Math 3c Solutions: Exam 2 Fall 2017

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1 Math 3c Solutions: Exam Fall points) The graph of a smooth vector-valued function is shown below except that your irresponsible teacher forgot to include the orientation!) Several points are indicated on the graph; they are shown as points, but you should: Draw all vectors below as arrows. When you are asked to sketch a vector in translated position, determine where to draw it so that its geometric significance becomes clear. Draw your vectors to approximately the right scale; each tick mark on the graph represents one unit. Everything should be drawn on the graph provided; if your picture gets too crowded, you may recopy the graph as needed. c) r0) r) d) e) r) b) is the circle a) Explain how you know that this graph was not parametrized using arc length. The distance between r0) and r) is more than, as is the distance between r) and r). Also, it appears that the distance between r) and r) is greater that that between r0) and r). If we use an arc length parametrization, the arc length between two points whose inputs are unit apart will always be. b) Sketch the osculating circle when t. The circle should touch the curve at r) and then align with the curve as closely as possible for as long as possible. c) Given that r 0), sketch r 0) in translated position. Note that: the tail of the vector is at r0) the length of the vector is approximately) the vector is tangent to the curve Page of

2 Math 3c Solutions: Exam Fall 07 the vector points left and down; this makes it consistent with the orientation. While the orientation was not shown, we know the orientation is the direction of increasing t, so we can determine the orientation by moving from r0) to r). d) Put a d on the graph at all points where you can see by inspection that the unit normal vector will not exist. We know that N always points to the concave side of the curve; this means that N will fail to exist at any point where the curve changes concavity. There is one such point on the curve above. e) Sketch 0 r t) dt in translated position. This is just r) r0), which is the vector that points from r0) to r). ; t 0. Find the arc length parametrization for this. 0 points) Let rt) t, t 6 6 curve that uses t 0 as its reference point and preserves the orientation. Introduce a dummy variable and calculate for r u). It s ok if you wait to introduce the dummy variable until the next step, when we integrate.) ru) r u) r u) u, u 6 6, u 0 u 3, u 5 u 3 ) + u 5 ) u 6 + u 0 u 6 + u ) u 6 + u u 3 + u }{{} u 0, so u 3 0 so we can drop the absolute value u 3 + u Page of

3 Math 3c Solutions: Exam Fall 07 Integrate to express s in terms of t. s }{{} multiply by in the form anticipating the w-sub below }{{} w + u dw u 3 dw t 0 t u 3 + u du 0 u 3 + u du +t w dw ) w t ) 3 6 +t ) Solve for t in terms of s. s 6s 6s + + t ) ) t ) 3 + t ) 3 6s + ) 3 + t 6s + ) 3 t 6s + ) 3 t 6s + ) 3 t 6s + ) 3 }{{} t t 0 so we can drop the absolute value Plug in this expression for t to get r in terms of s: Page 3 of

4 Math 3c Solutions: Exam Fall 07 rs) r ts)) 6s+) 3, ) 6 6s+) 3 6 Note that we could combine the th root and the 6th power in the y-component to write a fractional exponent of 6 3. Finally, record the restrictions on s. Since we preserved the orientation and t0 is our reference point and the s-value at the reference point is always 0), t 0 s 0. rs) 6s+) 3, ) 6 6s+) 3 6 ; s points) Suppose the acceleration of a particle is given by the function at) e t, e t. If the velocity at time t 0 is, and the position at time t 0 is 3,, find a formula for the position of the object as a function of time. at) vt) v0) }{{} e t, e t at) dt e t, e t dt e t, e t + c, + c, plug in 0 for t }{{}, + c set equal to the given value,, c 3, c vt) e t, e t + 3, Page of

5 Math 3c Solutions: Exam Fall 07 pt) vt) dt p0) }{{} et, e t + 3, ) dt e t, e t + 3 t, t + c, + 0, 0 + c 3, plug in 0 for t }{{}, + c set equal to the given value 3,, c 5, c pt) e t, e t + 3 t, t + 5, Note: it would also be fine to do the vector addition and write the position as a single vector.. 0 points) The following questions are grouped together because they are the same type: short answer. They are not necessarily related in topic. a) State the Extreme Value Theorem for functions of variables. If fx,y) is continuous on a region D that is closed and bounded, then f achieves an absolute max and an absolute min on D. b) Suppose we have a curve rt). If radius of curvature when t 0 is 5, what is κ0)? κ0) 5 The radius of curvature and the curvature are reciproacalls of each other.) c) Sketch the domain of f x, y) x y We need x y 0 x y Page 5 of

