Exam Review Solutions 1. True or False, an explain: (a) There exists a function f with continuous secon partial erivatives such that f x (x, y) = x + y f y = x y False. If the function has continuous secon partial erivatives, then Clairaut s Theorem woul apply (an f xy = f yx ). However, in this case: f xy = y f yx = 1 (b) The function f below is continuous at the origin. { xy f(x, y) = x +y, if (x, y) (0, 0) 0 if (x, y) = (0, 0) Check the limit- First, how about y = x versus y = x? x lim (x,x) (0,0) 3x = 3 x lim (x, x) (0,0) 3x = 3 Yep, that i it- The limit oes not exist at the origin, therefore the function is not continuous at the origin (it is continuous at all other points in the omain). (c) If r(t) is a ifferentiable vector function, then False. t r(t) = r (t) t r(t) = 1 (r(t) r(t)) 1/ (r (t) r(t) + r(t) r (t)) = r (t) r(t) r(t) () If z = 1 x y, then the linearization of z at (1, 1) is L(x, y) = x(x 1) y(y 1) False for two reasons. We have forgotten to evaluate the partial erivatives of f at the base point (1, 1) (an so the resulting formula is not linear). We have also forgotten to evaluate the function itself at (1, 1). The linearization shoul be: L(x, y) = f(a, b) + f x (a, b)(x a) + f y (a, b)(y b) = 1 (x 1) (y 1) (e) We can always use the formula: f(a, b) u to compute the irectional erivative at (a, b) in the irection of u. False. This formula only works if f is ifferentiable at (a, b) (See Exercise 4 below). (f) Different parameterizations of the same curve result in ientical tangent vectors at a given point on the curve. False. The magnitue of r(t) is the spee. For example, r(3t) will have a magnitue that is three times that of the original- If you want an actual example, consier r(t) = cos(t), sin(t) At the point on the unit circle (1/, 1/ ), the magnitue of r (π/4) = 1. Replace t by 3t (an evaluate at t = π/1 to have the same point on the curve), an the spee is 3 instea of 1. Why i we bring this up? If we re-parameterize with respect to arc length, the velocity is always 1 unit (so at s = 1, you ve travele one unit of length, etc). 1
(g) If u(t) an v(t) are ifferentiable vector functions, then t [ u(t) v(t)] = u (t) v (t) False. It looks like the prouct rule: t [ u(t) v(t)] = u (t) v(t) + u(t) v (t) (h) If f x (a, b) an f y (a, b) both exist, then f is ifferentiable at (a, b). False. Our theorem says that in orer to conclue that f is ifferentiable at (a, b), the partial erivatives must be continuous at (a, b). Just having the partial erivatives exist at a point is a weak conition- It is not enough to even have continuity. (i) At a given point on a curve (x(t 0 ), y(t 0 ), z 0 (t)), the osculating plane through that point is the plane through (x(t 0 ), y(t 0 ), z(t 0 )) with normal vector is B(t 0 ). True (by efinition).. Show that, if r(t) is a constant, then r (t) is orthogonal to r(t). (HINT: Differentiate r(t) = k) Using the hint an the fact that r(t) = r(t) r(t), we then ifferentiate both sies: r(t) = k r(t) r(t) = k r (t) r(t) + r(t) r (t) = 0 r(t) r (t) = 0 Therefore, the ot prouct is zero (an so r (t) an r(t) are orthogonal). 3. Reparameterize the curve with respect to arc length measuring from t = 0 in the irection of increasing t: r = ti + (1 3t)j + (5 + 4t)k Fin s as a function of t, invert it then substitute it back into the expression so that r is a function of s. In this case, Therefore, t = s/ 9, an r(s) = s = t 0 r (u) u = 9 t s, 1 3 s, 5 + 4 s 9 9 9 4. Is it possible for the irectional erivative to exist for every unit vector u at some point (a, b), but f is still not ifferentiable there? Consier the function f(x, y) = 3 x y. Show that the irectional erivative exists at the origin (by letting u = cos(θ), sin(θ) an using the efinition), BUT, f is not ifferentiable at the origin (because if it were, we coul use f u to compute D u f). Compute the irectional erivative at the origin by using the efinition: f(0 + h cos(θ), 0 + h sin(θ)) f(0, 0) h 3 cos (θ) sin(θ) D u f(0, 0) = lim = lim = 3 cos h 0 h h 0 h (θ) sin(θ) By using the efinition, the irectional erivative exists for all angles θ. In particular, θ = 0 correspons to f x, an θ = π/ correspons to f y, so that the partial erivatives also exist at the origin: f x (0, 0) = 0 an f y (0, 0) = 0. If we use our shortcut formula to compute D u f(0, 0), we get zero for every irection: D u f = f u = 0
