components of the airplane vector. In polar coordinates, our vector has angle 9 + 3 =. The x-component is 5 cos = 5( /) = 5. The y-component is 5 sin = 5( 3/) = 5 3. So the velocity relative to the air is 5, 5 3. The velocity of the air relative to the ground is,. Let me write v X/Y v plane/ground = v plane/air + v air/ground = for the speed of X relative to Y. As we saw in class, 5, 5 3 +, = 5, 5 3. Problem 4. A -pound weight is suspended from two strings, each making a 45 degree angle with the ceiling. How much force is exerted on the mass by each of the strings? This is a classic sort of problem. Since the situation is symmetric, the two forces from the strings are equal in magnitude, but in different directions. Let s say M is the answer. The two force vectors from the strings are F = M, M F = M, M (the first of those is M cos 45 for the left string, etc; similar to the previous problem) The gravitational force is g =, : a force of directly downward. Adding these up we should get : if the object isn t moving, the forces have to balance out: F + F + g =. So, M +, =,, which means M =, so M = / = 5. (This tells use how strong each string needs to be to keep the mass from falling: each individually is holding the same force as if it were supporting a single 5 7.7.

Math (Lesieutre).: Vectors in three dimensions January, 7 Problem. Consider the two points in three dimensions with coordinates P = (,, 3) and Q = (,, ). a) What is P Q? What is QP? How do these differ? P Q is given by (,, ) (,, 3) =,,. QP is given by (,, 3) (,, ) =,,. These differ by a factor of : they have the same length, but point in opposite directions. b) What is the length of P Q? It s given by ( ) + + ( ) = 8 =. c) Find a vector of length 3 parallel to P Q. We want 3 P Q/ P Q = 6,, 6 = 3,, 3 d) What is the midpoint of the line segment between P and Q? The midpoint formula says it s (( + ( ))/, ( + )/, (3 + )/) = (,, ).. Problem. Describe the sphere with equation (x ) + (y + ) + (z 3) = 5. The sphere has center (,, 3) and radius 5. Problem 3. Relative to the air, an airplane is flying 3 degrees west of north, with speed 5 MPH. The wind is traveling due north at MPH. What is the velocity vector of the airplane relative to the ground? First we need to translate to find the components of the airplane vector. In polar coordinates, our vector has angle 9 + 3 =. The x-component is 5 cos = 5( /) = 5. The y-component is 5 sin = 5( 3/) = 5 3. So the velocity relative to the air is 5, 5 3. The velocity of the air relative to the ground is,. Let me write v X/Y v plane/ground = v plane/air + v air/ground = for the speed of X relative to Y. As we saw in class, 5, 5 3 +, = 5, 5 3. Problem 4. A -pound weight is suspended from two strings, each making a 45 degree angle with the ceiling. How much force is exerted on the mass by each of the strings?

This is a classic sort of problem. Since the situation is symmetric, the two forces from the strings are equal in magnitude, but in different directions. Let s say M is the answer. The two force vectors from the strings are F = M, M F = M, M (the first of those is M cos 45 for the left string, etc; similar to the previous problem) The gravitational force is g =, : a force of directly downward. Adding these up we should get : if the object isn t moving, the forces have to balance out: F + F + g =. So, M +, =,, which means M =, so M = / = 5. (This tells use how strong each string needs to be to keep the mass from falling: each individually is holding the same force as if it were supporting a single 5 7.7. Problem 5. Let R be a parallelogram with legs given by the vectors u and v. Prove that both diagonals of R have the same midpoint. Place u and v with their tails at the origin. The diagonal from the origin to the far corner of the parallelogram is u + v, and so the position vector for the midpoint is (u + v)/. The second diagonal (oriented to start at u and end at v) is the vector v u. To get the position vector, we need to add half of this to the position of its tail: u+(v u)/ = (u+v)/. So the two answers match.

Math (Lesieutre).3: Dot products January 3, 7 Problem. For each of the following, try to estimate the dot product without making any calculations. Then check your answer using the formula. a) 3, 4 4, 3 These both have length 5, and point in roughly the same direction. That means cos θ is going to be positive and more or less close to. So we expect the dot product to be a little less than 5. The correct value is (3)(4) + (4)(3) = 4. b) 3,, 4 If you sketch these, you ll see that they re almost perpendicular, but not quite. The dot product will be small, at least compared to the lengths of the vectors. Positive or negative might be hard to tell unless you draw a good picture, but it should be slightly positive. The actual value is. c),,,, The first vector points along the x-axis, while the second lies in the yz-plane. The two vectors are orthogonal, and the dot product will be exactly. The formula confirms this. d),, 3,, Both vectors have length somewhere between 3 and 4. They point in basically opposite directions, so cos θ is going to be fairly close to. So the answer should be a negative number in the ballpark of. Multiplying it all out, you ll find the actual value is. Problem. Compute the angle between the two vectors in (a). (Your answer might be in terms of an cos ). We know that u v = u v cos θ, so cos θ = u v u v = 4 5. That means θ = cos (4/5), which is about 6. Problem 3. Suppose that v = a, b is a vector. What is v v? It s a, b a, b = a + b = v. This works for 3d vectors too!

Problem 4. Sketch the vectors u =, and v =,, and draw the vector proj v u. What is scal v u? (Sorry, I m omitting the drawing.) The formula for projection gives us proj v u = ( u v ) v = v v ( ) 3 3, =, 3. This is a vector in the same direction as v, and represents the part of u that s in the direction of v. For the scalar component, we use This is the length of proj v u. scal v u = u v v = 3 = 3. Problem 5. A force F =,, (Newtons) pushes an object from (,, ) to (3,, ) (meters). Calculate the work done (Joules). It s just W = F d =,,,, = 4. Problem 6. A pound block sits on a plane with slope 3 degrees. Compute the components of the gravitation force that are parallel and perpendicular to the plane. The force is F =,. The vector down the slope is v = 3,. The vector into the plane is u =, 3. We find and proj v F = proj u F = ( ) F v v = 5 3, 5 v v ( ) F u u = 5 3, 5. u u The one into the plane is usually called the normal component and written N. Does the object slide down the plane? It s a question of whether the frictional force is strong enough to overcome the force proj v F. This depends on the coefficient of friction, which takes us a little further into physics than we re really going to go... Problem 7. a) Describe the set of all vectors v for which v,, =. This is all vectors that are perpendicular to,,. These form the plane whose normal vector is,,.

b) For which vectors is proj,, = 5? This works for any vector 5, a, b ; the set of such vectors is the plane x = 5. 3

Math (Lesieutre).3: Dot products January 3, 7 Matt taught this day.

Math (Lesieutre).5: Lines and curves in space January, 7 Problem. Consider the line from (,, ) to (,, 3). a) Express a parametrization of the line in vector form. The direction vector is v = (,, 3) (,, ) =,,. Then we want the line r(t) =,, + t,,. This seems to work: when we plug in t =, we get,,, and when we plug in t =, we get,, 3. b) Express your parametrization as three functions x(t), y(t), and z(t). Our parametrization was r(t) = + t, + t, + t, so we want x(t) =, y(t) = + t, z(t) = + t. c) At what time t does the line cross the plane z =? We have z(t) = + t, so we just want to know when + t =, i.e. t = 9/. d) What is the projection of the line onto the xy-plane? It s x(t) = and y(t) = + t (we re in the xy-plane, so it s implicit that z = for all t). Problem. Sketch the parametrized curves indicated below. a) r(t) = cos t, sin t, for t 4π. This one is a circle contained in the horizontal plane z =. For t in the range in question, it goes around the circle twice, in a counterclockwise direction when viewed from above.

b) r(t) =,, t for t 5 This is just a straight line from (,, ) moving upwards. It ends at (,, 5). c) r(t) = cos t, sin t, t for t 4π. This one is a helix. It moves around in a circle, while simultaneously going upwards. Since we go to 4π, it makes two full cycles around. Problem 3. Do the two lines r (s) = s, + s, s 3, r (t) = t, t, t intersect?

To intersect, there must be values of s and t for which s = t, + s = t, and s 3 = t. This is an overdetermined system: there may or may not be any solutions. The first two are solved by s =, t =, but then the third equation isn t satisfied. So there is no s and t that meets both requirements. Problem 4. For what value of a do the two lines intersect? r (s) = 3 + s, + s, a + s, r (t) = t + a, t + a, What does it mean for them to intersect? It means that there is a value s and a value t for which r (s) = r (t) (they might pass through the same point, but at two different values of the parameter). So we need to solve the equations 3 + s = t, + s = t, and a + s = for s and t, in terms of a. For most values of a there are going to be no solutions, but for some special a there will be one. First we use the first two to solve for s and t (in principle our answer could involve a, but it s not going to this time). Subtracting the second equation from the first (or doing row reduction, or matrices,... ), we get s = + a and t =. Now we need to know whether the third coordinates are also equal for these values of s and t. Is a + s =? Well, s = + a, so we are asking whether a + =. This is the case if a = /. Again: if a has some other value, there is no s and t that make all the components equal. But if a = /, the lines do intersect. Problem 5. What is lim t + v(t), where v(t) = t, sin t t, e /t? We just take the limit of each one of the components, whcih gives us,,. 3

Math (Lesieutre).6: Calculus of vector valued functions January 3, 6 Problem. The position of a planet is given by r(t) = 3 cos t, sin t. Find a unit tangent vector to the path of the planet. First we find that This gives r (t) = 3 sin t, cos t. r (t) = 9 sin t + 4 cos t. Nothing to be done with that, really, but it gives r (t) r (t) = 3 cos t 9 sin t + 4 cos t, sin t. 9 sin t + 4 cos t Problem. Consider the function r(t) = cos t, sin t, t a) Sketch r(t) for t 4π. This one we did last time. It s the helix sketched below. b) Compute r (t) and r (t). Plug in t = and t = π/ to your answer. Do these make sense? Can you think of any other sanity checks? The derivatives are given by r (t) = sin t, cos t, r (t) = cos t, sin t,

