Winter 2012 Math 255 Section 006. Problem Set 7

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1 Problem Set 7 1 a) Carry out the partials with respect to t and x, substitute and check b) Use separation of varibles, i.e. write as dx/x 2 = dt, integrate both sides and observe that the solution also blows up in finite time. b) Section 15.6: 2) Near a buoy, the depth of a lake at the point with coordinates (x, y) is z = x y 3, where x, y and z are measured in meters. A fisherman in a small boat starts at the point (80, 60) and moves towards the buoy, which is located at (0, 0). Is the water under the boat getting deeper or shallower when he departs? Explain. We find that Dz = (0.04x, 0.003y 2 ) and hence that Dz(80, 60) = (3.2, 10.8). The direction of motion is ( 4, 3) and so (3.2, 10.8) ( 4, 3) = We conclude that the water under the boat is getting deeper. 3) If g(x, y) = x y 2, find the gradient vector g(3, 1) and use it to find the tangent line to the level curve g(x, y) = 2 at the point (3, 1). Sketch the level curve, tangent line and gradient vector. The gradient vector is given by g = (1, 2y) and so g(3, 1) = (1, 2). The corresponding tangent line 1

2 is given by (x(t), y(t)) = (3, 1) + t( 2, 1). Note that the tangent line is perpendicular to the gradient. The normal vector is along the line (3, 1, 2) + s(1, 2, 0). This are shown in Fig. 1 Figure 1: A computer generated drawing of the surface z = x y 2 along with level curves for z = 2 and the tangent and gradient vectors at (x, y) = (3, 1) 4) Show that the equation of the tangent plane to the elliptic parabloid z c = x2 a + y2 2 b 2 at the point (x 0, y 0, z 0 ) can be written as 2xx 0 + 2yy 0 = z + z 0. a 2 b 2 c 2

3 Now and hence z = z(x o, y o ) = ( 2xc a, 2yc ) b ( 2xo c a, 2y ) oc. b The equation of the tangent plane is (z z o ) = (x x o ) 2x oc a We can rearrange this to be + (y y o) 2y oc b. z z o c + 2x2 o a + 2y2 o b = 2xx o a + 2yy o b but we recall that 2x2 o a + 2y2 o b = 2 2z o c and so we get that z + z o c = 2xx o a + 2yy o b. Section ) a) Use a graph and/or level curves 1 to estimate the local maximum and minimum values and saddle points of f(x, y) = xy exp( x 2 y 2 ) The surface is shown in Fig. 2 b) Use calculus to find these values precisely 1 You may use a computer to generate the graphs. 3

4 Figure 2: A computer generated drawing of the surface z = xy exp( x 2 y 2 ). This problem is simpler in polar coordinates for which f(r, θ) = 0.5r 2 sin(2θ) exp( r 2 ). We let u = r 2 so that we get f(u, θ) = 0.5u sin(2θ) exp( u). Hence f u = 0 implies that u = 1 or sin(2θ) = 0 and f θ = 0 implies that u = 0 or cos(2θ) = 0. Hence we have a saddle at (x, y, f) = (0, 0, 0), local maxima at (x, y, f) = ( ±1 2, ±1 2, exp( 1) 4 ) (0.7, 0.7, 0.09) and local minima at (x, y, f) = ( ±1 2, 1 2, exp( 1) 4 ) (0.7, 0.7, 0.09). 6) a) Show that for all values of a, the function f(x, y) = x 3 3axy + y 3 has no global minimizers or global 4

5 maximizers. We find that f(x, y) for (x, y) (, ) and that f(x, y) for (x, y) (, ) and so we deduce that there are no global maximizers or minimizers. b) For each value of a, find all the critical point(s) of f and determine their nature, that is, determine whether each critical point is a local minimum, local maximum or saddle point. f x = 3x 2 3ay and f y = 3y 2 3ax. We solve these to find that at the critical point x = y = a. To determine the nature of the critical point, we can calculate the hessian, f xx f yy (f xy ) 2 = 36 9a 2. We find that since f xx = 6 > 0, we have a local minimum if a < 2 and a saddle point if a > 2. 5

6 Section ) The total production P of a certain product depends on the amount L of labor used and the amount K of capital investment. The cost of a unit of labor is m and the cost of a unit of capital is n. The productions is fixed at bl α K 1 α = Q, where Q is a constant. What values of L and K minimize the cost function C(L, K) = ml + nk? We find that L = ml + nk + λ(bl α K 1 α Q). At a critical point L L = m + λbαlα 1 K 1 α = 0 L K = n + λb(1 α)lα K α = 0 L λ = blα K 1 α Q = 0 We solve these equations to find that and L = Q b K = Q b ( α m(1 α) ) 1 α ( ) α m(1 α). α 8) The plane 4x 3y + 8z = 5 intersects the cone z 2 = x 2 + y 2 in an ellipse (a) Graph the cone, the plane and the ellipse. We can parameterize the surfaces in cylindrical 6