which is not true. Why i this happen? If we compute the first partial erivatives, f x = 3 x 1/3 y 1/3 f y = 1 3 x/3 y /3 we see that f x an f y are both iscontinuous where x = 0 an y = 0 respectively. 5. If f(x, y) = sin(x + 3y), then fin the linearization of f at ( 3, ). We have f( 3, ) = sin(0) = 0 an f x (x, y) = cos(x + 3y) f x ( 3, ) = Therefore, f y (x, y) = 3 cos(x + 3y) f y ( 3, ) = 3 L(x, y) = 0 + (x + 3) + 3(y ) = (x + 3) + 3(y ) 6. The raius of a right circular cone is increasing at a rate of 3.5 inches per secon while its height is ecreasing at a rate of 4.3 inches per secon. At what rate is the volume changing when the raius is 160 inches an the height is 00 inches? (V = 1 3 πr h) V t = V r r t + V h h V t t = 3 πrhr t + 1 h πr 3 t Use r = 160, h = 00, r/t = 3.5 an h/t = 4.3, V/t 37973.3π. I ll try to use nicer numbers for the exam. 7. Fin the ifferential of the function: v = y cos(xy) v = v x x + v y y = ( y sin(xy) x + (cos(xy) xy sin(xy)) y 8. Fin the maximum rate of change of f(x, y) = x y + y at the point (, 1), an the irection in which it occurs. The maximum rate of change occurs if we move in the irection of the graient. We see this by recalling that: D u f = f u = f cos(θ) so we fin the graient at the point (, 1) f = 4, 9/ So if we move in that irection, the we get the max rate of change, which is f = 4 + 81 145 4 = 6.0 9. Fin an expression for [u(t) (v(t) w(t)] t t [u(t) (v(t) w(t)] = u (v w) + u (v w) Taking this erivative, we see that t [u(t) (v(t) w(t)] = u (v w) + u (v w + v w ) 3
10. Use the efinition of the partial erivative to compute f x (x, y), if f(x, y) = x x+y. x+h f(x + h, y) f(x, y) x+h+y x lim = lim x+y h 0 h h 0 h If we work with the numerator a bit, we combine fractions by getting a common enominator: Put this back in, x + h x + h + y x x + y = (x + h)(x + y ) x(x + h + y ) (x + h + y )(x + y = ) 1 hy lim h 0 h (x + h + y )(x + y ) = y (x + y ) hy (x + h + y )(x + y ) 11. The curves below intersect at the origin. Fin the angle of intersection to the nearest egree: r 1 (t) = t, t, t 9 r (t) = sin(t), sin(5t), t The angle of intersection is the angle between the tangent vectors at the origin. First ifferentiate, then evaluate at t = 0: r 1 (t) = 1, t, 9t 8 r (t) = cos(t), 5 cos(5t), 1 1, 0, 0, 1, 5, 1 To fin the angle, we use the relationship: In our case: cos(θ) = u v = u v cos(θ) 1 1 + 5 + 1 θ = cos 1 (1/ 7) 79 1. Fin three positive numbers whose sum is 100 an whose prouct is a maximum. Let x, y, z be the three numbers. Then we want to fin the maximum of P (x, y, z) = xyz subject to the constraint that x + y + z = 100 an they are all positive. Let P (x, y) = xy(100 x y). The critical points are y(100 x y) xy = 0 x(100 x y) xy = 0 From the first equation, y = 100 x (we can throw out y = 0). Substitute this into the secon equation to fin that x = 100/3. Therefore, y = 100/3 an z = 100/3. These are the three numbers we wante. 13. Fin the equation of the tangent plane an normal line to the given surface at the specifie point: x + y 3z = 3 (, 1, 1) This is an implicitly efine surface of the form F (x, y, z) = k, therefore, we know that F is orthogonal to the tangent planes on the surface. Compute F at (, 1, 1), an construct the plane an line: Thus, the tangent plane is: F x = x F y = 4y F z = 6z F (, 1, 1) = 4, 4, 6 4(x ) 4(y + 1) 6(z 1) = 0 The normal line goes in the irection of the graient, starting at the given point. In parametric form, x(t) = + 4t y(t) = 1 4t z(t) = 1 6t 4