When t =, we have r () =,,. This is a vector whose projection to the xy-plane points in the y-direction, but also has z part equal to. This seems to match up with the picture. Similarly, we find that r () =,,, which points in the negative x-direction and upwards. Seems plausible. You might also notice that r (t) = for all t. This also makes sense: the rate at which the particle is moving upwards is not changing. c) Find a unit tangent vector for the parametrized curve, in terms of t. We know that the tangent vector is given by r (t) = sin t, cos t,, for any t. The problem is that this isn t a unit vector, since r (t) = sin t + cos t + =. To get a unit vector, we want r (t) sin t r (t) =, cos t,. Problem 3. Consider the path given by r(t) = t, t, t 3. a) Write down an equation for the tangent line to r(t) at t =. The direction of the line is given by r (t) =, t, 3t. At t = this is,, 3. It goes through the point r() =,,, so the equation is given by x(t) =,, +,, 3 t. b) What is d (r(t),, 3 )? dt Let s(t) =,, 3. Then d (r(t) s(t)) = dt r (t) s(t) + r(t) + s (t) =, t, 3t,, 3 + t, t, t 3,, = + 4t + 9t. Problem 4. a) What is d dt x(t), in terms of x(t) and its derivatives? This is d dt x(t) x(t) = x(t) x (t) + x (t) x(t) = x(t) x (t). b) Check your answer to (a) for x(t) = cos t, sin t. Does your answer make sense? According to the product rule, it s x(t) x (t) + x (t) x(t) = cos t, sin t sin t, cos t + sin t, cos t cos t, sin t =.

c) Suppose that x(t) has constant length. What does this tell you about the relationship between x(t) and x (t)? It tells you that they are orthogonal for all values of t! Problem 5. Let s(t) = sin t, cos t,. Compute s(t) dt. We just integrate each component: sin t, cos t, = cos t, sin t, t + c, c, c 3. This undoes what we did in the second problem. velocity of a particle, r(t) is its position. We ll see next time that if s(t) is the 3

Math (Lesieutre).7: Motion in space January 5, 7 Problem. Consider the path r(t) = cos t, sin t, t. acceleration, all as functions of t. Compute the velocity, speed, and First let s find the velocity and acceleration, which are just a matter of taking derivatives. We get r(t) = cos t, sin t, t, v(t) = sin t, cos t, t, a(t) = cos t, sin t,. The speed should be a scalar function, and to get it we take the length of velocity, as a function of time: v(t) = ( sin t) + (cos t) + (t) = + 4t. This seems plausible, I think. When t is big, it s moving upwards faster and faster, which is reflected in the speed increasing as t increases. Problem. A particle moves in a circular pattern, given by r(t) = cos t, sin t. a) Compute the velocity, acceleration, and speed, as functions of t. Taking derivatives, we get The speed is r(t) = cos t, sin t, v(t) = sin t, cos t, a(t) = cos t, sin t. v(t) = ( sin t) + ( cos t) = 4 =, which doesn t depend on t. This makes sense, since it s moving at a constant rate. Notice that the velocity vector v(t) does depend on t: this reflects the fact that the direction is changing. b) Sketch the path of the particle. For t = π/4, draw the vectors r(t), v(t), and a(t). Do these seem to make physical sense? The vectors are r(t) =,, v(t) =,, and a(t) =,. The velocity vector is tangent to the circle at the point, which makes sense: velocity is supposed to be the tangent vector. The acceleration points in to the circle, which may seem a little weird at first, but this is how things work for circular motion. This is consistent with Newton s law F = ma: because gravitation force pulls the particle inward, the acceleration should also be inward.

Problem 3. Suppose that a particle moves in a straight-line path r(t) = r + vt. What are the velocity and acceleration? The velocity will just be the vector v, while the acceleration is the zero vector. (Because the particle is moving at constant velocity, there is no acceleration.) Problem 4. A particle has acceleration a(t) = t,. Suppose that at time, it has position r() =, 3 and velocity v(t) =,. Find r(t). We know that v(t) = a(t) dt = t, dt = t, t + C Since v() =,, we get, + C =,, so C =,. That means v(t) = t, t +, = t, t. Then t r(t) = v(t) dt =, t t 3 dt = 3, t t + C Plugging in t =, we get, + C = r() =, 3, which means that C =, 3. So at last we have t 3 r(t) = 3, t t t 3 +, 3 = 3 +, t t + 3. Problem 5. A batter hits a baseball with initial velocity,, (a pop-up down the right field line; let s say the units are ft/sec). a) What is a(t)? Assume the only force acting on the ball is gravity. It s a constant downward force,,, 3. The 3 here is gravitation acceleration. b) Solve for v(t) and r(t). Well, v(t) = a(t) dt =,, 3t + C. Since v() =,,, plugging in t = we find that C =,,. This means that v(t) =,, 3t +,, =,, 3t. Then r(t) = v(t) dt = t, t, t 6t + C. Since r() =, the constant vector C is too. Thus r(t) = t, t, t 6t.

c) At what time t does the ball hit the ground? This will happen when t 6t =, so that t = /6 = 6.5 seconds. Problem 6. a) What is d dt x(t), in terms of x(t) and its derivatives? This is d dt x(t) x(t) = x(t) x (t) + x (t) x(t) = x(t) x (t). b) Suppose that x(t) has constant length. What does this tell you about the relationship between x(t) and x (t)? It tells you that they are orthogonal for all values of t! 3

Math (Lesieutre).8: Length of curves January 7, 7 Problem. Let R be a circle centered at with radius 5. Set up a parametrization r(t) for the circle, and use it to compute the circumference. This one is pretty painless. Use r(t) = 5 cos t, 5 sin t with t π. Then r (t) = 5 sin t, 5 cos t, which gives us r (t) = ( 5 sin t) + (5 cos t) = 5 = 5. This doesn t depend on t! So the arc length is π 5 dt = π. This matches up with the πr you probably learned before you took calculus. Problem. Consider the curve r(t) = cos t, sin t, 3 t3/. Compute the length of this curve between t = and t =. We have which gives r (t) = sin t, cos t, t /, r (t) = ( sin t) + (cos t) + t = + t. The formula for arc length then tells us that L = ( ) + t dt = 3 ( + t)3/ = 3 3/ 3 = 3 ( ). Sanity check? That s about 3.65, which seems in the right ballpark at least. Problem 3. Consider a cardioid r = + cos θ. Set up the integral for the arc length, and evaluate it if you can.

We have f(θ) = + cos θ and so f (θ) = sin θ. The formula we just saw gives L = = = = = β α π π π π f(θ) + f (θ) dθ ( + cos θ) + ( sin θ) dθ + cos θ + cos θ + sin θ dθ π + cos θ dθ = + cos θ dθ 4 cos (θ/) dθ. There is a subtle point here. It s pretty tempting to just write 4 cos (θ/) = cos(θ/). But here s the catch: it s really cos(θ/), which isn t the same thing if cos(θ/) is negative. So we can t make this simplification. However, notice that the cardioid is symmetrical: the length from to π is equal to the length from π to π, so it is true that π π 4 cos (θ/) dθ = 4 cos (θ/) dθ. Now, when θ π, we have θ/ π/, and so cos(θ/) is guaranteed to be positive. That means for θ in this range, it s actually true that 4 cos (θ/) = cos(θ/). So our integral becomes π 4 cos(θ/)dθ = 8 sin(θ/) π = 8. That was a little painful, but I think the result is interesting: the length of the cardioid is 8. No square roots, no π, just 8. Problem 4. The circle in the first problem was not parametrized by arc length. Give another parametrization in which it is. The issue is that the particle is moving too fast: it s going at a constant speed of 5. If we use r(t) = 5 cos(t/5), 5 sin(t/5) instead, it will slow down, and now is parametrized by arc length. Then r (t) = sin(t/5), cos(t/5), which gives us r (t) = ( sin(t/5)) + (cos(t/5)) = =, which is what it means to be parametrized by arc length. Problem 5. A batter hits a baseball with initial velocity,, (a pop-up down the right field line; let s say the units are ft/sec).

a) What is a(t)? Assume the only force acting on the ball is gravity. It s a constant downward force,,, 3. The 3 here is gravitational acceleration. b) Solve for v(t) and r(t). Well, v(t) = a(t) dt =,, 3t + C. Since v() =,,, plugging in t = we find that C =,,. This means that v(t) =,, 3t +,, =,, 3t. Then r(t) = v(t) dt = t, t, t 6t + C. Since r() =, the constant vector C is too. Thus r(t) = t, t, t 6t. c) At what time t does the ball hit the ground? This will happen when t 6t =, so that t = /6 = 6.5 seconds. d) Set up the integral for the length of the path of the ball between the moment it left the bat and the moment it hit the ground. We have r (t) =,, 3t, and so L = 6.5 = 65 6.5 6.5 r (t) dt = + + ( 3t) dt = + ( 3t) dt ( ( )) 3 + arcsinh 95.85. (I d say you don t need to worry about how to actually do that integral.) 3

Math (Lesieutre).: Planes and surfaces, part January 3, 7 Problem. a) What is the equation for a plane passing through (,, 3) and with normal vector,, 3? The equation is (x ) (y ) + 3(z 3) =, which simplifies to x y + 3z = 9. b) At what point does this plane meet the x-axis? A point on the x-axis has y = and z =. So we need (x ) ( ) + 3( 3) =, which means x = 9. So the point is (9/,, ). c) At what point does this plane meet the line r(t) = t, t, t +? We need to know for what value of t the equation of the plane is satisfied, which means that ( t) t + 3( + t) = 9. This gives t = 9, so t = 8. We are supposed to find the point where they intersect, rather than the time t. So we should plug this back in to our original equation: r( 8) = 7, 6, 7. This does indeed satisfy x y + 3z = 9 Problem. a) Find the equation of the plane containing the three points P = (, 3, ), Q = (,, 3), and R = (5, 4, ). We need to find the normal vector. It will be perpendicular to both P Q and P R, so we want to take the cross product to find n. We obtain P Q = (,, 3) (, 3, ) = (, 3, ) P R = (5, 4, ) (, 3, ) = (4,, ) i j k pq pr = 3 = 4i + 8j + k = 4, 8,. That s our normal vector. To get the equation, we just need to plug in a point it passes through. We might as well use P. You could also use Q or R; you ll get the same answer for the plane at the end of the day. So the equation is a(x x ) + b(y y ) + c(z z ) = 4(x ) + 8(y 3) + (z ) = 4x + 8y + z = 4 x + y + 3z =. b) Draw a sketch of the plane by computing its traces in the three coordinate planes. In the xy-plane, we plug in z = and get 4x + y = 3. In the xz-plane, we plug in y = and get 4x + 3z = 3. In the yz-plane, we plug in x = and get y + 3z = 3. These three planes are plotted below.

xy-plane: 5-5 - - -5 5 xz-plane: 5-5 - - -5 5 yz-plane: 5-5 - - -5 5 Putting it all together, we get the following plane. I m drawing a back view that makes it easier to see the axes. Problem 3. Graph the cylinder x + x + 4z = 5. This doesn t involve y, so we should try to graph the x and z part and then draw the walls by including all possible y-values.

