Calculus III Exam 2 Sample Solutions 2009

Exam 2 Sample Solutions Be sure to look over your old quizzes and homework as well. For limits, we will provide a graph and contours. No calculators will be allowed for this exam. 1. True or False, and explain: (a) There exists a function f with continuous second partial derivatives such that fx (x, y) = x + y 2 fy = x − y 2 SOLUTION: False. If the function has continuous second partial derivatives, then Clairaut’s Theorem would apply (and fxy = fyx ). However, in this case: fxy = 2y (b) The function f below is continuous at the origin.  2xy x2 +2y 2 , f (x, y) = 0 fyx = −2y if (x, y) 6= (0, 0) if (x, y) = (0, 0) SOLUTION: Check the limit- First, how about y = x versus y = −x? 2x2 2 = 3 (x,x)→(0,0) 3x2 lim −2x2 −2 = 3 (x,−x)→(0,0) 3x2 lim Yep, that did it- The limit does not exist at the origin, therefore the function is not continuous at the origin (it is continuous at all other points in the domain). (c) If ~r(t) is a differentiable vector function, then d |~r(t)| = |~r 0 (t)| dt SOLUTION: False. d 1 r0 (t) · r(t) |r(t)| = (r(t) · r(t))−1/2 (r0 (t) · r(t) + r(t) · r0 (t)) = dt 2 |r(t)| Extra: If you’re not sure about it, try to verify the formula with r(t) = h3t2 , 6t − 5i. (d) If z = 1 − x2 − y 2 , then the linearization of z at (1, 1) is L(x, y) = −2x(x − 1) − 2y(y − 1) SOLUTION: False for two reasons. We have forgotten to evaluate the partial derivatives of f at the base point (1, 1) (and so the resulting formula is not linear). We have also forgotten to evaluate the function itself at (1, 1). The linearization should be: L(x, y) = f (a, b) + fx (a, b)(x − a) + fy (a, b)(y − b) = −1 − 2(x − 1) − 2(y − 1) (e) We can always use the formula: ∇f (a, b) · ~u to compute the directional derivative at (a, b) in the direction of ~u. SOLUTION: False. This formula only works if f is differentiable at (a, b) (See Exercise 4 below). (f) Different parameterizations of the same curve result in identical tangent vectors at a given point on the curve. SOLUTION: False. The magnitude of r(t) is the velocity. For example, r(3t) will have a magnitude that is three times that of the original- If you want an actual example, consider r(t) = hcos(t), sin(t)i √ √ At the point on the unit circle (1/ 2, 1/ 2), the magnitude of r0 (π/4) = 1. Replace t by 3t (and evaluate at t = π/12 to have the same point on the curve), and the speed is 3 instead of 1. Why did 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 traveled one unit of length, etc). 1 (g) If ~u(t) and ~v (t) are differentiable vector functions, then d [~u(t) × ~v (t)] = ~u 0 (t) × ~v 0 (t) dt SOLUTION: False. It looks like the product rule: d [~u(t) × ~v (t)] = ~u 0 (t) × ~v (t) + ~u(t) × ~v 0 (t) dt (h) If fx (a, b) and fy (a, b) both exist, then f is differentiable at (a, b). SOLUTION: False. Our theorem says that in order to conclude that f is differentiable at (a, b), the partial derivatives must be continuous at (a, b). Just having the partial derivatives exist at a point is a weak condition- It is not enough to even have continuity. (i) At a given point on a curve (x(t0 ), y(t0 ), z0 (t)), the osculating plane through that point is the plane ~ 0 ). through (x(t0 ), y(t0 ), z(t0 )) with normal vector is B(t SOLUTION: True (by definition). 2. Show that, if |~r(t)| is a constant, then ~r 0 (t) is orthogonal to ~r(t). (HINT: Differentiate |~r(t)|2 = k) SOLUTION: Using the hint, 0=