14. If z = x y, x = w + 4t, y = w 5t + 4, w = r 5u, t = 3r + 5u, fin z/ r. This is simple if we use a chart (see below): z r = z x x w w r + z x x t t r + z y y w w r + z y y t t r With: z x = x z y = y x w = 1 x t = 4 y w = w y t = 5 w r = r t r = 3 so that: z r = 4xr + 4x 8ywr + 30y 15. If x + y + z = 3xyz an we treat z as an implicit function of x, y, then fin z/ x an z/ y. Let us efine F (x, y, z) = x + y + z 3xyz in keeping with the notation from the text. Then we compute: x F (x, y, z) = 0 F x x + F y y x + F z z x = 0 z x = F x F z = (x 3yz) z 3xy Similarly, we can show that z y = F y F z = (y 3xz) z 3xy 16. If a(t) = 10k an v(0) = i + j k, r(0) = i + 3j, fin the velocity an position vector functions. v(t) = a(t) t = 0, 0, 10t + v 0 = 1, 1, 10t 1 An antiifferentiate once more: r(t) = v(t) t = t, t, 5t t + r 0 = t +, t + 3, 5t t 17. Fin the equation of the normal line through the level curve 4 = 5x 4y at (4, 1) using a graient. The graient of g is orthogonal to its level curve at 5x 4y = 4. Fin the graient of g at (4, 1): g = 1 5, 4 8 For the line, we simply nee to move in the irection of the graient, so we can simplify the irection to 5, 4 (not necessary, but easier for the algebra). Therefore, the line (in parametric an symmetric form) is: x(t) = 4 + 5t y(t) = 1 4t or x 4 5 = y 1 4 Notice that the slope is 4/5. If we wante to check our answer, we coul fin the slope of the tangent line: 5x 4y = 16 5 4 y x = 0 y x = 5 4 5
18. Fin all points at which the irection of fastest change in the function f(x, y) = x + y x 4y is i + j. The irection of the fastest increase is in the irection of the graient. Therefore, another way to phrase this question is: When is the graient pointing in the irection of 1, 1 : f = k 1, 1 x, y 4 = k, k So k = x an k = y 4, therefore, the points are on the line x = y 4, or y = x + 1. Our conclusion: There are an infinite number of possibilities- All of the form (a, a + 1), which result in the graient: (a ) 1, 1, a > 1 19. Fin the vectors T an N if r(t) = cos(t), sin(t), ln(cos(t)) at the point (1, 0, 0). Recall the efinitions: With the given function r, we have: T(t) = r (t) r (t) r = N(t) = T (t) T (t) sin(t), cos(t), sin(t) cos(t) An the magnitue is (we re assuming cos(t) > 0 for the log to be efine): r (t) = sin (t) + cos (t) + tan (t) = 1 + tan (t) = sec (t) = sec(t) = sec(t) = 1 cos(t) Now For N, we ifferentiate T an normalize it: with T(t) = sin(t) cos(t), cos (t), sin(t) T = cos (t) + sin (t), sin(t) cos(t), cos(t) T = ( cos (t) + sin (t)) + 4 sin (t) cos (t) + cos (t) If you square this out an combine the first two terms, we get: so we en up with: cos 4 (t) cos (t) sin (t) + sin 4 (t) + 4 sin (t) cos (t) = cos 4 (t) + cos (t) sin (t) + sin 4 (t) = (cos (t) + sin (t)) = 1 T = 1 + cos (t) (You wouln t have to go this far to simplify on an exam, but its goo practice). Now we can write N(t) = 1 cos (t) + sin (t), sin(t) cos(t), cos(t) 1 + cos (t) 0. Fin an classify the critical points: f(x, y) = 4 + x 3 + y 3 3xy 6
The partial erivatives are: f x = 3x 3y f y = 3y 3x f xx = 6x f yy = 6y f xy = 3 The critical points are where x = y an y = x, so x, y are both zero or positive: x 4 = x x 4 x = 0 x(x 3 1) = 0 so x = 0, y = 0 or x = 1, y = 1. Put these points into the Secon Derivatives Test: f xx (0, 0)f yy (0, 0) f xy(0, 0) = 9 < 0 The origin is a SADDLE f xx (1, 1)f yy (1, 1) f xy(1, 1) = 36 9 > 0 f xx (1, 1) > 0 Local MIN 1. Let z = x + y. (a) Draw the level curves for z =, 4, 6. See the sketch. curves are circles. (b) Calculate the graient at (, 1). f(, 1) = 4,. The level (c) Plot the graient vector you compute in the previous problem, along with the earlier level