First, complete the square to obtain (x ) + 4z = 6, or perhaps easier, ((x )/4) + (z/z) = This you should recognize: it s an ellipse. It looks like this: 5-5 -5 5 Problem 4. Consider the three planes given by the following equations: x + y + 3z = x y + z = x + y + 3z = 6. a) Two of these planes are parallel: which two? Describe the intersection of these planes. It s the first and the third: the way you can tell is that they both have the same normal vector n =,, 3. b) The first and second planes intersect in a line. Give a parametrization of this line, and check that your line is actually contained in the first plane. First we need to find a point on the line. There are many points, and you can solve for one however you want. One way is just plug in z = and then figure out what x and y have to be. You can also just guess one, which is what I am about to do: (3,, ). Now we need to find the direction vector. How to do that? Well, it s perpendicular to both of the normal vectors, which are,, 3 and,,. So we can use the cross product v =,, 3,, as our normal vector! This guy is v = 5,, 3. So our line is r(t) = r + tv = 3,, + t 5,, 3 = 3 + 5t, t, 3t. To see whether it indeed line in the plane, just plug this in to the equation for the plane. You get x + y + 3z = (3 + 5t) + (t) + 3( 3t) =, so it works as it should. You could check the same thing with the second plane if you were so inclined. 3

Math (Lesieutre).: Planes and surfaces, part February, 7 Problem. Consider the two planes given by the following equations: x + y + 3z = x y + z = These planes intersect in a line. Give a parametrization of this line, and check that your line is actually contained in the first plane. First we need to find a point on the line. There are many points, and you can solve for one however you want. One way is just plug in z = and then figure out what x and y have to be. You can also just guess one, which is what I am about to do: (3,, ). Now we need to find the direction vector. How to do that? Well, it s perpendicular to both of the normal vectors, which are,, 3 and,,. So we can use the cross product v =,, 3,, as our normal vector! This guy is v = 5,, 3. So our line is r(t) = r + tv = 3,, + t 5,, 3 = 3 + 5t, t, 3t. To see whether it indeed line in the plane, just plug this in to the equation for the plane. You get x + y + 3z = (3 + 5t) + (t) + 3( 3t) =, so it works as it should. You could check the same thing with the second plane if you were so inclined. Problem. a) Graph the cylinder x + x + 4z = 5. This doesn t involve y, so we should try to graph the x and z part and then draw the walls by including all possible y-values. First, complete the square to obtain (x ) + 4z = 6, or perhaps easier, ((x )/4) + (z/z) = This you should recognize: it s an ellipse. It looks like this: Problem 3. Consider the surface given by x 5 + y 5 + z =. a) Sketch the xy-, xz-, and yz- traces of this figure. Use these to guide a drawing of the entire surface. xy-plane:

-6-4 - 4 6 6 4 - -4-6 xz-plane: 6 4 - -4-6 -6-4 - 4 6 yz-plane: 6 4 - -4-6 -6-4 - 4 6 Putting it all together, we get the following figure. I m drawing a back view that makes it easier to see the axes. b) Now consider the line r(t) = 3t, 4t, t. At what points does the line intersect the surface? We do the same procedure we used on Monday: plug in x(t) = 3 3t, y(t) = 4 4t, and

z(t) = 5t to our equation and solve for t. This gives x 5 + y 5 + z = x + y + 5z = 5 (3 3t) + (4 4t) + 5( t) = 5 5 5t + 5t = 5 5t + 5t =, which means either t = or t =. The points are then r() =,, and r() = 3, 4,. Problem 4. Consider the figure x + x + y 4y z + 4 =. a) First, complete the square to simplify the equation for the surface. x + x + y 4y z = 4 x + x+ + y 4y+4 z = 5 4 (x + ) + (y ) z = b) Next, sketch some traces of the figure. You can use the coordinate planes, but it might be better to use planes parallel to coordinate planes (for example, y = might be a good one) The plane z =. The trace is (x + ) + (y ) =, a circle of radius. 4 - -4-4 - 4 The plane y =. The trace is (x + ) z =, a hyperbola. 4 - -4-4 - 4 the plane x = : 3

-4-4 4 - -4 c) Use your traces to sketch the 3D surface. Putting it all together, we get the following figure. I m drawing a back view that makes it easier to see the axes. 4

Math (Lesieutre).3: Limits and continuity February 6, 6 Problem. a) Evaluate the limit: This one we can do using the rules for limits. b) Evaluate the limit: lim x + xy = (x,y) (,) lim (x,y) (,) x + xy lim (x,y) (,) x + lim ( ) = + lim (x,y) (,) x 3x + y lim (x,y) (3,) x + y (x,y) (,) xy lim xy = +. (x,y) (,) This is another one where you can just plug in and things will work OK. The important thing to remember is that to take a limit of a quotient, you can just take quotient of the limits, unless the limit of the denominator is! Since we re looking at the point (3, ), the limit is just 9 (what you get when you plug in x = 3 and y = ). Problem. Evaluate the following limits along the path y = mx. Does the limit exist? a) This gives us x 4 + y 4 lim (x,y) (,) x + y x 4 + y 4 lim (x,y) (,) x + y = lim x 4 + (mx) 4 x x + (mx) along y = mx ( + m 4 )x 4 = lim x ( + m )x = lim + m 4 x + m x =, no matter what m is. So at least for these paths, the limit doesn t depend on the path. We can t really be sure from this that the limit actually exists, but in fact it does. b) (x + y) lim (x,y) (,) x + y

Let s again use y = mx as our path. (x + y) lim (x,y) (,) x + y along y = mx = lim (x + mx) x x + (mx) = lim x ( + m) + m = ( + m) + m. This answer depends on m, which means that the limit doesn t exist. Problem 3. a) Give an example of a function f(x, y) that is not continuous at (, ). Can you come up with an example that is continuous for all (x, y) other than (, )? One function you could use is f(x, y) = (x ) + (y ). As (x, y) gets closer and closer to (, ), this gets closer and closer to infinity, and the function is not defined there. b) Give an example of a function g(x, y) that is not continuous for any (x, y) satisfying y = x. You can use a similar idea here as on the last one: think about the function g(x, y) = y x. If y = x, you can t plug it in to this equation. Problem 4. Is the function continuous at (, )? p(x, y) = { x y x 4 +y 4 if (x, y) (, ), if (x, y) = (, ). Being continuous means three things: the function is defined at (, ), the limit exists, and the limit is equal to the function. It is defined there (p(, ) = by fiat). So we need to think about the limit. Let s try the limit along the line y = mx. lim (x,y) (,) along y = mx x y x 4 + y 4 = lim x x (mx) x 4 + (mx) 4 = lim x (m ) ( + m 4 ) = m + m 4, so the direction limit depends on the path, which means that the limit doesn t exist, and the function isn t continuous.

Problem 5. Set up the integral for the arc length of r(t) = cos t, sin t for t π. We have r (t) = sin t, cos t r (t) = ( sin t) + (cos t) =, and so L = π π r (t) dt = dt = π. 3

Math (Lesieutre).4/.5: Partial derivatives and the chain rule April 8, 7 Problem. Compute the indicated partial derivatives: a) x (x y) It s xy: just think of this as being a constant times x, and the derivative is the constant (i.e. y) times x. b) y (x y) This one is x, just as the derivative of 7y is 7 you have to get used to thinking of the x s as constants. c) f y, where f(x, y, z) = e xyz (xz)e xyz : treat the x and z in the exponent as a constant, and use the chain rule. d) g y, where g(x, y) = y cos x It s just cos x. Problem. Consider the function f(x, y) = x cos(xy). Compute the four second partial derivatives. The first partials are: f x = cos(xy) xy sin(xy) f y = x sin(xy) That gives the second partials as: f xx = y sin(xy) y(xy cos(xy) + sin(xy)) = xy cos(xy) y sin(xy) f xy = x sin(xy) x(xy cos(xy) + sin(xy)) = x y cos(xy) x sin(xy) f yx = (x (y cos(xy)) + x sin(xy)) = x y cos(xy) x sin(xy) f yy = x 3 cos(xy) The thing I want you to notice here is that f xy = f yx. This is always happens, and it s called Clairaut s theorem.

Problem 3. Suppose that f(x, y) = x + y, and that x(t) = t, y(t) = sin t. What is df dt? Here we have to use the chain rule. The chain rule for one independent variable tells us that df dt = f dx x dt + f dy y dt = (x)() + (y)(cos t) = ( t)() + ( sin t)(cos t) = 8t + 4 sin t cos t = 8t + sin(t). Problem 4. Suppose that f(x, y) = xy, and x(s, t) = s + t and y(s, t) = s cos t. Compute the partial derivative f s. This is a little more painful. Again we have to use the chain rule. f s = f x x s + f y y s = (y )() + (xy)(cos t) = (s cos t) + ()(s + t)(s cos t)(cos t) = s cos t + (4s + st) cos t = 6s cos t + st cos t. Problem 5. Consider the ellipse defined by F (x, y) =, where F (x, y) = x + xy + y. Compute dy dx. The formula for implicit differentiation gives dy dx = F x F y = x + y y + x. Let s make sure we actually understand what this is saying. Here s the ellipse: 3 - - -3-3 - - 3 (It s tilted because of the xy-term; you probably haven t plotted one like that, and it s tough until you ve taken Math 3.)

