Math 207 Honors Calculus III Final Exam Solutions

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1 Math 207 Honors Calculus III Final Exam Solutions PART I. Problem 1. A particle moves in the 3-dimensional space so that its velocity v(t) and acceleration a(t) satisfy v(0) = 3j and v(t) a(t) = t 3 for all t. (i) What is the speed of the particle at time t? We have d dt v(t) 2 = d (v(t) v(t)) = a(t) v(t) + v(t) a(t) = 2v(t) a(t) = 2(t 3). dt Hence v(t) 2 = 2(t 3) dt = (t 3) 2 + k = v(t) = (t 3) 2 + k for some constant k. Since v(0) = 3j = 3, we have (0 3) 2 + k = 3 = k = 0 = v(t) = (t 3) 2 = t 3. (ii) What is the length of the path traced out by the particle during the time interval 0 t 7? By the arclength formula, the required length is 7 0 v(t) dt = Problem 2. The equations 7 0 t 3 dt = = 1 2 (3 t)2 3 = = (3 t) dt (t 3) y e x p 2 q = p and x 2 p + y q 3 = 2 3 (t 3) dt define p and q implicitly as smooth functions of x and y near (x, y, p, q) = (0, 2, 1, 1), so we have p = p(x, y) and q = q(x, y), with p(0, 2) = q(0, 2) = 1. Compute the partial derivatives p x, p y, q x, q y at the point (x, y) = (0, 2). Take the partial derivatives of each side of the given equations with respect to x, y to obtain { { y e x 2pp x q p 2 q x = p x e x 2pp y q p 2 q y = p y 2x p + x 2 p x + 3y q 2 and q x = 0 x 2 p y + q 3 + 3y q 2 q y = 0. Substituting the values x = 0, y = 2, p = 1, q = 1, these equations simplify to { { 2 2p x q x = p x 1 2p y q y = p y and 6 q x = q y = 0.

2 It follows that p x = 2 3, p y = 7 18, q x = 0, q y = 1 6. Problem 3. Find the global maximum and minimum of the function f (x, y) = 2x + 2y x 2 y 2 on the triangle in the first quadrant bounded by the lines x = 0, y = 0, y = 9 x. The global extrema of f over the triangle occur either at a critical point inside or along the boundary. We have D f (x, y) = [ f x f y ] = [2 2x 2 2y] = [0 0] = 2 2x = 2 2y = 0 = x = y = 1. This shows that f has a single critical point at (1, 1) which is inside the triangle, with f (1, 1) = 2. To study the extrema of f along the boundary, we consider the three edges of the triangle separately: Along the horizontal edge, we have y = 0 and 0 x 9 and the function reduces to f (x, 0) = 2x x 2 which has a maximum of f (1, 0) = 1 and a minimum of f (9, 0) = 63. Similarly, along the vertical edge, we have x = 0 and 0 y 9 and the function reduces to f (0, y) = 2y y 2 which has a maximum of f (0, 1) = 1 and a minimum of f (0, 9) = 63. Finally, along the slanted edge, we have y = 9 x and 0 x 9 and the function reduces to f (x, 9 x) = 2x + 2(9 x) x 2 (9 x) 2 = 2x x 63. An easy freshman calculus exercise shows that this function has a maximum of f (9/2, 9/2) = 45/2 and a minimum of f (9, 0) = f (0, 9) = 63 (0,9) (0,1) (1,1) (0,0) (1,0) (9/2,9/2) (9,0) We conclude that over this triangle, f reaches a global maximum of 2 at the critical point (1, 1) and a global minimum of 63 at the two corners (9, 0) and (0, 9). Problem 4. Use Lagrange multipliers to show that among all rectangular boxes of a given fixed volume, the cube has the smallest surface area. The surface area and volume of an x y z rectangular box are given by the formulas 2(xy + yz + zx) and xyz. The problem is then to show that the minimum value of the function f (x, y, z) = 2(xy + yz + zx)