d d d k= |~r(t)|2 = (~r(t) · ~r(t)) = ~r 0 (t) · ~r(t) + ~r(t) · ~r 0 (t) = 2~r 0 (t) · ~r(t) dt dt dt Therefore, the dot product is zero (and so ~r 0 (t) and ~r(t) are orthogonal). 3. Reparameterize the curve with respect to arc length measuring from t = 0 in the direction of increasing t: r = 2ti + (1 − 3t)j + (5 + 4t)k SOLUTION: Find 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, Z t √ s= |r0 (u)| du = 29 t 0 √ Therefore, t = s/ 29, and  r(s) =  2 3 4 √ s, 1 − √ s, 5 + √ s 29 29 29 4. Is it possible for the directional derivative to exist for every unit vector ~u at some point (a, b), but f is still not differentiable there? p Consider the function f (x, y) = 3 x2 y. Show that the directional derivative exists at the origin (by letting ~u = hcos(θ), sin(θ)i and using the definition), BUT, f is not differentiable at the origin (because if it were, we could use ∇f · ~u to compute D~u f ). SOLUTION: Compute the directional derivative at the origin by using the definition: p p h 3 cos2 (θ) sin(θ) f (0 + h cos(θ), 0 + h sin(θ)) − f (0, 0) Du f (0, 0) = lim = lim = 3 cos2 (θ) sin(θ) h→0 h→0 h h Notice that by using the definition, θ = 0 corresponds to the rate of change parallel to the x−axis, and θ = π/2 is the rate of change parallel to the y−axis: fx (0, 0) = 0 fy (0, 0) = 0 so that, if f were differentiable at the origin, we could use Du f = ∇f · ~u = 0 for every vector ~u, but that is not the case (our directional derivative is not always zero). 2 5. If f (x, y) = sin(2x + 3y), then find the linearization of f at (−3, 2). SOLUTION: We have f (−3, 2) = sin(0) = 0 and fx (x, y) = 2 cos(2x + 3y) ⇒ fx (−3, 2) = 2 fy (x, y) = 3 cos(2x + 3y) ⇒ fy (−3, 2) = 3 Therefore, L(x, y) = 0 + 2(x + 3) + 3(y − 2) = 2(x + 3) + 3(y − 2) 6. The radius of a right circular cone is increasing at a rate of 3.5 inches per second while its height is decreasing at a rate of 4.3 inches per second. At what rate is the volume changing when the radius is 160 inches and the height is 200 inches? (V = 31 πr2 h) SOLUTION: dV 2 dr 1 dh = πrh + πr2 dt 3 dt 3 dt Use r = 160, h = 200, dr/dt = 3.5 and dh/dt = −4.3, dV /dt ≈ 37973.3π 7. Find the differential of the function: v = y cos(xy) SOLUTION: dv = vx dx + vy dy = (−y 2 sin(xy) dx + (cos(xy) − xy sin(xy)) dy 8. Find the maximum rate of change of f (x, y) = x2 y + it occurs. √ y at the point (2, 1), and the direction in which SOLUTION: The maximum rate of change occurs if we move in the direction of the gradient. We see this by recalling that: Du f = ∇f · ~u = |∇f | cos(θ) so we find the gradient at the point (2, 1) ∇f = h4, 9/2i So if we move in that direction, the we get the max rate of change, which is r √ 81 145 2 |∇f | = 4 + = ≈ 6.02 4 2 9. Find an expression for d [u(t) · (v(t) × w(t)] dt SOLUTION: d [u(t) · (v(t) × w(t)] = u0 · (v × w) + u · (v × w)0 dt Taking this derivative, we see that d [u(t) · (v(t) × w(t)] = u0 · (v × w) + u · (v0 × w + v × w0 ) dt 10. Use Lagrange Multipliers to find the maximum and minimum of f subject to the given constraints: f (x, y) = x2 y x2 + y 2 = 1 SOLUTION: At optimality, the gradients are parallel, so the system of equations we are solving is given below: 2xy x2 2 x + y2 3 = 2λx = 2λy =1 If x 6= 0 in the first equation, then λ = y. Going to the second equation, that implies that x2 = 2y 2 . Now to the third equation, we can solve for y, and therefore also x: r r 1 2 2 3y = 1 ⇒ y = ± ⇒ x=± 3 3 Are there any other solutions? If x = 0 in the first equation, then y = ±1 from the third equation- in that case, f (0, ±1) = 0. Substitute into f and we find the max and min: r r ! 2 1 2 f ± , = √ 3 3 3 3 r ± f 2 ,− 3 r ! 