One point on the ellipse is (x, y) = (, ). Our formula is telling us that dy, which is the dx slope of the tanegnt line, is given by x+y = 3 = at this point. That appears to y+x 3 match up with the figure. So we were able to find the slope of the tangent line, even though we don t actually have a formula for y as a function of x. 3

Math (Lesieutre).6: Directional derivatives and the gradient April 8, 7 Problem. Suppose that f(x, y) = xy, and x(s, t) = s + t and y(s, t) = s cos t. Compute the partial derivative f s. This is a little more painful. Again we have to use the chain rule. f s = f x x s + f y y s = (y )() + (xy)(cos t) = (s cos t) + ()(s + t)(s cos t)(cos t) = s cos t + (4s + st) cos t = 6s cos t + st cos t. Problem. Consider the ellipse defined by F (x, y) =, where F (x, y) = x + xy + y. Compute dy dx. The formula for implicit differentiation gives dy dx = F x F y = x + y y + x. Let s make sure we actually understand what this is saying. Here s the ellipse: 3 - - -3-3 - - 3 (It s tilted because of the xy-term; you probably haven t plotted one like that, and it s tough until you ve taken Math 3.) One point on the ellipse is (x, y) = (, ). Our formula is telling us that dy, which is the dx slope of the tanegnt line, is given by x+y = 3 = at this point. That appears to y+x 3 match up with the figure. So we were able to find the slope of the tangent line, even though we don t actually have a formula for y as a function of x.

Problem 3. Consider the function f(x, y) = x 4y. a) Compute the gradient f(x, y). The gradient is given by f x, f y = x, 8y. b) Find the derivative in the direction of the vector v =, at the point (, ). (Watch out! This isn t a unit vector.). First we need to know a unit vector in the direction of v. That s given by u = Now the directional derivative is D u f(a, b) = f(a, b) u =, 8, = 4 = 5. c) Find the directional derivative in the direction of the vector u =, at (, ). Same as strategy as above: D u f(a, b) = f(a, b) u =, 8, = 8. d) Sketch some level curves of f(x, y), including the level curve with z = 5. The level curves look like x + 4y = C, which are ellipses. Here are a few., 3-3 -5-5 - -5 - -3 5 - - -5 - - -5-3 -3-3 -5-3 - - 3 z = 5 is the innermost one that s marked. Notice that the point (, ) is contained in this level curve. Here s a graph of the whole surface, for what it s worth:

e) Find the unit vectors in the directions of steepest ascent and descent at the point (, ). Do your answers make sense? The gradient is given by, 8, which means steepest ascent is in the direction of, 8. A unit vector in this direction is / 68, 8/ 68 (a mess, sorry). Looking at this on the plot, this vector points inwards on the ellipse. Makes sense: the whole surface is shaped like a hill, and the inwards direction is uphill. Steepest descent is in the opposite direction, which is / 68, 8/ 68. This is downhill, the direction a ball would roll if placed on the graph. f) Find the directional derivative in the direction of steepest ascent. Is this steeper than the answers you got for directional derivatives earlier in the problem? The directional derivative is given by D u f(a, b) = f(a, b) u =, 8 / 68, 8/ 68 = 68/ 68 = 68. This is slightly more than the 8 that we got in part (c), so it s plausible that this is indeed the direction of steepest ascent. g) Find a direction that is tangent to the level curve z = 5 at the point (, ). What is the directional derivative in this direction? We want a direction that s perpendicular to the gradient. One option is 8/ 68, / 68, which will work because the dot product is. This points to the right and slightly down, which looks plausible for a tangent direction to the level curve at (, ), based on the picture. The directional derivative is, because it s tangent to a level curve: the function doesn t change in this direction. Taking the dot product with the gradient confirms this. Problem 4. Suppose that a function has gradient f(, ) = (, ). a) What is the directional derivative of this function in a direction with angle θ? The unit vector we want is cos θ, sin θ, and so the directional derivative is f(, ) u = cos θ + sin θ. 3

b) Plot the directional derivative for θ π. For what θ is it maximized? Zero? Here s a plot:.5..5 -.5 3 4 5 6 -. -.5 It s maximized at θ = π/4, which is the direction parallel to f. It s at 3π/4 and 7π/4, which is orthogonal to the gradient. 4

Math (Lesieutre).6 Directional derivatives, and some review April 8, 7 Problem. Suppose that a function has gradient f(, ) = (, ). a) What is the directional derivative of this function in a direction with angle θ? The unit vector we want is cos θ, sin θ, and so the directional derivative is f(, ) u = cos θ + sin θ. b) Plot the directional derivative for θ π. For what θ is it maximized? Zero? Here s a plot:.5..5 -.5 3 4 5 6 -. -.5 It s maximized at θ = π/4, which is the direction parallel to f. It s at 3π/4 and 7π/4, which is orthogonal to the gradient. Problem. Consider the function f(x, y) = x 4y. a) Sketch some level curves of f(x, y), including the level curve with z = 5. The level curves look like x + 4y = C, which are ellipses. Here are a few. 3-3 -5-5 - -3 - -5 5 - - -5 - - -5-3 -3-3 -5-3 - - 3

z = 5 is the innermost one that s marked. Notice that the point (, ) is contained in this level curve. Here s a graph of the whole surface, for what it s worth: b) Find the unit vectors in the directions of steepest ascent and descent at the point (, ). Do your answers make sense? The gradient is given by, 8, which means steepest ascent is in the direction of, 8. A unit vector in this direction is / 68, 8/ 68 (a mess, sorry). Looking at this on the plot, this vector points inwards on the ellipse. Makes sense: the whole surface is shaped like a hill, and the inwards direction is uphill. Steepest descent is in the opposite direction, which is / 68, 8/ 68. This is downhill, the direction a ball would roll if placed on the graph. c) Find the directional derivative in the direction of steepest ascent. Is this steeper than the answers you got for directional derivatives earlier in the problem? The directional derivative is given by D u f(a, b) = f(a, b) u =, 8 / 68, 8/ 68 = 68/ 68 = 68. This is slightly more than the 8 that we got in part (c), so it s plausible that this is indeed the direction of steepest ascent. d) Find a direction that is tangent to the level curve z = 5 at the point (, ). What is the directional derivative in this direction? We want a direction that s perpendicular to the gradient. One option is 8/ 68, / 68, which will work because the dot product is. This points to the right and slightly down, which looks plausible for a tangent direction to the level curve at (, ), based on the picture. The directional derivative is, because it s tangent to a level curve: the function doesn t change in this direction. Taking the dot product with the gradient confirms this. Problem 3. Find the parametrization for a line which...

a) Has r() = (,, 3) and r() = (,, ). For the point r, use,, 3). For the direction, use (,, ) (,, 3) = (, 4, ). So the equation is r(t) = r + vt =,, 3 + t, 4, =, 4t, 3 + t. b) Normal to the plane 3x y + z = and passes through the origin. We want the line to be in the direction of the normal vector, so v = 3,,. The point is r =,,, and so: r(t) =,, + 3,, t = 3t, t, t. c) The intersection of the planes x + y + z = 3 and x y + z =. This time the direction should be the cross product of the normal vectors, which is,,,,. This comes out to i j k,,,, = = 3,,. We also need to find a point on the line. There are many, so to narrow down our search and make it so there s only one answer (which is then easy to find), let s try to find a point with z =. Then we need x + y = 3 and x y =. The solution is x =, y =, and so our point is,,. That means the line in question is given by r(t) =, 4, 8 +, 4, t = + t, 4 + 4t, 8 + t. d) Tangent to the curve r(t) = t, t, t 3 at t =. The direction is given by the derivative, which is r (t) =, t, 3t. We want to plug in t = to get the direction vector at that time, and so the direction of the tangent line is v =, 4,. What point does the tangent line need to go through? It s r() itself, which is, 4, 8. So our equation for the line is l(t) =, 4, 8 +, 4, t = + t, 4 + 4t, 8 + t. (Note: I m calling the line l(t) since r(t) was already in use for the original curve. It doesn t matter what you call it, as long as you re clear about what you re doing.) Problem 4. Consider the two vectors u =,, 3 and v =,,. a) What is the angle between u and v? We know u v = u v cos θ, and so the angle is given by ( ) ( ) u v 6 θ = cos = cos. u v 4 3 3

b) If a triangle has (,, ) as a vertex, with u and v the two edges from this vertex, what is the vector for the third edge? It s u v =, 3, 4 (or the other way, depending on which direction we want the edge vector to point). c) What is the area of this triangle? Remember that magnitude of the cross product gives the area of the parallelogram spanned by the vectors, and we want half of that. i j k u v = 3 =,,. So area =,, = 6. 4

Math (Lesieutre) Exam review April 8, 7 Problem. Find the parametrization for a line which... a)... has r() = (,, 3) and r() = (,, ). For the point r, use,, 3). For the direction, use (,, ) (,, 3) = (, 4, ). So the equation is r(t) = r + vt =,, 3 + t, 4, =, 4t, 3 + t. b)... is normal to the plane 3x y + z = and passes through the origin. We want the line to be in the direction of the normal vector, so v = 3,,. The point is r =,,, and so: r(t) =,, + 3,, t = 3t, t, t. c)... is the intersection of the planes x + y + z = 3 and x y + z =. This time the direction should be the cross product of the normal vectors, which is,,,,. This comes out to i j k,,,, = = 3,,. We also need to find a point on the line. There are many, so to narrow down our search and make it so there s only one answer (which is then easy to find), let s try to find a point with z =. Then we need x + y = 3 and x y =. The solution is x =, y =, and so our point is,,. That means the line in question is given by r(t) =, 4, 8 +, 4, t = + t, 4 + 4t, 8 + t. d)... is tangent to the curve r(t) = t, t, t 3 at t =. The direction is given by the derivative, which is r (t) =, t, 3t. We want to plug in t = to get the direction vector at that time, and so the direction of the tangent line is v =, 4,. What point does the tangent line need to go through? It s r() itself, which is, 4, 8. So our equation for the line is l(t) =, 4, 8 +, 4, t = + t, 4 + 4t, 8 + t. (Note: I m calling the line l(t) since r(t) was already in use for the original curve. It doesn t matter what you call it, as long as you re clear about what you re doing.) Problem. Consider the two vectors u =,, 3 and v =,,.

a) What is the angle between u and v? We know u v = u v cos θ, and so the angle is given by ( ) ( ) u v 6 θ = cos = cos. u v 4 3 b) If a triangle has (,, ) as a vertex, with u and v the two edges from this vertex, what is the vector for the third edge? It s u v =, 3, 4 (or the other way, depending on which direction we want the edge vector to point). c) What is the area of this triangle? Remember that magnitude of the cross product gives the area of the parallelogram spanned by the vectors, and we want half of that. i j k u v = 3 =,,. So area =,, = 6. Problem 3. Let u = 3,, and v =,,. In what general direction does u v point? (Use the right-hand rule.) It s more or less straight down: when we use the right hand rule, the first vector points to the left in the xy-plane (it is almost parallel to i), while the second points straight ahead (roughly j). The right hand rule says the cross product points down. Problem 4. Use the two-path test to explain why the limit does not exist. y 3x lim (x,y) (,) y x I ve trained you to always try the path y = mx first. In this case we get (mx) 3x lim x (mx) x = lim m x 3 x x x = 3, which doesn t depend on m. So it seems like the test isn t going to help us. But there are other paths we might want to try too. In this case, we can use the path x =, and we get y lim y y =, which isn t the same. So the answer actually does depend on the path, even though y = mx all give the same answer. The two-path test tells us the limit doesn t exist.

