Calculus III Chris Herald Math 283 Fall 2014 Review #2 Practice Problems 1. Let f (x, y) = x2 y 3 . What is the linear approximation L(x, y) for this function centered at (2, 1)? Use the linear approximation to obtain an approximate value for f (1.99, 1.01). Solution. L(x, y) = f (2, 1) + fx (2, 1)(x − 2) + fy (2, 1)(y − 1) = 4 + 4(x − 2) + 12(y − 1). Thus f (1.99, 1.01) ≈ 4 + 4(−0.01) + 12(0.01) = 4.08. 2. If g(x, y, z) = zex−xy , compute gyy and gyz . Solution. gy = zex−xy (−x) = −xzex−xy . Thus, gyy = x2 zex−xy and gyz = −xex−xy . 3. With a coordinate system on a region of Nevada such that North corresponds to ˆj and East corresponds to ˆi, the elevation is given by a function e(x, y) with the property that ∇e(x, y) = h−0.3, 0.4i everywhere. (a) What implications does this have for someone hiking North? (Be as specific as you can). Is it uphill or downhill? How steep is it (i.e. how many feet does one climb per foot traveled North)? Solution. Uphill, 4 feet up per 10 feet North. (b) Describe the implications does this have for someone hiking West. Solution. Note that West is the -ˆi = h−1, 0i direction. Uphill, 3 feet up per 10 feet West. (c) If you like to hike uphill at a rate of 1 foot of elevation gain per 10 feet of horizontal travel, then in which direction or directions should you hike? Solution. If you hike at an angle of θ from the gradient, the directional derivative (rate of climb) is |∇e| |ˆ u| cos θ = 0.5 cos θ. We want this to equal 0.1, so we need cos θ = 0.1/0.5 = 0.2. Therefore, the two directions arccos(0.2) = 1.369 radians to the right or left of the gradient direction will have the prefered climb rate. (d) If, instead of elevation, e(x, y) represents the temperature, what is the corresponding question (a) above? Note that “steep” is a peculiar word to use in this context. Find a better way to express what the gradient tells you about someone hiking North. Solution. If e(x, y) is the temperature at the point (x, y), then the directional derivative is a rate of change of temperature per unit distance traveled in the specified direction. As one hikes North, since the directional derivative is .4 in this direction, the temperature is increasing at the rate of .4 degrees per foot traveled in the Northern direction. (Most people would say this is a rapid change or a quick change, when one moves North, but would not use the word steep to describe temperature change unless there was a graph of the temperature and the graph was steep.) 4. Let f (x, y) = x3 y + 2 xy . (a) Find the directional derivative Duˆ f (2, 2) where u ˆ= √1 h−1, 4i. 17 Solution. Note that ∇f (x, y) = h3x2 y + y2 , x3 − 2x i, so ∇f (2, 2) = h25, 7i. We compute y2 3 Duˆ f (2, 2) = ∇f (2, 2) · u ˆ= √ . 17 (b) What is the directional derivative at (2, 2) in the direction of θ = counter-clockwise from the x axis? √ Solution. h25, 7i · h √12 , √12 i = 16 2. π 4 radians around (c) What is the direction of fastest increase of f at (2, 2)? At what rate does f increase in this direction? 1 1 Solution. The direction of greatest increase is √ h25, 7i = √ h25, 7i. The 2 2 674 25 + 7 √ rate at which f changes in this direction is 674, the length of the gradient. (That is, |∇f | |ˆ u| cos 0 = |∇f |.) (d) What is the tangent direction of the level curve of f through (2, 2)? Solution. The level curve direction would be perpendicular to gradient, so its tangent direction would be h−7, 25i . One might come up with this by simply noting that taking the vector h25, 7i and interchanging the components and making one of them negative, produces a vector that has dot product zero with h25, 7i. Another way to see this is to view that gradient, for the moment, as a horizontal 3D vector, h25, 7, 0i and cross with h0, 0, 1i. 5. Determine whether the curve x4 + y 3 + xy = 3 intersects the curve x2 + y 2 = 2 orthogonally at (x, y) = (1, 1). Justify your answer. Solution. ∇(x4 +y 3 +xy) = h4x3 +y, 3y 2 +xi = h5, 4i at (1, 1). ∇(x2 +y 2 ) = h2x, 2yi = h2, 2i at this point. Since the level curves are perpendicular to these gradients, if the level curves were perpendicular to each other, then their normal vectors (i.e. the gradient vectors) would be perpendicular to each other. They aren’t, since their dot product is nonzero, so the curves don’t meet perpendicularly. 