3 subject to the constraint g(x, y, z) = xyz V = 0 occurs when x = y = z = V 1/3. Here V > 0 is the given fixed volume. By Lagrange multipliers, at the minimum point we must have f = λ g or [2y + 2z 2x + 2z 2x + 2y] = λ[yz xz xy] for some λ R. This, together with the constraint g = 0 leads to the system of equations 2(y + z) = λyz 2(x + z) = λxz 2(x + y) = λxy xyz = V. If any one of x, y, z is zero, the first three equations imply x = y = z = 0, which is inconsistent with the last equation. Thus, x, y, z are all non-zero. We can then express λ/2 from the first equation as (y + z)/(yz) = 1/y + 1/z. Doing the same on the second and third equations, we obtain 1 y + 1 z = 1 x + 1 z = 1 x + 1 y = 1 x = 1 y = 1 z The last equation then shows x = y = z = V 1/3. = x = y = z. How do we know that f indeed has a minimum for x 0 = y 0 = z 0 = V 1/3 and not, say, a maximum or saddle point? It is intuitively clear that when either of x, y, z gets too small or too large, f gets very large. So f must reach a minimum somewhere in the octant x > 0, y > 0, z > 0, which by the above analysis can only be (x 0, y 0, z 0 ). Below we verify this rigorously. Of course this wasn t required on the test, but aren t you curious to know?! Using the constraint, the surface area of the box can be expressed as a function of x, y only: S(x, y) = f ( x, y, V ) ( V = 2 xy + xy x + V ). y Here x, y vary in the quadrant x > 0, y > 0. The trouble is this quadrant is neither bounded nor closed, so we cannot apply the standard recipe to find the global minimum. But we can apply a cutting off trick to reduce the problem to the standard case. Take a large number c > 0 and consider the bounded closed region K = { (x, y) : x 1 c, y 1, xy c} c (see the figure). Take any point (x, y) outside or on the boundary of K. There are three possibilities: 0 < x 1/c. Then S(x, y) > 2V/x 2Vc. 0 < y 1/c. Then S(x, y) > 2V/y 2Vc. xy c. Then S(x, y) > 2xy 2c.

4 It follows that by choosing c > 0 large enough (depending on V), we can guarantee that S(x, y) > S(x 0, y 0 ) = 6V 2/3 for every (x, y) outside or on the boundary of K (in particular, (x 0, y 0 ) itself lies in the interior of K). By the extreme value theorem, S must reach a global minimum on K. By the choice of c, this minimum cannot be attained on the boundary of K, so it must occur at a critical point inside, which can only be (x 0, y 0 ). It follows that the global minimum of S on the whole quadrant x > 0, y > 0 occurs at (x 0, y 0 ). 2 (1/c,c ) K (1/c,1/c) xy=c 2 (c,1/c) Problem 5. The velocity of a moving fluid in 3-dimensional space is described by the vector field F(x, y, z) = 2x i + (e z cos y 3y) j (e z sin y + 3z) k. (i) Compute curl(f). i j k curl(f) = F = x y z 2x e z cos y 3y e z sin y 3z = ( e z cos y + e z cos y) i (0) j + (0) k = 0. (ii) Show that F is a gradient vector field, i.e., find a scalar function f such that F = f. The condition F = f is equivalent to (1) f x = 2x f (2) y = e z cos y 3y f (3) z = e z sin y 3z. Integrating (1) with respect to x gives f (x, y, z) = x 2 + C(y, z), where C is a differentiable function of y, z. To find C, take the partial derivative of this equation with respect to y and set it equal to (2): Integrating this with respect to y gives f y = C y = e z cos y 3y. C(y, z) = e z sin y 3 2 y2 + D(z),

5 where D is a differentiable function of z only. Hence, f (x, y, z) = x 2 + e z sin y 3 2 y2 + D(z). To find D, take the partial derivative of this equation with respect to z and set it equal to (3): f z = e z sin y + dd dz = e z sin y 3z. This gives dd dz = 3z = D(z) = 3 2 z2 + E for some constant E. It follows that f (x, y, z) = x 2 + e z sin y 3 2 y2 3 2 z2 + E. It is easy to check that this f indeed satisfies f = F. (iii) Let B 0 be a ball of radius 1 in space and B t be its image under the flow of F after time t. Find the volume V(t) of B t as a function of t. What happens to V(t) as t +? We have div(f) = F = x (2x) + y (e z cos y 3y) + z ( e z sin y 3z) By Liouville s theorem, Solving this first order linear ODE gives = 2 e z sin y 3 + e z sin y 3 = 4. V (t) V(t) = div(f) = V (t) = 4V(t). V(t) = V(0)e 4t. But V(0) is the volume of the ball B 0, which is 4π/3. It follows that Evidently V(t) 0 as t +. V(t) = 4π 3 e 4t. Bonus Problem. Every 2 2 matrix with real entries can be identified with a vector in R 4 in a natural way: a b (a, b, c, d). c d This allows us to think of the determinant as a function det : R 4 R. Use this identification to write the second order Taylor formula for the function det about the identity matrix I. Under the above identification, det(a, b, c, d) = ad bc,