Problem 5. A moving particle has v(t) = sin t, cos t,. a) What is the tangent vector to its path at t =? That s just v() =,, (if I d given you r(t) instead, you d first calculate v(t) = r (t), and then plug in to that). b) Find the length of the curve from t = to t = π. What is the tangent vector at t =? The length is l = π v(t) dt = π c) Suppose that r() =,,. Find a formula for r(t). We know r(t) = v(t) + C, and so π ( sin t) + (cos t) + () dt = = 4π. r(t) = cos t + c, sin t + c, t + c 3. Plugging in t =, we get,, + c, c, c 3, and so c =, c =, and c 3 =. Thus r(t) = cos t, sin t, t. Problem 6. An object sits at (, ) on a surface sloped 45. g =,. The gravitational force is a) What is the component of the gravitational force parallel to the surface? We want the component in the direction u =,. This is given by ( g u ) proj u g = u =, = 5, 5. u u b) The object slides down the slope to the point (, ). Find the work done by gravity. It s just W = F d, where d is the displacement. This is,, = (Joules, perhaps). Problem 7. Let f(x, y) = x y + 3x. a) Find the gradient f. I got f = f x, f y = xy + 3, x. b) Find the direction of fastest increase at the point (x, y) = (, ). What is the rate of increase? Fastest increase occurs in the direction of the gradient, which is f(, ) = 5,. The rate of increase (i.e. the directional derivative in this direction) is given by the length of the gradient, which is 6. 3

Problem 8. What is the domain of the function f(x, y) = ln(x + y 4)? Sketch some level curves. The domain is everywhere that the formula is going to make sense. The only thing that can go wrong is that we try to take the logarithm of a negative number, which will happen if x + y 4 <. This means that x + y < 4, which is the inside of a circle of radius. That s where the function isn t defined the domain is everywhere that it is, which means it s the area outside of a circle of radius centered at the origin. The level curve at z = is all (x, y) with ln(x + y 4) =, so x + y 4 = e. This is a circle of radius 4 + e ; all other level curves are also circles too. 4

Math (Lesieutre).7: Tangent planes and linear approximation April 8, 7 Problem. Let s start with a tangent plane to a sphere. a) Write down the equation f(x, y, z) for a sphere with center (,, 3) and radius 3. We want the sphere to be (x ) + (y ) + (z 3) = 5. b) One point on the sphere is (a, b, c) = (,, 5). Compute the gradient f, and evaluate f(a, b, c). The gradient is just f = (x ), (y ), (z 3). Plugging in the points in question, we get f(,, 5) =, 4, 4. c) Use your answer to write down the equation for the tangent plane to the sphere at (a, b, c). The normal direction is, 4, 4, and it goes through the point (,, 5). Using the regular old formula for the equation of a sphere, we get (x ) 4(y ) + 4(z 5) =. d) Try to plot the sphere and the plane and convince yourself that this answer is reasonable. I cheated and used a computer. Things look good, however: (Note: I ve rotated the picture somewhat to give us a better view, so yours might look different.)

Problem. Let f(x, y) = x y. a) Find the tangent plane to the graph at (x, y) = (, ). Does your answer make sense? The formula is z = f x (a, b)(x a) + f y (a, b)(y b) + f(a, b) In our situation, f(x, y) = x y f(, ) = f x (x, y) = x f y (x, y) = y f x (, ) = f y (, ) = So the plane is z = (x ) + (y ) + i.e. a horizontal plane z =. This makes sense, based on the graph: it s the tangent plane at the very top of the graph. b) Find the tangent plane to the graph at (x, y) = (, ). Does your answer make sense?

This time we get f(, ) = f x (, ) = f y (, ) = The plane is Seems plausible. Here s a graph: z = (x ) + (y + ) Problem 3. Consider the function f(x, y) = x +y. Use a linear approximation to approximate the value of f(.,.9). (.,.9) is close to the much cleaner value (, ), so we re going to use (a, b) = (, ) in the approximation formula. First, let s figure out what the formula actually gives is in this case. The derivative needs a little thought. To get f x, we treat y as a constant C, and then our function is g(h(x)) where g(x) = /x and h(x) = x + C. Notice that g (x) = /x and h (x) = x, and so according to the chain rule the derivative is g (h(x))h (x) = /(x + C) x = x (x +y ). f(x, y) = x + y x f x = (x + y ) y f y = (x + y ) 3

Now plug in the values (, ): f(, ) = 5 f x (, ) = 5 f y (, ) = 4 5. The linear approximation is then L(x, y) 5 5 (x ) 4 (y ). 5 We want x =. and y =.9, in which case this gives L(x, y) 5 5 (.) 4 5 (.) = 5 5 5 + 4 5 = 5 5 =.8. How did we do? The true value is about.7469, so we re off by.53. Problem 4. A cylinder has radius and height 3. Suppose that the radius and height each increase by.. Approximately how much will the volume change? The volume is given by V (r, h) = πr h. We have dv = V r (a, b) dr + V h (a, b) dh. In this case V r = πrh so V r (, 3) = π and V h = πr so V h (, 3) = 4π. Then dv = V r (, b) dr + V h (a, b) dh = (π)(.) + (4π)(.) =.6π. Let s check it. In fact the cylinder has volume π, and the new cylinder has volume 3.67π. The increase is.67π, which is just about what we expected. (This true increase is what the book calls z.) 4

Math (Lesieutre).8: Maxima and minima April 8, 7 Problem. A cylinder has radius and height 3. a) Suppose that the radius and height each increase by.. Approximately how much will the volume change? The volume is given by V (r, h) = πr h. We have dv = V r (a, b) dr + V h (a, b) dh. In this case V r = πrh so V r (, 3) = π and V h = πr so V h (, 3) = 4π. Then dv = V r (a, b) dr + V h (a, b) dh = (π)(.) + (4π)(.) =.6π. Let s check it. In fact the cylinder has volume π, and the new cylinder has volume 3.67π. The increase is.67π, which is just about what we expected. (This true increase is what the book calls z.) b) Give a formula for the linear approximation to V (r, h) near (r, h) = (, 3). This one s supposed to be a review of last time. The formula for the linear approximation is L(r, h) = V (, 3) + V r (, 3)(r ) + V h (, 3)(h 3) = (π)(r ) + 4π(h 3). Problem. Here is a graph of the function. f(x, y) = e (x ) (y/3) + e (x+) (y/3). How many critical points can you identify on the graph? saddle points? Are they maxima, minima, or I see three: there are two maxima, which are the two mountain peaks, and there is a saddle point, which is the point directly between them. (One way to think about this: a critical point is a point where you could balance a marble on the graph and it wouldn t roll in any direction.) There are no local minima.

Problem 3. For each of the following functions, compute all four second derivatives. Check that each function has a critical point at (, ), and classify it as a maximum, a minimum, or a saddle point. Let me preface these solutions with a warning. I made all the functions in this example very simple (just quadratics), which has the side effect that the second derivatives are all constant. This is not how things work in general! For other functions, you will need to plug in the (a, b) for your critical points into f xx (a, b) etc. before using the test. a) f(x, y) = x + y Here we get f x = x f y = y f xx = f xy = f yx = f yy = Plugging in (x, y) =, we get f x (, ) = f y (, ) =, as needed, so it is a critical point, as claimed by the problem statement. To figure out whether it s a max or a min, we need to use the d version of the second derivative test. We have D(x, y) = f xx f xy = = 4, f yx f yy which is positive. So it s either a maximum or a minimum, and since f xx >, it s a minimum. b) f(x, y) = x y This time, f x = x f y = y f xx = f xy = f yx = f yy = As before, plugging in (x, y) =, we get f x (, ) = f y (, ) =, so this is also a critical point. Now we compute the determinant of the second partials. This time, D(x, y) = f xx f xy = = 4. f yx f yy This is negative, so it s a saddle point. Roughly, what s going on is: if you fix y = and just think of it as a function of x, it s a minimum (the function is just x, after all). If you fix x = and think of it as a function of y, it s a maximum (since the function is y ). So whether the function increases or decreases depends on which direction you move.

c) f(x, y) = x y This time, f x = x f y = 4y f xx = f xy = f yx = f yy = 4 As before, plugging in (x, y) =, we get f x (, ) = f y (, ) =, so this is also a critical point. Now we compute the determinant of the second partials. This time, D(x, y) = f xx f xy = 4 = 8. f yx f yy positive, so it s either a max or a min. since f xx <, in this case we know it has to be a maximum. d) f(x, y) = x + xy + y This one s a little more interesting. f yy = f x = x + y f y = x + y f xx = f xy = f yx = As before, plugging in (x, y) =, we get f x (, ) = f y (, ) =, so this is also a critical point. Now we compute the determinant of the second partials. This time, D(x, y) = f xx f xy = = 3. f yx f yy Positive, so max or min. To tell which, we check f xx. Since it s positive, the point is a minimum. Problem 4. For each of the following functions, find the critical points. determine whether it is a maximum, minimum, or saddle point. Pick one, and a) x 4 + y 4 6xy We have f x = 4x 3 6x and f y = 4y 3 6y. To get both, we need x 3 4x = and y 3 4y =, so x =, ± and y =, ±. Any combination of these works, so the function has nine critical points. 3

