Name: PID:Circle your section time/TA:Christine James James3pm (A01) 4pm (A02) 5pm (A03)Math 10CMidterm Examination 2Nov 15, 2010Turn off and put away your cell phone.No calculators or any other devices are allowed on this exam.You may use one page of notes, but no books or other assistance on this exam.Read each question carefully, answer each question completely, and show all of your work.Write your solutions clearly and legibly; no credit will be given for illegible solutions.If any question is not clear, ask for clarification.# Points Score1 72 123 94 85 8Σ 441. Suppose ~w =~ı − 2~ + 3~k and ~v = −3~ı + ~ +~k.(a) (3 points) Find ~v × ~w.Solution: We use the formula~v × ~w = (v2w3− v3w2)~ı − (v1w3− v3w1)~ + (v1w2− v2w1)~k.Therefore we obtain~v × ~w = (v2w3− v3w2)~ı − (v1w3− v3w1)~ + (v1w2− v2w1)~k= (1 · 3 − 1 · (−2))~ı − ((−3) · 3 − 1 · 1)~ + ((−3) · (−2) − 1 · 1)~k= 5~ı + 10~ + 5~k.(b) (4 points) Find the equation of the plane parallel to both ~v and ~w through thepoint (0, −1, 1).Solution: The plane parallel to both ~v and ~w has normal vector ~v × ~w. Thus,the plane has normal vector ~n = 5~ı + 10~ + 5~k, and goes through the pointP0= (0, −1, 1). Therefore, the equation of the plane is given by0 = 5(x − 0) + 10(y − (−1)) + 5(z − 1)or, if we simplify,−1 = x + 2y + z.2. Let f(x, y) = ex+yy + 2x2.(a) (4 points) Find the gradient of f .Solution: The gradient of f is−→∇f = fx~ı + fy~. Taking derivatives, we havefx(x, y) = ex+yy + 4x and fy(x, y) = ex+y+ ex+yy. Therefore, we have−→∇f(x, y) = (ex+yy + 4x)~ı + (ex+y+ ex+yy)~.(b) (4 points) Find the equation for the tangent plane to f at the point (−1, 1, 3).Solution: The tangent plane at (−1, 1) is given by z = f(−1, 1) +fx(−1, 1)(x − (−1)) + fy(−1, 1)(y − 1). Here, f(−1, 1) = 3. From part (a),fx(−1, 1) = e−1+1· 1 + 4(−1) = −3 and fy(−1, 1) = e−1+1+ e−1+1· 1 = 2.Thus, we have the tangent plane at (−1, 1, 3) isz = 3 − 3(x + 1) + 2(y − 1).(c) (4 points) Verify that ~u =12~ı −√32~ is a unit vector. Calculate f~u(2, −2).Solution: To verify that ~u is a unit vector, we take its magnitude:||~u|| =qu21+ u22=r14+34= 1.Therefore, ~u is a unit vector. Moreover, we havef~u(2, −2) = ~u ·−→∇f(2, −2)= (12~ı −√32~) · ((e0(−2) + 4(2))~ı + (e0+ e0(−2))~)= (12~ı −√32~) · (6~ı −~)=12· 6 −√32· (−1)= 3 +√323. Consider the following contour diagram for z = f (x, y):(a) (2 points) Estimate the partial derivatives with respect to x and y at the point(1, 0).Solution: From (1, 0), traveling to the right, there is a distance of 0.4 to thenext contour, so we have fx(1, 0) ≈10.4=52. Similarly, traveling up, there is adistance of about 0.5 to the next contour, so fy(1, 0) ≈10.5= 2.(b) (2 points) Is the directional derivative at (−1, 0) in the direction −~ı + ~ positiveor negative? Explain.Solution: Going from (−1, 0) in the −~ı + ~ direction takes us from the z =1 contour to the z = 2 contour, so the function is increasing. Thus, thedirectional derivative is positive.(c) (2 points) Draw vectors in the (approximate) direction of the gradient vectorsat points A and B. (Do not worry about the length of the vectors, just thedirection).Solution: Both vectors should be normal to the curves at the given points.At A, the gradient west, at B, it goes slightly east of north.(d) (3 points) Is the length of the gradient larger at A or at B? Explain.Solution: The length of the gradient should be larger at B. The contoursare much closer together at B, meaning the function is changing much morequickly. A faster change means a larger derivative, so the gradient vector islonger at B than at A.4. Suppose z = f (x, y) = e2xy3.(a) (5 points) Find the Taylor polynomial of degree 2 at the point (0, −2).Solution: We first need all first and second partials:fx(x, y) = 2e2xy3, fxx(x, y) = 4e2xy3, fxy(x, y) = 6e2xy2,fy(x, y) = 3e2xy2, fyx(x, y) = fxy(x, y), fyy(x, y) = 6e2xyWe then use the formula for a Taylor polynomial of degree 2 at a point (a, b):P2(x, y) = f (a, b) + fx(a, b)(x − a) + fy(a, b)(y − b) +12fxx(a, b)(x − a)2+12fxy(a, b)(x − a)(y − b) +12fyy(a, b)(y − b)212fyx(a, b)(x − a)(y − b).Here, we have (a, b) = (0, −2), and f(0, −2) = −8, and the following:fx(0, −2) = −16, fxx(0, −2) = −32, fxy(0, −2) = 24fy(0, −2) = 12, fyx(0, −2) = 24, fyy(0, −2) = −12Therefore, we have the Taylor polynomialP2(x, y) = −8 − 16x + 12(y + 2) − 16x2+ 12x(y + 2) − 6(y + 2)2+ 12x(y + 2)= −8 − 16x + 12(y + 2) − 16x2+ 24x(y + 2) − 6(y + 2)2(b) (3 points) Use the Taylor polynomial to estimate f(1, −1.5).Solution: To estimate f(1, −1.5), we use P2(1, −1.5). Thus we obtainf(1, −1.5) ≈ P2(1, −1.5) = −8 − 16 + 12(0.5) − 16 + 24(0.5) − 6(0.5)2= −8 − 16 + 6 − 16 + 12 − 1.5= −23.55. (a) (4 points) You are traveling in a car. The car is going 30◦north of east at aconstant speed of 50 miles per hour. Setting north to be the positive y direction,find an expression for the (x, y)-position of the car after t hours. (Hint: Thinkabout the displacement vector after t hours.)Solution: After t hours, the displacement vector has length 50t in thedirection indicated. Therefore, we can write the displacement vector as(50t cos 30)~ı + (50t sin 30)~ = (25√3t)~ı + (25t)~. Thus, the (x, y)-position ofthe car after t hours is x = 25√3t and y = 25t.(b) (4 points) The landscape over which you are traveling is not flat. In fact, ithas a large hill! The height above sea level can be described by the functionz = f (x, y) = e−(x+y)2. Using the information from part (a), calculate the rate ofchange of height above sea level as a function of t.Solution: We may assume here that the change in height does not affectour travel in the x and y directions. So we have the following movement:z = f(x, y) = e−(x+y)2, where x = 25√3t and y = 25t. Thus, by the chainrule, we obtaindzdt=∂z∂xdxdt+∂z∂ydydt= (−2(x + y)e−(x+y)2)(25√3) + (−2(x + y)e−(x+y)2)(25)= (25√3 + 25)(−2(x + y)e−(x+y)2)= (25√3 + 25)(−2(25√3t + 25t)e−(25√3t+25t)2)= (−50√3 − 50)(t(25√3 +