1 −2 = √ 3 3 3 11. The curves below intersect at the origin. Find the angle of intersection to the nearest degree: ~r1 (t) = ht, t2 , t9 i ~r2 (t) = hsin(t), sin(5t), ti SOLUTION: The angle of intersection is the angle between the tangent vectors at the origin. First differentiate, then evaluate at t = 0: ~r1 0 (t) = h1, 2t, 9t8 i ~r2 0 (t) = hcos(t), 5 cos(5t), 1i ⇒ h1, 0, 0i, h1, 5, 1i To find the angle, we use the relationship: u · v = |u| |v| cos(θ) In our case: cos(θ) = √ 1 12 + 52 + 12 √ θ = cos−1 (1/ 27) ≈ 79◦ ⇒ 12. Find three positive numbers whose sum is 100 and whose product is a maximum. SOLUTION: Let x, y, z be the three numbers. Then we want to find the maximum of P (x, y, z) = xyz subject to the constraint that x + y + z = 100 and they are all positive. Alternative 1: Let P (x, y) = xy(100 − x − y). The critical points are y(100 − x − y) − xy x(100 − x − y) − xy =0 =0 From the first equation, y = 100 − 2x (we can throw out y = 0). Substitute this into the second equation to find that x = 100/3. Therefore, y = 100/3 and z = 100/3. These are the three numbers we wanted. Alternative 2: Using Lagrange Multipliers, yz xz xy x+y+z =λ =λ =λ = 100 From the first three equations, if we do not allow zero (then P = 0), we have x = y = z. Substitute into the fourth equation to see that x = y = z = 100/3. 13. Find the equation of the tangent plane and normal line to the given surface at the specified point: x2 + 2y 2 − 3z 2 = 3 (2, −1, 1) SOLUTION: This is an implicitly defined 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 (2, −1, 1), and construct the plane and line: Fx = 2x Fy = 4y Fz = −6z 4 ⇒ ∇F (2, −1, 1) = h4, −4, −6i Thus, the tangent plane is: 4(x − 2) − 4(y + 1) − 6(z − 1) = 0 The normal line goes in the direction of the gradient, starting at the given point. In parametric form, x(t) = 2 + 4t y(t) = −1 − 4t z(t) = 1 − 6t 14. If z = x2 − y 2 , x = w + 4t, y = w2 − 5t + 4, w = r2 − 5u, t = 3r + 5u, find ∂z/∂r. SOLUTION: Use a chart to keep track of the variables; see the solution attached. 15. If x2 + y 2 + z 2 = 3xyz and we treat z as an implicit function of x, y, then find ∂z/∂x and ∂z/∂y. SOLUTION: Let us define F (x, y, z) = x2 + y 2 + z 2 − 3xyz in keeping with the notation from the text. Then we compute: F (x, y, z) = 0 ⇒ Fx ∂x ∂y ∂z ∂z −Fx −(2x − 3yz) + Fy + Fz =0⇒ = = ∂x ∂x ∂x ∂x Fz 2z − 3xy Similarly, we can show that −Fy ∂z −(2y − 3xz) = = ∂y Fz 2z − 3xy 16. If a(t) = −10k and v(0) = i + j − k, r(0) = 2i + 3j, find the velocity and position vector functions. SOLUTION: Z v(t) = a(t) dt = h0, 0, −10ti + v0 = h1, 1, −10t − 1i And antidifferentiate once more: Z r(t) = v(t) dt = ht, t, −5t2 − ti + r0 = ht + 2, t + 3, −5t2 − ti √ 5x − 4y at (4, 1) using a gradient. √ SOLUTION: The gradient of g is orthogonal to its level curve at 5x − 4y = 4. Find the gradient of g at (4, 1): 1 ∇g = h5, −4i 8 For the line, we simply need to move in the direction of the gradient, so we can simplify the direction to h5, −4i (not necessary, but easier for the algebra). 17. Find the equation of the normal line through the level curve 4 = Therefore, the line (in parametric and symmetric form) is: x(t) = 4 + 5t y(t) = 1 − 4t or y−1 x−4 = 5 −4 Notice that the slope is −4/5. If we wanted to check our answer, we could find the slope of the tangent line: dy dy 5 5x − 4y = 16 ⇒ 5 − 4 =0 ⇒ = dx dx 4 18. Find all points at which the direction of fastest change in the function f (x, y) = x2 + y 2 − 2x − 4y is ~i + ~j. SOLUTION: The direction of the fastest increase is in the direction of the gradient. Therefore, another way to phrase this question is: When is the gradient pointing in the direction of h1, 1i (very reminiscent of the Lagrange Multiplier): ∇f = kh1, 1i ⇒ h2x − 2, 2y − 