The problem only asks us to look at one of these points, so let s use (, ). We have f xx = x 6 f xy = f yy = y 6. At (, ), this gives f xx (, ) = 8 and f yy (, ) = 8 D(, ) = f xx f yy fxy = 64, which is positive. That means it s a max or a min. Since f xx >, it s a minimum. b) f(x, y) = x x + y 4y + xy + 5 We have f x = x + y, f y = y 4 + x. We need both of these to be, so x + y = and x + y = 4. Solving the linear system using whatever your favorite method is (probably just eliminating one of the variables), we get x = and y =. This is the only critical point. Is it a maximum or a minimum? Well, f xx = f xy = f yx = f yy = look at the determinant: = 3. It s either or maximum or a minimum. To know which, we have f xx =, so it s a minimum. c) g(x, y) = x ye x y. This is going to be fun! Let s get all of the derivatives out of the way right off the bat. g x = x ye x y ( x) + xye x y = (xy x 3 y)e x y, g y = x e x y + (x y)e x y ( y) = (x x y )e x y. So critical points happen whenever g x = g y =. Since e x y is never, we need to solve (xy x 3 y) = and x x y =. So xy( x ) = and x ( y ) =. There are two possibilities: either x = and y is another, or y = ± and x = ±. I m not actually going to figure out what are the max and mins and saddle points. The point of this problem is simply to scare you: sometimes there are infinitely many critical points! 4

Math (Lesieutre).8: Max/min, continued April 8, 7 Problem. For each of the following functions, find the critical points. determine whether it is a maximum, minimum, or saddle point. Pick one, and a) x 4 + y 4 6xy We have f x = 4x 3 6y and f y = 4y 3 6x. The first of these then gives y = x 3 /4, and plugging this into the second, we obtain ( ) x 3 3 4 6x = 4 4 x9 64 6x = x 9 6 6x = x 9 56x = x(x 8 56) = The second factor x 8 56 is if x = or x = (there are six other solutions over the complex numbers, but we re only worried about the real solutions). If x =, then y = x 3 /4 = ; if x =, then y = ; if x =, then y =. So there are three critical points: (, ), (, ), and (, ). We need to figure out which of these are maxes and which are mins. To do that we re going to need the second derivatives, so we might as well get it over with and calculate them now. f xx = x, f xy = 6, f yy = y. At (, ), this gives f xx (, ) = 48, f yy (, ) = 48, and f xy (, ) = 6. Thus D(, ) = (48)(48) ( 6) = 48. The test tells us that this is a minimum. At (, ), this gives f xx (, ) = 48, f yy (, ) = 48, and f xy (, ) = 6. Thus D(, ) = (48)(48) ( 6) = 48. The test tells us that this is a minimum. At (, ), this gives f xx (, ) =, f yy (, ) =, and f xy (, ) = 6. Thus D(, ) = ()() ( 6) = 56. So that s a saddle point. b) f(x, y) = x x + y 4y + xy + 5 We have f x = x + y, f y = y 4 + x.

We need both of these to be, so x + y = and x + y = 4. Solving the linear system using whatever your favorite method is (probably just eliminating one of the variables), we get x = and y =. This is the only critical point. Is it a maximum or a minimum? Well, f xx = f xy = f yx = f yy = look at the determinant: = 3. It s either or maximum or a minimum. To know which, we have f xx =, so it s a minimum. c) g(x, y) = x ye x y. This is going to be fun (not really)! Let s get all of the derivatives out of the way right off the bat. g x = x ye x y ( x) + xye x y = (xy x 3 y)e x y, g y = x e x y + (x y)e x y ( y) = (x x y )e x y. So critical points happen whenever g x = g y =. Since e x y is never, we need to solve (xy x 3 y) = and x x y =. So xy( x ) = and x ( y ) =. There are two possibilities: either x = and y is another, or y = ± and x = ±. I m not actually going to figure out what are the max and mins and saddle points. The point of this problem is mostly to warn you: sometimes there are infinitely many critical points! Problem. Find the point on the surface x yz = which is closest to the origin. (Hint: minimize the square of the distance, instead of the distance itself). The square of the distance is x + y + z. Since x = + yz on the surface, we want to minimize + yz + y + z (a function of the independent variables y and z). So we ll use the second-derivative test: f y = y + z, f z = y + z. These vanish simultaneously at (, ), where x = ±. So the two critical points of the distance are (,, ) and (,, ). The second derivatives are D(y, z) = f yy f yz = = 3 As this is positive, and f yy >, this is a minimum as expected. f zy f zz

Problem 3. Find the maximum value of the function f(x, y) = (x ) + y on or inside a circle of radius centered at (, ). First we find the critical points on the interior, i.e. on the inside of the circle. f x = (x ) and f y = y, so the unique critical point is (, ). By inspection it is a minimum, and so the maximum of the circle is going to be on the boundary. To find the maximum value on the boundary, use a parametrization r(t) = ( cos t, sin t). This parametrizes the circle, and so we just need to find the t for which f(r(t)) is as large as possible. Plugging in, we want to maximize ( cos t ) + ( sin t) = 4 cos t 4 cos t + + 4 sin t = 5 4 cos t. This has a maximum at t = π, when cos t =. The corresponding point is γ(π) = (, ) and the value of f there is 3. Problem 4. Find the maximum and minimum values of the function f(x, y) = 5 (x ) (y ) on a triangle with vertices (, ), (8, ), and (, 4). First we need to find the critical points. That s the easy part. We have f x = (x ), f y = (y ). The critical points are where both of these vanish. There is only one: the point (x, y) = (, ). This is inside the triangle in question, as needed. Is that point a maximum or a minimum? Maybe you can guess from the formula (it sure looks like a max... ), but the way to check is to use the second derivative test with the Hessian. The second derivatives are f xx =, f xy =, f yy =. At (, ), we have f xx (, ) =, f xy (, ) =, f yy (, ) =. Thus D(, ) = ( )( ) ()() = 4, which is positive. Since f xx (, ) >, the point is a maximum. For future comparison, notice that f(, ) = 5. However, we still don t actually know that it s a global maximum: maybe the function achieves an even bigger value somewhere, but it s on the edge, and so it isn t a critical point. So we need to find the maxima and minima going around the outside of the triangle. We have to check each edge. To do that, we ll parametrize it, and then solve for the value(s) of t that maximize and minimize the function in question. Let s start with the edge from (, 4) to (8, ). It s parametrized by r (t) = 8t, 4 4t, where t (here I used our usual strategy for parametrizing a line. Then we get: 3

f(r (t)) = 5 (x ) (y ) = 5 (3 4t) ( + 8t) = 8t + 4t 5. We now do max/min, math 8-style. The derivative is 6t+4, which is when t = /4. This corresponds to the point (, 3), for which f(r (/4)) =. Since f (r (t)) <, it s a local max, and we find f(r (5/)) =. We also need to check the two endpoints: t = and t = 4. At t =, we re at the point r () = (, 4), and f(r ()) = 5. At t = 4, it s (4, ), and f(r (4)) = 5 too. That s only one edge. There are two more. Let s go from (, ) to (8, ) now. This one has r (t) = 8t, with t. Then f(r (t)) = 5 (8t ) ( ) = 4 (8t ) = 64t d + 6t + 3. What are the extrema? Well, f(r dt (t)) = 6 8t, which vanishes for t = /8, which is the point (, ). Here the function has the value 4. There are also the endpoints r () = (, ) and r () = (8, ), where the values are respectively 3 and 45. The last edge goes from (, ) to (, 4), and is parametrized by r 3 (t) =, 4t. Then f(r 3 (t)) = 5 (4t ) ( ) = 6t + 8t + 3. Then d f(r dt 3(t)) = 8 3t, which vanishes for t = /4, corresponding to the point r 3 (/4) = (, ), for which f(r 3 (/4)) = 4. The two endpoints are (, ) and (, 4), which are already on the list. All told, the possible extrema come from the edges and the critical points of the D function. They re all listed below. Part of R Possible extremum Value Type Interior (, ) 5 Max Edge # (, 3) Min (8, ) 45? (, 4) 5? Edge # (, ) 4 Min (, ) 3? (8, ) 45? Edge #3 (, ) 4 Min (, ) 3? (, 4) 5? (The? s indicate that at the corners of the region, we don t really know if the points are going to be maxima or minima, because we can t use the second derivative test. We just have to compare the values against the other values. Those are the candidates. To see the actual global max and min on the region, look for the biggest and smallest numbers. The global max occurs at the critical point (, ), which the value is 5. The global min is at (8, ), where the value is 45. 4