6 so D det(a, b, c, d) = [ d c b a ] and H det(a, b, c, d) = a b Hence, for a matrix A =, c d D det(i)a = [ ] a b c = a + d = tr(a) d and a 1 2 At H det(i)a = 1 a b c d b c d d = 1 a b c d c 2 b = ad bc = det(a). a Thus, the second order Taylor formula for det at I is det(i + A) = det(i) + D det(i)a At H det(i)a + E 2 (A) = 1 + tr(a) + det(a) + E 2 (A). Remark. Here the second order error term E 2 (A) is in fact zero since det(i + A) is already a quadratic polynomial in entries a, b, c, d of A: 1 + a b det(i + A) = det c 1 + d = (1 + a)(1 + d) bc = 1 + (a + d) + (ad bc) = 1 + tr(a) + det(a). 1. THEOREMS AND DEFINITIONS PART II. (i) What does it mean for a map f : R n R m to be continuous at a R n? f is continuous at a if lim x a f (x) = f (a). Specifically, this means for every ε > 0 there is a δ > 0 such that x a < δ implies f (x) f (a) < ε.

7 (ii) What does it mean for a linear map T : R n R m to be the derivative of f : R n R m at a R n? It means f (a + h) f (a) T(h) lim h 0 h = 0 Such T, if it exists, is unique and it is often denote by D f (a). (iii) Let f : R n R be a C 2 scalar function and a R n. State the 2nd order Taylor formula on quadratic approximation of f near a. Taylor s formula is f (a + h) = f (a) + D f (a) h ht H f (a) h + E 2 (h). Here D f (a) = [ f (a) ] is the derivative matrix of f at a, H f (a) = [ 2 f (a) ] is the x i x i x j Hessian matrix of f at a, and the error term satisfies E 2 (h)/ h 2 0 as h MULTIPLE CHOICE 1. For a path c : R R 3, if the velocity v(t) is orthogonal to the acceleration a(t) for all t, then the path (A) has constant speed (B) has constant curvature (C) traces out a line (D) traces out a circle or helix Answer: A. In fact, d dt v(t)2 = d dt v(t) 2 = d (v(t) v(t)) = 2v(t) a(t) = 0, dt from which it follows that v 2, hence v, is constant. Alternatively, you could use the decomposition a = v T + κv 2 N of the acceleration into tangential and normal components. If a is always orthogonal to v, there is no tangential component, hence v = 0, showing that v must be constant. 2. The curvature κ(t) of the path c(t) = (sin t t cos t, cos t + t sin t, 4) is (A) t (B) t 2 (C) 1 t (D) 1 t 2 Answer: C. We compute v(t) = c (t) = (t sin t, t cos t, 0) a(t) = c (t) = (sin t + t cos t, cos t t sin t, 0),

8 so It follows that i j k v a = det t sin t t cos t 0 = t 2 k. sin t + t cos t cos t t sin t 0 κ(t) = v a v 3 3. The function f (x, y) = x 3 + y 3 3xy has (A) a local maximum and a local minimum (B) a local maximum and a saddle point (C) a local minimum and a saddle point (D) two saddle points = t2 t 3 = 1 t. Answer: C. The derivative of f is D f (x, y) = [3x 2 3y 3y 2 3x], from which it follows that f has critical points at (0, 0) and (1, 1). We have 6x 3 H f (x, y) =, 3 6y so H f (0, 0) = H f (1, 1) = Since det H f (0, 0) = 9 < 0, by the second derivative test (0, 0) is a saddle point. Similarly, since det H f (1, 1) = 27 > 0 and a 11 = 6 > 0, by the second derivative test (1, 1) is a local minimum. 4. Let a, b be non-zero vectors in R 3. The function f : R 3 R defined by f (x) = x a 2 + x b 2 reaches a minimum at (A) x = 0 (B) x = 1 (a + b) (C) x = a b (D) nowhere 2 Answer: B. We have f (x) = ( x a 2 ) + ( x b 2 ) = 2(x a) + 2(x b) = 2(2x (a + b)). Setting f (x) equal to 0, we obtain the unique critical point x 0 = (a + b)/2. It is intuitively clear that since f (x) gets arbitrarily large as x gets large, f must reach a global minimum at this critical point x 0. Here is a more precise reasoning in case you re interested: By the triangle inequality, x a x a and x b x b. Hence, f (x) ( x a ) 2 + ( x b ) 2, which shows f (x) as x. Choose a radius R > 0 so large that f (x) > f (x 0 ) whenever x R. By the extreme value theorem, f must reach a global minimum on the closed ball x R. By the choice of R, this minimum cannot be attained on