4i = hk, ki So k = 2x − 2 and k = 2y − 4, therefore, the points are on the line 2x − 2 = 2y − 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 gradient: (2a − 2)h1, 1, i a>1 5 19. Find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in the plane x + 2y + 3z = 6. SOLUTION: The volume is V = xyz subject to the constraint that x + 2y + 3z = 6. Alternative 1: Take V = 13 xy(6 − x − 2y), then find the CPs. The only CP with no zeros is x = 2, y = 1, z = 2/3. Alternative 2: Use Lagrange Multipliers to get the same answer (probably much quicker). 20. Find and classify the critical points: f (x, y) = 4 + x3 + y 3 − 3xy SOLUTION: The partial derivatives are: fx = 3x2 − 3y fy = 3y 2 − 3x fxx = 6x fyy = 6y fxy = −3 The critical points are where x2 = y and y 2 = x, so x, y are both zero or positive: x4 = x ⇒ x4 − x = 0 x(x3 − 1) = 0 ⇒ so x = 0, y = 0 or x = 1, y = 1. Put these points into the Second Derivatives Test: 2 fxx (0, 0)fyy (0, 0) − fxy (0, 0) = −9 < 0 ⇒ 2 fxx (1, 1)fyy (1, 1) − fxy (1, 1) = 36 − 9 > 0 The origin is a SADDLE fxx (1, 1) > 0 ⇒ Local MIN 21. Let f (x, y) = x − y 2 . Find the gradient of f at (3, −1). We said that this gradient was perpendicular to a level curve of f - Which one? Draw a sketch showing the level curve and the gradient vector, then find the equation of the tangent line to the level curve and the equation of the normal line. SOLUTION: The level curve is the one that contains the given point (in this case, (3, −1)): Substitute to get 3 − 1 = 2, so the curve is x − y 2 = 2 or x − 2 = y 2 , which is a sideways parabola shifted to the right two units. The gradient at (3, −1) is h1, 2i. The tangent line has slope −1/2: 1 Tangent Line: 1(x + 3) + 2(y + 1) = 0 or y + 1 = − (x − 3) 2 Normal Line:x = 3 + t, y = −1 + 2t or y+1 = (x − 3) or y + 1 = 2(x − 3) 2 22. Find the equation of the tangent plane to the surface implicitly defined below at the point (1, 1, 1): x3 + y 3 + z 3 = 9 − 6xyz SOLUTION: First write this as F (x, y, z) = 0, then the gradient of F is the normal vector for the plane (evaluated at (1, 1, 1)): F (x, y, z) = x3 + y 3 + z 3 + 6xyz − 9 = 0 Then, at (1, 1, 1) Fx = 3x2 + 6yz Fy = 3y 2 + 6xz Fz = 3z 2 + 6xy ⇒ ∇F = h9, 9, 9i Therefore, the tangent plane is: 9(x − 1) + 9(y − 1) + 9(z − 1) = 0 Alternate and Note: We could have used the following formulas to compute the tangent plane: zx = −Fx = −1 Fz zy = −Fy = −1 Fz to write the tangent plane as: z − 1 = −(x − 1) − (y − 1) 6 23. Find parametric equations of the tangent line at the point (−2, 2, 4) to the curve of intersection of the surface z = 2x2 − y 2 and z = 4. (Hint: In which direction should the tangent line go?) The curve is 2x2 − y 2 = 4, which is an ellipse (at height 4 in 3-d). The gradient is h4x, −2yi so at (−2, 2), the gradient is < −8, −4 >, so the tangent line (in the xy plane) is: −8(x + 2) − 4(y − 2) = 0 or y = −2(x + 2) + 2 The tangent line can be expressed as: ht, −2(t + 2) + 2, 4i (But there are multiple ways of expressing it). 24. Find and classify the critical points: f (x, y) = x3 − 3x + y 4 − 2y 2 SOLUTION: We use the second derivatives test to classify the critical points as local min, local max or saddle. Solving for the CPs, we get: fx (x, y) = 3x2 − 3 = 0 fy (x, y) = 4y 3 − 4y = 0 from which we get x = ±1, y = ±1 and x = ±1, y = 0 Continuing with second derivatives,   6x 0 fx x fx y = 24x(3y 2 − 1) = D(x, y) = 0 12y 2 − 4 fy x fy y We’ll arrange the results in a list: Point (1, 1) (1, −1) (1, 0) (−1, 1) (−1, −1) (−1, 0) D and Classification 48 : Local Min 48 : Local Min −24 : Saddle −48 : Saddle −48 : Saddle 24 : Local Max 7