curves. I m mainly looking to see that you rew it perpenicular to the level curves, an the graient vector is fairly long at that point. () Fin the equation of the tangent line to the curve at (, 1). The slope is f x /f y, or from the graient,. y 1 = (x ) (e) Fin the equation of the normal line to the curve at (, 1). The slope is the negative reciprocal, or 1/. y 1 = 1 (x ) You coul write this in parametric vector form using the graient vector: + 4t, 1 + t. Fin the equation of the tangent plane to the surface implicitly efine below at the point (1, 1, 1): First write this as F (x, y, z) = 0 x 3 + y 3 + z 3 = 9 6xyz F (x, y, z) = x 3 + y 3 + z 3 + 6xyz 9 = 0 Now, for z x at (1, 1, 1): 3x + 3z z ( x + 6y z + x z ) = 0 x 7
At the point (1, 1, 1) we have: Similarly, for z y : 3 + 3z x + 6(1 + z x ) = 0 z x = 9 9 = 1 3y + 3z z ( y + 6x z + y z ) = 0 y so that z y = 1 as well. For the tangent plane, compute the graient for the normal vector: F x = 3x + 6yz F y = 3y + 6xz F z = 3z + 6xy F = 9, 9, 9 Therefore, the tangent plane is: (You coul ivie by 9 if you like). 9(x 1) + 9(y 1) + 9(z 1) = 0 3. Fin parametric equations of the tangent line at the point (,, 4) to the curve of intersection of the surface z = x y an z = 4. (Hint: In which irection shoul the tangent line go?) The curve is x y = 4, which is an ellipse (at height 4 in 3-). The graient is 4x, y so at (, ), the graient is < 8, 4 >, so the tangent line (in the xy plane) is: 8(x + ) 4(y ) = 0 or y = (x + ) + We shoul express the line in three imensions, since the question was phrase that way. Here s one way: t, (t + ) +, 4 (But there are multiple ways of expressing it). 4. Fin an classify the critical points: f(x, y) = x 3 3x + y 4 y We use the secon erivatives test to classify the critical points as local min, local max or sale. Solving for the CPs, we get: f x (x, y) = 3x 3 = 0 f y (x, y) = 4y 3 4y = 0 from which we get x = ±1, y = ±1 an x = ±1, y = 0 Continuing with secon erivatives, D(x, y) = f ] xx f x y = f y x f y y 6x 0 ] 0 1y = 4x(3y 1) 4 We ll arrange the results in a list. The plot is inclue just for fun: Point D an Classification (1, 1) 48 : Local Min (1, 1) 48 : Local Min (1, 0) 4 : Sale ( 1, 1) 48 : Sale ( 1, 1) 48 : Sale ( 1, 0) 4 : Local Max 8
5. Fin the curve of intersection between the plane y + z = 3 an x + y = 5 (in parametric form). SOLUTION 1: Since we have a circle, let with 0 t π. x(t) = 5 cos(t), y(t) = 5 sin(t), z(t) = 3 5 sin(t) SOLUTION : An alternative that s not quite as nice woul be to write x an z as functions of y. It s not as nice because x is not a function of t, so you nee to write two curves: x = 5 y y = y z = 3 y x = 5 y y = y z = 3 y 6. Fin parametric equations for the tangent line to the curve of intersection of the paraboloi z = x +y an the ellipsoi 4x + y + z = 9 at ( 1, 1, ). We coul go through an fin the curve of intersection, but notice that we are not aske to o that. We are given the point of intersection, so we just nee a tangent vector. We make the observation that the tangent line will lie in the tangent plane to both surfaces, so that the irection we nee is orthogonal to the normal vector for both tangent planes. The normal vector for the tangent plane to x + y z = 0 is x, y, 1 at ( 1, 1, ), so n 1 =,, 1 Similarly, the normal vector for the tangent plane to 4x + y + z evaluate at ( 1, 1, ): n = 8,, 4 n = 4, 1, (You in t nee to scale this, but it may simplify your work later) = 9 will be the graient Taking the cross prouct, we get the irection for the tangent line, so that the equation in parametric vector form is: 1, 1, + t 5, 8, 6 7. The following table gives numerical values of z = g(x, y). Use these to estimate g x (, 5) an g y (, 5) by taking an average. Also estimate the irectional erivative of g at (, 5) in the irection of 1, 1 Sie Remark: I woul give you numbers that woul be easier to work with on the exam, since you wouln t have a calculator. Go ahea an use a calculator for this problem. x = 1.5 x = x =.5 y = 5. 16.0 17. 18.4 y = 5.0 0.0 1..3 y = 4.8 4. 5.3 6.6 TYPO for the irectional erivative. We can t really go in the irection of 1, 1 using the table, so that really shoul have been 0.5, 0. or something scaling like that. So we ll take that as the irection below. Continuing, for g x (, 5), take the average change horizontally: 1. 0.0 1.5 +.3 1..5.0 =.4 +. =.3 9