9 the boundary, so it must occur at a critical point inside the ball, which can only be x 0. Since the values of f are larger than f (x 0 ) outside this ball, it easily follows that the global minimum of f on the entire R 3 occurs at x Let f : R 2 R be a C 1 scalar function and c : R R 2 be a C 1 path. The composition f c : R R has a critical point at t if and only if the vectors f (c(t)) and c (t) are (A) both zero (B) equal (C) parallel (D) orthogonal Answer: D. Use the chain rule to write ( f c) (t) = f (c(t)) c (t). Now t is a critical point of f c if and only if ( f c) (t) = 0, which by the above equation happens when f (c(t)) c (t) = Suppose the partial derivatives of a C 1 function f : R 2 R are non-zero and satisfy f x = 5 f y everywhere. Then the level sets of f are lines of the form (A) y = 5x + b (B) y = 5x + b (C) y = 1 5 x + b (D) y = 1 5 x + b Answer: B. The assumption says that the gradient vector f at every point is a nonzero multiple of 5i + j: f = f x i + f y j = 5 f y i + f y j = f y (5i + j). The level sets of f are everywhere orthogonal to f, so they must be everywhere orthogonal to the vector 5i + j. This shows that these level sets are lines of the form y = 5x + b. To see this differently, we can parametrize the line y = 5x + b by the path c(t) = (t, 5t + b). By the chain rule, ( f c) (t) = f (c(t)) c (t) = f (c(t)) (i 5j) = f y (5i + j) (i 5j) = 0. This shows that f c is a constant function, which means f is constant on the line y = 5x + b. As an exercise, prove that under these assumptions, there is a C 1 function g : R R with non-zero derivative such that f (x, y) = g(5x + y). 7. The isotherms (level sets of temperature) in the plane are shown, where the arrows indicate the gradient of temperature. The hottest and coldest points along the circle, respectively, are a (A) a and d (B) d and a (C) b and c (D) c and b b d c

10 Answer: B. The temperature is constant along each curve and the direction of arrows at a, b, c, d shows that it gets hotter as we go from top to the bottom. This suggests, intuitively, that the hottest and coldest points on the circle are d and a, respectively. To make this precise, observe that by Lagrange multipliers, the extrema of the temperature T on the circle occur at the points where T is parallel to g, where g(x, y) = (x x 0 ) 2 + (y y 0 ) 2 is the quadratic function whose level set g(x, y) = R 2 describes the circle. Since the level sets of T and g are orthogonal to T and g, it follows that these level sets must be tangent to one another at these extrema. Looking at the plot, we see that a and d are the only points where the isotherms are tangent to the circle. The direction of the arrows then shows that T must have a maximum at d and a minimum at a. 8. Let c(t) be the trajectory of the vector field F(x, y) = y i + x 2 j in the plane with the initial condition c(0) = 2i j. The initial velocity and acceleration of this trajectory are (A) v(0) = i 4j, a(0) = 4i 4j (B) v(0) = i 4j, a(0) = 4i + 4j (C) v(0) = i + 4j, a(0) = 4i 4j (D) v(0) = i + 4j, a(0) = 4i + 4j Answer: C. Let c(t) = (x(t), y(t)). The condition of being a trajectory means c (t) = F(c(t)) which translates into the system of ODE s { x (t) = y(t) (4) y (t) = x 2 (t). Setting t = 0, we obtain x (0) = y(0) = 1 and y (0) = x 2 (0) = 4. Thus, the initial velocity is v(0) = i + 4j. Differentiating (4) with respect to t gives { x (t) = y (t) = x 2 (t) y (t) = 2x(t)x (t) = 2x(t)y(t). Setting t = 0, we obtain x (0) = x 2 (0) = 4 and y (0) = 2x(0)y(0) = 4. Thus, the initial acceleration is a(0) = 4i 4j. 9. The vector field F(x, y, z) = x cos z i + y j sin z k (A) is the curl of some vector field (B) is the gradient of some scalar function (C) is both (D) is neither Answer: D. F is neither a curl nor a gradient. In fact, if there were a vector field G such that F = curl(g), we would have div(f) = div(curl(g)) = 0. But div(f) = (x cos z) + x y (y) (sin z) = 1. z

11 Similarly, if there were a scalar function f such that F = grad( f ), we would have curl(f) = curl(grad( f )) = 0. But i j k curl(f) = det x y z = x sin z j. x cos z y sin z 10. Let F be a vector field in R 3 for which the derivative DF is a symmetric 3 3 matrix. Then (A) curl(f) = 0 (B) div(f) = 0 (C) both (A) and (B) (D) nothing can be said in general Answer: A. Let F = P i + Q j + R k. Then P x P y P z DF = Q x Q y Q z. Since DF is assumed to be symmetric, R x R y R z P y = Q x P z = R x Q z = R y. Hence i j k curl(f) = x y z = (R y Q z ) i (R x P z ) j + (Q x P y ) k = 0. P Q R The divergence of F in this case does not have to be zero, as the example F(x, y, z) = x i + y j + z k shows.