Math (Lesieutre).8: Max/min, continued April 8, 7 Problem. An airline will let you carry on any rectangular bag for which the sum of the dimensions x, y, and z is less than 6. Suppose that you want to bring a bag with the largest possible volume. To find the appropriate x, y, and z, what function should you maximize, and on what region? The volume is xyz. We can assume that z = 6 x y, since otherwise we could make z bigger without breaking the rules. Then the function we should maximize is xy(6 x y). The region is where x, y, and x + y 6, which is a triangle. Problem. Find the maximum and minimum values of the function f(x, y) = 5 (x ) (y ) on a triangle with vertices (, ), (8, ), and (, 4). First we need to find the critical points. That s the easy part. We have f x = (x ), f y = (y ). The critical points are where both of these vanish. There is only one: the point (x, y) = (, ). This is inside the triangle in question, as needed. Is that point a maximum or a minimum? Maybe you can guess from the formula (it sure looks like a max... ), but the way to check is to use the second derivative test with the Hessian. The second derivatives are f xx =, f xy =, f yy =. At (, ), we have f xx (, ) =, f xy (, ) =, f yy (, ) =. Thus D(, ) = ( )( ) ()() = 4, which is positive. Since f xx (, ) >, the point is a maximum. For future comparison, notice that f(, ) = 5. However, we still don t actually know that it s a global maximum: maybe the function achieves an even bigger value somewhere, but it s on the edge, and so it isn t a critical point. So we need to find the maxima and minima going around the outside of the triangle. We have to check each edge. To do that, we ll parametrize it, and then solve for the value(s) of t that maximize and minimize the function in question. Let s start with the edge from (, 4) to (8, ). It s parametrized by r (t) = 8t, 4 4t, where t (here I used our usual strategy for parametrizing a line. Then we get:

f(r (t)) = 5 (x ) (y ) = 5 (3 4t) ( + 8t) = 8t + 4t 5. We now do max/min, math 8-style. The derivative is 6t+4, which is when t = /4. This corresponds to the point (, 3), for which f(r (/4)) =. Since f (r (t)) <, it s a local max, and we find f(r (5/)) =. We also need to check the two endpoints: t = and t = 4. At t =, we re at the point r () = (, 4), and f(r ()) = 5. At t = 4, it s (4, ), and f(r (4)) = 5 too. That s only one edge. There are two more. Let s go from (, ) to (8, ) now. This one has r (t) = 8t, with t. Then f(r (t)) = 5 (8t ) ( ) = 4 (8t ) = 64t d + 6t + 3. What are the extrema? Well, f(r dt (t)) = 6 8t, which vanishes for t = /8, which is the point (, ). Here the function has the value 4. There are also the endpoints r () = (, ) and r () = (8, ), where the values are respectively 3 and 45. The last edge goes from (, ) to (, 4), and is parametrized by r 3 (t) =, 4t. Then f(r 3 (t)) = 5 (4t ) ( ) = 6t + 8t + 3. Then d f(r dt 3(t)) = 8 3t, which vanishes for t = /4, corresponding to the point r 3 (/4) = (, ), for which f(r 3 (/4)) = 4. The two endpoints are (, ) and (, 4), which are already on the list. All told, the possible extrema come from the edges and the critical points of the D function. They re all listed below. Part of R Possible extremum Value Type Interior (, ) 5 Max Edge # (, 3) Min (8, ) 45? (, 4) 5? Edge # (, ) 4 Min (, ) 3? (8, ) 45? Edge #3 (, ) 4 Min (, ) 3? (, 4) 5? (The? s indicate that at the corners of the region, we don t really know if the points are going to be maxima or minima, because we can t use the second derivative test. We just have to compare the values against the other values. Those are the candidates. To see the actual global max and min on the region, look for the biggest and smallest numbers. The global max occurs at the critical point (, ), which the value is 5. The global min is at (8, ), where the value is 45. Problem 3. Consider the function f(x, y) = e x y.

a) Find the maximum and minimum on or inside a square with vertices (±, ±). Step : Find the critical points. We have f x (x, y) = e x y ( x) = xe x y, f y (x, y) = e x y ( 4y) = 4ye x y. How can the first of these be? It s only if x =, since e x y can t possibly be. How about the second? Only if 4y =, for the same reason. So the only critical point is (, ). We ll need to know the second partials to determine the types of the critical point. This is a little more messy, since we have to use the product rule. f xx (x, y) = e x y + 4x e x y x f xy (x, y) = 8xye x y f yy (x, y) = 4e x y + 6y e x y At (, ), the values of these things are respectively,, and 4. Then D(, ) = ( )( 4) = 8, which is positive, and since f xx (, ) <, we conclude that the point is a maximum. The annoying part is that the maximum or minimum could be on an edge, and there are four of them. Here are the parametrizations: r (t) =, + 4t r (t) =, + 4t r 3 (t) = + 4t, r 4 (t) = + 4t, (top) (bottom) (left) (right) Let s do r (t) in detail. We need to plug this in to f(t) and think of it as a function of the single variable t, which we then find max/min for using the usual single-variable calculus strategies. Plugging in, we get f(r (t)) = e 4 ( +4t) = e +3t 3t. The derivative is d f(r dt (t)) = ( 64t + 3)e +3t 3t, which is when t = /. This corresponds to the point (, ), and we find f(, ) = /e 4. This is a local maximum (on the edge). We also need to check the endpoints, which are (, ) and (, ). Both of these give /e. Now we need to do the second edge. We find that f(r (t)) = e +3t 3t, which is the same thing that we got with r (t). There is again a maximum at t = /, which is the point (, ), and the value is again /e 4. The two endpoints (, ) and (, ) give values of /e. 3

Now the third edge r 3 (t) = + 4t, gives us f(r 3 (t)) = e +6t 6t. The derivative is d dt f(r 3(t)) = e +6t 6t (6 3t), which again has a max at t = /, which this time is the point (, ), where we get /e 8. It s getting very late, so I m not going to write down the fourth edge. Sorry! Take my word for it that it works the same as the third one, or email me if you want me to add some more details. Part of R Possible extremum Value Type Interior (, ) Max Edge # (, ) /e 4 Max (, ) /e? (, ) /e? Edge # (, ) /e 4 Max (, ) /e? (, ) /e? Edge #3 (, ) /e 8 Max (, ) /e? (, ) /e? Edge #4 (, ) /e 8 Max (, ) /e? (, ) /e? The global max is (, ), where the function is f(, ) =. The global min is at each of the four corners (±, ±), where f takes the value /e. b) Find the maximum of the same function on the unit circle. The strategy is the same, and this one is actually a little easier, since a circle only has one edge. Parametrize it by r(t) = cos t, sin t. Plugging that in, we get f(r(t)) = e cos t sin t = e 3 + cos(t) (I used a trig identity here, but you d get the same answer even without it.) We have d 3 f(r(t)) = e + cos(t) ( sin(t)). dt The critical points are t =, π/, π, and 3π/. The values of the function are t f(r(t)) e π/ e π e 3π/ e The critical points are all either on the edge, or in the interior of the unit disk, where we already found that the only critical point is (, ). Hence the max is at (, ), where f(, ) =, and the minimum occurs on the boundary at r(π/) = (, ) and r(3π/) = (, ). At these points, the value of the function is e. 4

Math (Lesieutre).9: Lagrange multipliers April 8, 7 Problem. You want to find a rectangle with perimeter 6 and area as large as possible. a) Convert this into a constrained optimization problem: let x and y be the two sides of the rectangle. What function f(x, y) are you trying to maximize? What constraint must be satisfied by x and y? We are trying to maximize f(x, y) = xy subject to g(x, y) = x + y = 6. b) Use the method of Lagrange multipliers to find the maximum value of the function. We have f = y, x g =, The equations are f(x, y) = λ g(x, y) and g(x, y) =, which gives three equations in three variables: y = λ x = λ x + y = 6 The second equation becomes (λ)+(λ) = 6, so that λ =. Then x = 4 and y = 4 must be the maximum. The value of the function is given by f(4, 4) = 6. Hence the maximum possible area is achieved when the function is a square. Problem. Find the maximum value of the function f(x, y) = y 4x constraint g(x, y) = x + y = 4. In this case, we have subject to the f(x, y) = 8x, y g(x, y) = x, 4y Our three equations come from f(x, y) = λ g(x, y), so that 8x = λx y = λ4y x + y = 4 This is a fairly typical set-up for one of these problems. The equations aren t so hard to solve, but you need to be extremely careful not to forget that the variables can be.

If x is not, then dividing the first equation through by x gives λ = 4. Then the second equation gives y = 6y, which means that y =. The third then says that x = 4, so x = or x =. This gives two points: (x, y) = (, ) and (x, y) = (, ). We have f(, ) = 6 and f(, ) = 6. If x is, then the first equation is true no matter what λ is. The third equation reads y = 4, so y = ±. Since y is nonzero, the second equation just turns into λ = / (but again, we don t really care what λ is). So we have two more solutions: (x, y) = (, ) and (x, y) = (, ). Plugging in we get f(, ) = and f(, ) =. So the function is maximized at (, ) and (, ), where the value is, and minimized at (, ) and (, ), where the value is 6. Problem 3. Use Lagrange multipliers to find the point on the parabola y = x which is closest to the origin. We want to minimize the distance function h(x, y) = x + y. Nobody likes square roots, so notice that we can just as well minimize the distance squared, which is f(x, y) = x + y. This is subject to the constraint y = x. This is better written as g(x, y) = y x + =. We have f(x, y) = x, y g(x, y) = x, Our equations are thus x = xλ, y = λ, and y x + =. Assume first that x. Then λ = from the first equation, so y = /. Then ( /) x + = so that x = ± ( ) /. So we get two points:,. The square of the distance at either one is ( ) f, = 3. 4 If x =, then the third equation gives y =. The distance here is. So the closest points were the first two that we found. Problem 4. You are making a open-top drawer out of wood. The material for the sides and back costs $ per square foot, and material for the bottom costs $, and the material for the front costs $4. a) Suppose that the dimensions of the drawer are x (side to side), y (top to bottom), and z (front to back). What is the total cost of the materials? It should be cost = cost front + cost back + cost sides + cost bottom = 4xy + xy + (yz) + xz = 6xy + 4yz + xz.

b) What are the dimensions of the cheapest drawer with volume 4 cubic feet? We want to minimize f(x, y, z) = 6xy + 4yz + xz subject to the constraint xyz = 4. The gradients are Our four equations are f(x, y, z) = 6y + z, 6x + 4z, 4y + x g(x, y, z) = yz, xz, xy (6y + z) = λ(yz) (6x + 4z) = λ(xz) (4y + x) = λ(xy) xyz = 4 After some painful algebra, you ll get that x = 4, y =, z = 6, and λ =. So the minimum is a drawer that s 4 6. 3

Math (Lesieutre).9, Lagrange multipliers, and a little 3. April 8, 7 Problem. Find the maximum value of the function f(x, y) = x+y subject to the constraint x + y =. First let s come up with the equations. The functions are f(x, y) = x + y and g(x, y) = x + y. f(x, y) =, The equations are then, = λ x, y, so g(x, y) = x, y. xλ =, yλ =, x + y =. Notice that λ can t possibly be : then the first equation wouldn t have any solution. Assuming λ, the first two equations give x = y, from which the third gives x =, i.e. x = ± (. Thus there are two possibilities: (x, y) =, ) ( and (x, y) =, ). ( (x, y) f(x, y), ) (, ) The first point is the max, and the second point is the min. Problem. You are making a open-top drawer out of wood. The material for the sides and back costs $ per square foot, and material for the bottom costs $, and the material for the front costs $4. a) Suppose that the dimensions of the drawer are x (side to side), y (top to bottom), and z (front to back). What is the total cost of the materials? It should be cost = cost front + cost back + cost sides + cost bottom = 4xy + xy + (yz) + xz = 6xy + 4yz + xz. b) What are the dimensions of the cheapest drawer with volume 4 cubic feet?