6 Math 3c Solutions: Exam Fall 07 Note that the boundary parabola is included, so it is drawn as a solid curve. 5. Evaluate each limit or show it does not exist. If you use paths, be sure to explicitly write the paths you are using. a) lim x,y), ) xy+ y+ Hmmm. Top and bottom both go to 0. I don t see a promising substitution and we are not approaching the origin so that switching to polar is unlikely to help. I will hope the limit doesn t exist and try paths. path : r t), t lim x, y), ) along r t) xy + y + lim t t + t + t + ) lim t t + Path : r t) t, t Note: we can t use the path parallel to the x-axis holding y constant at ) since this path is not in the domain of r. lim x, y), ) alongr t) xy + y t + lim t t + t + )t ) lim t t ) lim t + ) t ) so the limit does not exist. b) lim x,y),) tanxy ) sinxy )+xy Hmmmm. Top and bottom both go to 0. The same expression, xy, appears in 3 places. I ll try a substitution, letting u xy. Note that as x, y), ), u 0. Page 6 of

7 Math 3c Solutions: Exam Fall 07 lim x,y),) tanxy ) sinxy ) + xy tan u lim u 0 sin u + u }{{} LH lim u 0 sec u cos u + sec 0 cos points) Let f x, y, z) e xyz x sin z. a) Find an equation for the local linear approximation to f at the point P 0,, 0). Lx, y, z) 3z Lx, y, z) f P 0 ) + f P 0 ) x, y +, z f P 0 ) e 0 sin 0 0 f yze xyz sin z, xze xyz, xye xyz x cos z f P 0 ) 0, 0, 3 Lx, y, z) + 0, 0, 3 x, y +, z b) Circle the correct words from the underlined options : The graph of Lx,y,z) is a line/plane/ hyperplane in R, R, R 3, R. c) Write an equation for the level surface of f that contains the point P 0. The level surface will be the set of all points in the domain of f that map to the same w-value as P 0. Since f P 0 ), this makes the equation for the level surface: e xyz x sin z d) Find an equation for the plane that is tangent to the level surface in part c). For this, we use point-normal form. point: P 0,, 0) normal: n f P 0 ) 0, 0, 3 point-normal form: If Qx,y,z) represents an arbitrary point on the plane, then P 0 Q x, y, z will be parallel to the plane and thus orthogonal to the plane s normal. Page 7 of

8 Math 3c Solutions: Exam Fall 07 n P0 Q 0 0, 0, 3 x, y, z 0 3z 0 z points) Let z f x, y) xy + y Use the chain rule for functions of more than one variable to receive credit, you must do it this way!) to find a formula for θ z where θ is a polar coordinate. Your answer should be given in polar coordinates. The paths in red below show how the value of θ influences the value of z. x z x x θ z z y y y θ r θ r θ z xy + y ) z x z y x r cos θ x θ r sin θ y r sin θ y θ r cos θ z θ }{{} chain rule, see diagram above xy + y ) y) xy + y ) x + y) z x x θ + z y y θ }{{} combine as one fraction and write in terms of r and θ xy + y ) y) r sin θ) + xy + y ) x + y)r cos θ) y r sin θ) x + y)r cos θ) + xy + y xy + y r sin θ + r cos θ + r sin θ)r cos θ) r cos θ sin θ + r sin θ Page 8 of

9 Math 3c Solutions: Exam Fall points) Let f x, y) x y. a) Sketch the level curve that passes through the point,). f, ) so the level curve in question is x y x y; y 0 y x ; y 0 c),) b) Find the unit vector that points in the direction in which f decreases most rapidly at,). The direction in which f decreases most rapidly at a point is the opposite direction of the gradient at that point. So we start by finding f, ). f f x, f y y, xy y, x y f, ),, Since we are asked for a unit vector, we need to normalize: Page 9 of