7 coordinates so that the cone becomes the surface r = z and the plane becomes the surface 5 = r(4 cos θ 3 sin θ + 8). The curve of intersection 5(cos θ,sin θ,1) is 4 cos θ 3 sin θ+8. A plot of the surfaces and the curve of intersection is shown in Fig. 3. (b) Use Lagrange multipliers to find the highest and lowest points on the ellipse. It is easiest to use the parametric form of the curve to find that we need θ ( This implies that 5 4 cos θ 3 sin θ + 8 ) = 0. θ = atan( 0.75) 0.644, , 143. We find that the corresponding heights are and We also want to find these values by maximizing x 2 +y 2 subject to the constraint (4x 3y 5) = 8 x 2 + y 2. We can do thi in polar coordinates. We get that L = r 2 + λ(4r cos θ 3r sin θ 5 8r) so that we need to solve L = 2r + λ(4 cos θ 3 sin θ 8) = 0 r L = 4 sin θ 3 cos θ = 0 θ L = 4r cos θ 3r sin θ 5 8r = 0 λ 7

8 Figure 3: A computer generated drawing of the surfaces 4x 3y + 8z = 5, z 2 = x 2 + y 2 5(cos θ,sin θ,1) and the curve 4 cos θ 3 sin θ+8. 8

9 These are the same equations we had before, and so we get the same values. We can now also calcluate the value of the Lagrange multiplier, which from the Fig. 3 is non-zero since it is binding. Section ) a) Estimate the volume of the solid that lies below the surface z = x 2 + 2y 2 and above the rectangle R = [0, 1] [0, 4]. Use a Riemann sum with m = n = 2 and choose the sample points to be lower right corners. V i,j f(x i, y j ) x y = [f(0, 0) + f(0, 2) + f(1/2, 0) + f(1/2, 2)] (1/2) 2 = = 16.5 b) Use the Midpoint Rule to estimate the volume in part (a). V i,j f(x i+1/2, y j+1/2 ) x y = [f(0.25, 1) + f(0.25, 3) + f(0.75, 1) + f(0.75, 3)] (1/2) 2 = c) Explain why the two sums are so different. 9

10 There are not enough grid points used to obtain a good approximation of the integral. If more grid points are used, then the two sums should approach the same limiting value. 10) Evaluate the integral where R (5 x)da R = {(x, y) 0 x 5, 0 y 3} by identifying it as the volume of a solid. The solid is a triangular prism, which has volume = ) If R = [0, 1] [0, 1] show that 0 sin(x + y)da < 1. R For the domain of interest 0 sin(x + y) < 1. The domain has area equal to one. Hence using the volume interpretation, since the height is always positive and less than one, we deduce that the integral must also be less than one. 12) Consider the problem subject to the constraint min x 2 + y 2 x 2 (y 1) 3 = 0. 10

11 a) Solve the problem geometrically Figure 4 shows that the minimum value is approximately 0.4 which occurs for (x, y) (±0.4, 0.45). Figure 4: A computer generated drawing of the level curves of x 2 + y 2 and the constraint curve x 2 (y 1) 3 = 0. b) Try and solve the problem using Lagrange multipliers L = x 2 + y 2 + λ(x 2 (y 1) 3 ) 11

12 and so the critical point equations are 2x + 2λx = 0 λ = 1 2y 3λ(y 1) 2 = 0 2y + 3(y 1) 2 = 0 x 2 + (y 1) 2 = 0 A solution of 2y 3(y 1) 2 = 0 gives that y = (4± 7/3) 0.451, 21, 5. The corresponding x values are ( ) 2/3 (±0.406) and ( ) (±1.749i). The second set of solutions are not useful and so we find that our solutions are (x, y) = (±0.406, 0.451) with corresponding minimum value c) What is wrong with substituting x 2 = (y 1) 3 in the objective and then trying to solve the unconstrained optimization problem min y 2 + (y 1) 3. Is there any way to fix this approach? The functional y 2 +(y 1) 3 does not have a global minimum and has no real local critical points, since we find the critical point equations are only satisfied for y = 2+i 5 3, this function is shown in Fig. 5. The appropriate unconstrained optimization problem is y 2 (y 1) 3, which has critical points as shown in Fig. 6. The negative sign is obtained from the Lagrange multiplier. 12

13 Figure 5: A computer generated drawing of y 2 + (y 1) 3. Figure 6: A computer generated drawing of y 2 (y 1) 3. 13