6. Sketch the gradient vector at the indicated point on the contour diagram in Figure 1. Pay attention to both the direction, and the length. Solution. The dotted line added to the figure indicates the direction of the gradient, because it it perpendicular to the level curve on which P lies. As far as which direction it points, nearby level curve values indicate f increases as we move down and left, so the gradient points down and left along this line. To estimate the length of the gradient, the function goes up 4 units between P and the next place this dotted line hits the next contour line. Comparing the distance between these intersections to the tick marks on the axes, it appears these points are one unit apart, so f increases 4 units when we move one unit in this direction. Therefore, the gradient vector should be 4 units long (making it stick well outside this picture, something I can’t do with my graphics program). Figure 1: The contour diagram for Problem 6. 7. Find the lowest value of x2 − y 2 on the circle x2 + y 2 = 1. Solution. Set g(x, y) = x2 + y 2 − 1 = 0 and f (x, y) = x2 − y 2 . fx = λgx fy = λgy g(x, y) = 0 2x = λ2x 2y = −λ2y 2 2 x +y −1 = 0 (λ − 1)2x = 0 (λ + 1)2y = 0 x2 + y 2 = 0 Case 1. λ = 1. Then 4y = 0, so y = 0. Then the last equation implies that x = ±1. So we get candidates (1, 0) and (−1, 0). Case 2. x = 0. Then the last equation implies y = ±1 (and λ = −1, but we are not so interested in this value), so we get candidates (0, 1) and (0, −1). Evaluating f at these points, we can see that the minimum value is -1, and it occurs at (0, ±1). 8. Find and classify the critical points of f (x, y) = x3 − 3xy + y 3 . Solution. fx = 3x2 − 3y = 0 x2 = y fy = −3x + 3y 2 = 0 (y 2 )2 = y y2 = x y 4 − y = 0 so y(y 3 − 1) = 0 y = 0 and so x = 02 = 0, or y = 1 and so x = 12 = 1. This gives critical points (0, 0) and (1, 1). Perform the second derivative test, using 2 D = fxx fyy − fxy = (6x)(6y) − (−3)2 . (0, 0) is a saddle since D < 0 there, and (1, 1) is a local min because D > 0 and fxx > 0 there. 9. Write down the equation of the tangent plane to the cylinder y 2 + z 2 = 25 at the point (3, −3, 4). Is it parallel to the plane −3x + 3y + 4z = 2? Explain briefly. Solution. To find the normal vector to the tangent plane, use ∇(y 2 + z 2 ) = h0, 2y, 2zi. At (3, −3, 4), the normal vector to the tangent plane is h0, −6, 8i. Thus the equation of the tangent plane is 0(x − 3) − 6(y + 3) + 8(z − 4) = 0. The normal vector to the other plane is h−3, 3, 4i, which is not a multiple of h0, −6, 8i, so the normal vectors are not parallel. Therefore, the planes are not parallel. (You might also ask if they are perpendicular planes. Since the normal vectors are not perpendicular, they are not that either.) 10. What is the tangent plane to the graph of f (x, y) = cos x sin y at the point ( π2 , π4 , 0)? Solution. The normal vector is given by h−fx , −fy , +1i = hsin x sin y, − cos x cos y, 1i evaluated at x = π2 , y = 1 1 · √ , 0, 1 . 2 π 4 = Thus the tangent plane equation is 1 π π 1 π √ x− +0 y− + (z − 0) = 0 or z = − √ x − . 2 4 2 2 2 11. Suppose f (x, y, z) = x2 y + z, and a point is moving through space with position vector ~r(t) = hx(t), y(t), z(t)i. At one instant in time, t = 0, the point has coordinates x(0) = 1, y(0) = 2, z(0) = 1, and the velocity at that instant is hx0 (0), y 0 (0), z 0 (0)i = h1, −1, 2i. What d is dt f (x(t), y(t), z(t)) at that instant? dy dz = fx dx pardt + fy dt + fz dt (where the tial deriatives are evaluated at the current x, y, z values, so we obtain 2xy|(1,2,1) (1) + Solution. The chain rule says x2 (1,2,1) (−1) + 1(2) = 5. d dt f (x(t), y(t), z(t))