10 Math 3c Solutions: Exam Fall 07 f, ) ) + ) f, ) f, ) 5 5, 5, 5 5 5, 5 5 c) Draw the vector from part b) in translated position put it where its geometric significance becomes clear) on the graph from part a). See graph. Note that the vector is perpendicular to the level curve at,). d) What is the rate of f at,) in the direction you found in part b)? The rate is f, ) 5 Note that the norm of the gradient is the same as the norm of its opposite, which we calculated above.) e) Calculate the directional derivative of f at,) in the direction of v i 3j. We need to find the unit vector in the direction of v: v + 3) 3 u ˆv v 3, , Now we can find the directional derivative using the dot product: Page 0 of

11 Math 3c Solutions: Exam Fall 07 D u f, ) f, ) 3 3, 3 3 3, 3 3, points) Let f x, y) x x + y y and let D be the triangular region shown below the boundary is included). Find the location and value of the absolute max and the absolute min of f on D.,) 3,0) We know an absolute max and an absolute min occur by the Extreme Value Theorem. We need to check critical points and boundary points. A master table is given at the end of this problem, in which I record all the points we find below. Critical points: f f x, f y x, y f is always defined. f 0 x 0 }{{} x and y 0 }{{} y ) Since the point, is in D, we put this critical point on our master table. Boundary, part : r t) t, 0 ; 0 t f r t)) t t d dt f r t))) t Page of

12 Math 3c Solutions: Exam Fall 07 t is always defined, and it is equal to 0 when t. This is in the domain, so it is our only critical point on the boundary, part. Applying the Extreme Value Theorem for functions of one variable to f, restricted to this part of the boundary, we check this critical point and the endpoints. { t x, y) ) f x, y) critical point, 0 { min 0 0, 0) 0 endpoints, 0) max ) We add the points, 0 and,0) to our master table. Boundary, part : r t), t ; 0 t f r t)) + t t t t + d dt f r t))) t + t is always defined, and it is equal to 0 when t. This is in the domain, so it is our only critical point on the boundary, part. Applying the Extreme Value Theorem for functions of one variable to f, restricted to this part of the boundary, we check this critical point and the endpoints. We add the point Boundary, part 3: { t x, y) f x, y) critical point, ) 7 min { 0, 0) endpoints, ) max ), to our master table;,) is already there. r 3 t) t, t ; 0 t f r 3 t)) t t + t t t t d dt f r 3t))) t t is always defined, and it is 0 when t. This is in the domain, so it is our only critical point on the boundary, part 3. Page of

13 Math 3c Solutions: Exam Fall 07 Applying the Extreme Value Theorem for functions of one variable to f, restricted to this part of the boundary, we check this critical point and the endpoints. Both the points nothing new to add to it. { t x, y) ) f x, y) critical point, { min 0 0, 0) 0 endpoints, ) max, ) and,) are already on the master table, so there is Master Table { x, y) ) f x, y) critical points, min, 0) boundary points, 0) ),, ) 7 max The absolute max is ; it occurs ) at,). The absolute min is ; it occurs at, points) Let f x, y) x y and let D be the unit disk. That is, D { x, y) x + y } Find the location and value of the absolute max and absolute min of f on D. Use the method of Lagrange multipliers on the boundary. critical points: f is always defined. f 0 x 0 and y 0 0, 0) is a critical point. f f x, f y x, y boundary points: The boundary is x + y }{{} gx,y) g g x, g y x, y Page 3 of

14 Math 3c Solutions: Exam Fall 07 g 0 on the boundary, so we can proceed. We are looking for points where f λ g x, y λ x, y We set up the following systerm of equations: st components equal : x λx) ) nd components equal : y λy) ) constraint : x + y 3) Case : x 0 This would prevent us from solving for λ in ). ) ) 0 0 3) 0 + y y y ± ) λ See note below.) If y ±, then y 0 so we can divide by it to get λ. Thus we have found two solutions to our system: x 0, y ±, λ Case : y This would prevent us from solving for λ in ). ) 0 0 3) x + 0 x As in case, since we know x 0 we can x ± ) λ See note below.) divide by it to solve for λ. We have two more solutions to our system: x ±, y 0, λ Case 3: x 0, y 0 In this case we are free to divide by either of these variables. ) λ x x ) λ y y λ λ }{{} contradiction Thus we get no points from case 3. x, y) f x, y) critical point {0, 0) 0 0, ) min 0, ) min boundary points, 0) max, 0) max The absolute max is ; it occurs at,0) and, 0). The absolute min is ; it occurs at 0,) and 0, ). Page of