Similarly, we estimate g y the same way, except go vertically: 1. 5.3 5 4.8 + 17. 1. 5. 5.0 = 0.5 + ( 0) = 0.5 An we estimate the irectional erivative going iagonally. The esimation is a bit ifferent, we are taking the change in height over the change in the istance in the omain, which is the istance between the points (.5, 5.) an (.0, 5.0), which is approximately 0.54: 18.4 1. 0.54 + 1. 4. 0.54 5.37 8. If P = u + v + w, where u = xe y, v = ye x an w = e xy, then fin P x, P y when x = 0, y =. Using a chart can help with the notation: P x = P u u x + P v v x + P w w x P y = P u u y + P v v y + P w w y Taking all of the partial erivatives an evaluating, we shoul get that P x = 8 5 P y = 5 9. Fin the rate of change of f(x, y) = xy at P (, 8) in the irection of Q(5, 4). We nee the partial erivatives to compute the graient: y f x = x = 1 f y = x=,y=8 x y = 1 4 x=,y=8 We note that the partial erivatives are continuous at P, so f is ifferentiable an our shortcut formula for the irectional erivative will work. We now nee a unit vector in the right irection. The irection we want is: P Q = 5, 4 8 = 3, 4 u = 1 3, 4 5 The answer is now the irectional erivative: D u f = 1, 1 3 4 5, 4 = 5 5 30. Fin the points on the surface of y = 9 + xz that are closest to the origin. When we optimize the istance, we can instea optimize the square of the istance- The location where the max or min occurs will not change, although you ll nee to take the square root of the max or min at the en. In this case, we want to minimize: (x 0) + (y 0) + (z 0) where the point (x, y, z) must lie on the surface: y = 9 + xz. Substituting this in, we want to fin the minimum of: We ll fin the critical points first: F (x, z) = x + (9 + xz) + z x + z = 0 x + z = 0 From these, the only critical point is the origin (0, 0). Is this a min or a max? We shoul o a quick check using the secon erivatives test- F xx = F xz = 1 F zx = 1 F zz = D = 3 > 0 an F xx > 0 Therefore, this is a local min, so from that we can conclue that we have the minimum. The points on the surface giving us the minimum are (0, ±3, 0), which each has a istance of 3. 10
31. The figure below shows level curves for the temperature z = T (x, y) on a square plate. First, estimate the values of both partial erivatives of T at (3, ), then for all of the secon partial erivatives, just estimate whether they are positive or negative. For T x, a rough estimate woul be about 0. Similarly, a rough estimate for T y woul be about +0. For T xx, T x goes from about 5 to about 14 as we increase x, so that woul mean that T xx > 0. For T yy, as we go up, T y increases, so T yy > 0. For T xy, we might estimate T x (3, ) 0, an T x (3,.5) 40, so T xy < 0. 3. The figure below shows the level curves of z = h(x, y). Fin whether each is positive or negative: (i) h x (50, 30) (ii) h y (50, 30) (iii) h xx (50, 30) h is ecreasing as we increase x, so h x < 0. h is ecreasing as we increase y (by a small amount), so h y < 0. The level curves are getting wier apart as we move in the positive x, so h x is getting less negative- h xx > 0. 33. Using the previous graph, if we make a path from the center of the number 8 always going in the irection of h, raw the result. The main point here is that whenever you cross a level curve, the path is orthogonal to the level curve. 11