We want to minimize f(x, y, z) = 6xy + 4yz + xz subject to the constraint xyz = 4. The gradients are Our four equations are f(x, y, z) = 6y + z, 6x + 4z, 4y + x g(x, y, z) = yz, xz, xy (6y + z) = λ(yz) (6x + 4z) = λ(xz) (4y + x) = λ(xy) xyz = 4 After some painful algebra, you ll get that x = 4, y =, z = 6, and λ =. So the minimum is a drawer that s 4 6. Problem 3. Compute the following double integrals. a) xy dy dx The inner integral is everything from the second integral sign to the dwhatever. The outer integral is then That s our final answer. b) The inner integral is given by The final answer is e y dy = xy dy = y x 3 x dx = 3 y= x e xy dx = e xy x= = x x = 3 x. = 3 x= 4 = 3 4. ye xy dx dy = e y e y = e y. ( ) ( ) ( ) ey y = y= e = e 3. Notice that if we tried to do it in the other order, it would be tough going: ye xy dy would require an integration by parts. In some cases, it s even worse: the integral simply can t be done unless the order is right.

Math (Lesieutre) 3.: Double integrals over rectangular regions April 8, 7 Problem. Compute the following double integrals. Sketch the region R over which the integral is being taken. a) xy dy dx The inner integral is everything from the second integral sign to the dwhatever. The outer integral is then That s our final answer. b) xy dx dy xy dy = y x Inner: 3 x dx = 3 Outer: y= x xy dx = y x y = x x = 3 x. = 3 x= 4 = 3 4. dy = y 4 = y x=. = 3 4. Same answer as before. Coincidence? No Fubini s theorem. Changing the order of the variables doesn t change the value of the integral. Problem. a) The inner integral is ye xy dy dx ye xy dy, which is moderately unpleasant, since we have to integrate by parts. If we change the order of integration, it s a little better: The inner integral is given by ye xy dx = e xy x= ye xy dx dy = e y e y = e y.

The final answer is e y dy = ( ) ( ) ( ) ey y = y= e = e 3. Notice that if we tried to do it in the other order, it would be tough going: ye xy dy would require an integration by parts. In some cases, it s even worse: the integral simply can t be done unless the order is right. b) What is the average value of f(x, y) = ye xy on the region R? It s given by ye xy dy dx = ( area(r) e 3 ) = 4 e 3 4. Problem 3. a) Sketch the region of integration for x y dy dx. It s the region between y = and y = x with x. b) Evaluate the integral. Inner: x y dy = y x = x4. Outer: x 4 x5 dx = = 3 = 6 5. c) Rewrite the integral with the variables in the opposite order. This time y is going to go on the outside. Based on the picture, we need to go from y = to y = 4. For a given value of y, what x s do we want? It s frmo x = to y to. d) Evaluate the integral. Inner: 4 y y dx dy. y y dx = y( y) = y y 3/. Outer: 4 y y 3/ dy = (y 5 y5/ ) 4 = 6 6 3 = 5 5.

Math (Lesieutre) 3.3: Double integrals in other coordinate systems April 8, 7 Problem. Evaluate the integral x xy dy dx Inner: Outer: xy dy = x y x = x x x5. ( ) ( ) x x5 x ( dx = 4 x6 = 4 ) (, ) = 6. Problem. Sketch the region of integration for each of the following double integrals. a) x f(x, y) dy dx Here x goes from to, and for a given value of x, y goes from to x. This makes a triangle...8.6.4...4.6.8. b) f(x, y) dx dy y Here y goes from to, and for a given value of y, x goes from y to. Remembering that x = y is a sideways parabola, here s the picture we get:

..8.6.4. c) 4x x 3 f(x, y) dy dx..4.6.8. Here x goes from to, and for a given value of x, y goes from x 3 to 4x. Here it is: 8 6 4.5..5. Problem 3. For each of the double integrals in the previous problem, write bounds in new coordinates to reverse the order of integration. a) x f(x, y) dy dx Based on the picture, the new bounds are y f(x, y) dx dy b) y dx dy

Again, go to the picture. x is going to run from to, and for a given x, y goes from to x. x f(x, y) dy dx. c) 4x x 3 dy dx Now we need to go in the other direction. We can see that y goes from to 8. What are the bounds for a given value of y? The lower bound is the orange curve, which is y = 4x, aka x = y/4. The upper bound is the blue curve, y = x 3, so x = 3 y. Hence the integral is 8 3 y y/4 f(x, y) dx dy. Problem 4. Find the area of the region in (b) by integrating the function f(x, y) =. Check that you get the same answer for either order of integration. First, we want to evaluate y dx dy. Inner: dx = x = y y y Outer: y dy = Let s try to do the same thing the other way: x ) (y y3 = 3 3. dy dx. Inner: x dy = y x = x. Outer: x dx = 3 x3/ = 3. That s the same answer we got before, as it should be. 3

Math (Lesieutre) 3.3: Double integrals in polar coordinates April 8, 7 Problem. Sketch the regions corresponding to the following double integrals. a) π π/ f(r, θ) r dr dθ It s the region between orange and blue curves in the following illustration, which is a sort of quarter-ring...5..5 b) +cos θ π f(r, θ) r dr dθ -. -.5 -. -.5 This is the inside of the cardioid pictured inthe following picture...5.5..5. -.5 -. Problem. Find the area inside the cardioid r(θ) = + cos θ.

This is a classic one. To find the area, we have to integrate da = r drdθ on the region in question. Don t forget the r!. This is The inner integral is: +cos θ The full integral is now given by π +cos θ r dr dθ. r dr = r +cos θ ( + cos θ) = = + cos θ + cos θ = + cos θ + ( ) + cos θ = 3 4 + cos θ + cos θ. 4 Seems plausible. π 3 4 + cos θ + 4 cos θ dθ = 3 4 (π) + + = 3π. Problem 3. Find the volume of the region under the graph f(x, y) = x y above the unit circle R. The volume is going to be given by f(r, θ) da. There are three things we need to do: (i) R express the bounds on the integral in polar (ii) express the function in question in polar (iii) express the area element da in polar. Then we need (iv): actually compute the integral. The function f(x, y) = x y is given in terms of θ by f(r, θ) = r. The area element is da = r dr dθ (that s what you always want to use here). The volume is then given by The inner integral is π ( r ) r dr dθ. ( r ) r dr = (r r 3 ) dr = ( r r4 4 ) = 4. The outer integral is π That s the volume. Seems plausible. 4 dθ = (π) 4 = π.

Problem 4. a) Rewrite the following integral in Cartesian coordinates: xy da, where R R is part of a circle of radius 4 that lies in the first quadrant. Again, we need to get the bounds in polar, the function in polar, and da is polar. The hardest part is the function. Remember that x = r cos θ and y = r sin θ, and so f(x, y) = xy = 4r cos θ sin θ = r sin θ. This gives us π/ 4 b) Evaluate the integral. This one is π/ 4 r sin(θ) r dr dθ = π/ 4 ( ) π/ ( 4 r 3 sin(θ) dr dθ = sin θ dθ ( cos θ = π/ ) ( r 4 ) 4 = r 3 sin(θ) dr dθ ) r 3 dr ( ) ( ) 56 = 8. 3

Math (Lesieutre) 3.4: Triple integrals April 8, 7 Problem. Compute the following triple integrals over rectangular regions. a) xyz dz dy dx Inner: xyz dz = xy z Middle: xy dy = xy Outer: That s our final answer. b) xyz dx dy dz Inner: Middle: yz y= 3x dx = 3x xyz dx = yz x dy = zy 4 Outer: 3z 4 y= dz = 3z 8 = xy. = (4x) (x) = 3x. = 3. x= = yz. = z 4 z 4 = 3z 4. = 3. Of course, you re supposed to notice that this is the same integral as the one from part (a), but with the order changed. We got the same answer, which is reassuring. In fact, there are four other ways to rearrange it, and they ll all give you the same thing. Problem. Set up bounds for integrating a function f(x, y, z) on a cylinder of height 3 and radius, with base centered at (,, ). This one we want to use 3 z= x= 4 x y= 4 x f(x, y, z) dx dy dz (Note that sometimes people like to skip the x =, y =, z = at the bottom of the integral, since you can surmise which is which from the order of the d s. But I think this makes it easier to keep things straight.)

Problem 3. Evaluate the triple integral x x yz dz dy dx Inner: Middle: x x yz dz = y z ( x) y dy = y x z= ( x) = y x y= ( x). = ( x )( x). 4 Outer: 4 ( x )( x) 4 dx = 44 5. The last integral is a hassle: I don t see any better way to do it than to just expand it the whole way. Problem 4. Change the bounds on the following integral from dx dy dz to dy dx dz. 4 y f(x, y, z) dx dy dz. This is a triangular prism. The base is a triangle: y goes from to, and x from to y, which gives the triangle with vertices at (, ), (, ), and (, ). In the other order, we get 4 x/ f(x, y, z) dy dx dz.

Math (Lesieutre) 3.5: Triple integrals in cylindrical coordinates April 8, 7 Problem. a) Find the cylindrical coordinates for the point (x, y, z) = (,, ). We have r = x + y = + =, z =, and θ = π/ (same as polar). b) Find the rectangular coordinates for the point (, π/4, 3). We get x = r cos θ = ()( /) = /, y = r sin θ = ()( /) = /, z = 3, and so the point is ( /, /, 3). c) Sketch the cylindrical surface defined by z = r. This is a cone with vertex at the origin: Problem. Use a triple integral in cylindrical coordinates to compute the volume of a cylinder with radius R and height h. To find the area of a region, you integrate the function. Ditto to find the volume of a region in 3d: we need to integrate the function over this region. We are going to need r R, θ π, and z h. The integral then becomes V = = R π h R π h dv r dz dθ dr Inner: h r dz = rh