Math 10C - Fall Final Exam

Transcription

1 Math 1C - Fall Final Exam Problem 1. Consider the function f(x, y) = 1 x 2 (y 1) 2. (i) Draw the level curve through the point P (1, 2). Find the gradient of f at the point P and draw the gradient vector on the level curve. (ii) Draw the graph of f. (iii) Explain why the function f admits a global minimum over the rectangle x 2, 1 y 1. Determine the minimum value and the point(s) where it occurs. (i) We have f(1, 2) = 1 so that the level is 1. The level curve is f(x, y) = 1 = x 2 + (y 1) 2 = 2 which is a circle of center (, 1) and radius 2. We have f = ( 2x, 2(y 1)) so that f(p ) = ( 2, 2). The vector should be shown normal to the level curve in (i) and pointing inwards. (ii) The graph of f can be obtained from that of z = x 2 + y 2 by shifting y to y 1, shifting the sign of z and then translating z by 1. The graph is a paraboloid pointing downwards. The highest point is (, 1, 1). (iii) The set x 2, 1 y 1 is compact so the function achieves a global minimum. The minimum value occurs when x 2 + (y 1) 2 is maximal. The maximum value of x 2 occurs when x has the largest magnitude in absolute value, namely at x = 2. For (y 1) 2 we need to take the largest magnitude of y 1 in absolute value. Note that y 2 gives 2 y 1 so the largest y 1 can be is 2 achieved for y = 1. The minimum occurs at (2, 1). The minimum value is if f(2, 1) = ( 2) 2 = 7.

2 Problem 2. Evaluate the following integral by changing the order of integration: 4 2 y ye x5 dx dy. The region of integration is y x 2, y 4. This can be rewritten as x 2, y x 2. The region is contained below the parabola y = x 2, above the x-axis and to the left of x = 2. Changing the order of integration, the integral becomes The inner integral is The outer integral is 4 2 x 2 y ye x5 dx dy = 2 x 2 ye x5 dy dx. ye x5 dy = y2 2 ex5 y=x2 y= = x4 2 ex5. 2 x 4 2 ex5 dx. Writing u = x 5 so that du = 5x 4 dx, the integral becomes 2 x 4 2 ex5 dx = 32 du 1 eu = eu 1 u=32 u= = e

3 Problem 3. (i) Find the critical points of the function (ii) Indicate the type of the critical points. (i) We compute Substituting we obtain f(x, y) = x 3 + 3xy y 3. f x = 3x 2 + 3y = = x 2 + y = f y = 3x 3y 2 = = x = y 2. x 2 = y 4 = y = y = or y 3 = 1 giving y = or y = 1. Thus x = y = or x = 1, y = 1. There are two critical points (ii) We run the second derivative test: At (, ) we have At (1, 1) we have (, ) and (1, 1). A = f xx = 6x B = f xy = 3 C = f y y = 6y. A = C =, B = 3 = AC B 2 = 9 < = A = 6, B = 3, C = 6 = AC B 2 = 27 >, A > = saddle. local minimum.

4 Problem 4. Find the global minimum and global maximum of the function f(x, y) = x 2 + 2y 2 2x 8y + 9 over the closed disk x 2 + 2y Over the interior x 2 + 2y 2 < 36 we find the critical points of f: f x = 2x 2 = = x = 1 f y = 4y 8 = = y = 2. The only critical point is (1, 2) with f(1, 2) =. Over the boundary g = x 2 + 2y 2 = 36, we use Lagrange multipliers to find f = λ g. Since f = (2x 2, 4y 8) and g = (2x, 4y) we obtain 2x 2 = 2λx = x 1 = λx = x(1 λ) = 1 = x = 1 1 λ 4y 8 = 4λy = y 2 = λy = y = 2 1 λ. Thus y = 2x. Substituting into x 2 + 2y 2 = 36 we obtain x 2 + 2(2x) 2 = 36 = 9x 2 = 36 = x 2 = 4 = x = ±2 = y = ±4. We obtain the critical points ±(2, 4). We compute f(2, 4) = 9, f( 2, 4) = 81. Thus the minimum is at (1, 2) while the maximum is 81 achieved at ( 2, 4).

5 Problem 5. Assume that z = ye x, x = u uv, y = u v, u = 3s + t, v = t2. Compute z s at the point where s = t = 1. We compute z s = z x x u u s + z y y u u s. When s = t = 1 we obtain u = 4, v = 1 and x =, y = 4. We compute z x = ye x = 4 e = 2 Substituting we find z y = 1 2 y ex = e = 1 4 x u = 1 v = y u = 1 v = 1 u s = 3. z s = = 3 4.

6 Problem 6. Consider the points P (1,, 1), Q( 2, 1, 3), R(1, 1, ). (i) Find the equation of the plane through P, Q, R. (ii) Find the cosine of the angle between P Q and P R. (i) We have P Q = ( 3, 1, 2), P R = (, 1, 1). We find P Q P R = ( 3, 1, 2) (, 1, 1) = (1, 3, 3). The equation of the tangent plane has coefficient (1, 3, 3). Requiring that the plane pass through P, we find the equation of the plane to be (ii) We compute Thus cos θ = (x 1) 3y + 3(z 1) = = x 3y + 3z = 4. cos θ = P Q P R P Q P R. ( 3, 1, 2) (, 1, 1) ( 3) ( 3) + 1 ( 1) + 2 ( 1) = = ( 1) 2 + ( 1)

7 Problem 7. Consider the function f(x, y) = 2y ln(x y). (i) Find the direction of steepest increase for f at the point P (2, 1). (ii) Find the directional derivative of f at P in the direction of the vector u = 3 i 4 j (iii) Approximate linearly the value f((2, 1) u). (i) We compute 1 f x = 2y x y = f x(2, 1) = 2 f y = 2 ln(x 1 y) + 2y x y 1 2 y = f y(2, 1) = 1. Therefore f(2, 1) = (2, 1). The unit direction of steepest increase is (ii) We have v = f f = (2, 1) (2, 1) = ( 1) 2 5 ( ) 3 f u = f u = (2, 1) 5, 4 = ( 1) = 2. (iii) We have f(2, 1) = 2 1 ln 1 =. Thus ( f (2, 1) + 1 ) 2 u 1 = f(2, 1) + f u (2, 1) 2 = = 1 1 =.1. 5.

8 Problem 8. Consider the function f(x, y) = x 2 y 3 e xy 1. (i) Find the tangent plane to the graph of f at (1, 1, ). (ii) Find the quadratic approximation of the function f near (1, 1). (i) We find f x = 2xy 3 ye xy 1 = f x (1, 1) = 2 e = 2 1 = 1 f y = 3x 2 y 2 xe xy 1 = f y (1, 1) = 3 e = 2. The tangent plane is (ii) We compute z = (x 1) + 2(y 1) = z = x + 2y 3. f xx = 2y 3 y 2 e xy 1 = f xx (1, 1) = 1 f xy = 6xy 2 xye xy 1 e xy 1 = f xy (1, 1) = 4 f yy = 6x 2 y x 2 e xy 1 = f yy (1, 1) = 5. The quadratic approximation is f f(1, 1)+f x (1, 1)(x 1)+f y (1, 1)(y 1)+ f xx 2 (x 1)2 +f xy (1, 1)(x 1)(y 1)+ f yy(1, 1) (y 1) 2 2 which becomes z + (x 1) + 2(y 1) (x 1)2 + 4(x 1)(y 1) (y 1)2.

9 Problem 9. Consider the triangular T region bounded by Calculate x, y, x + 2y 2. T x + y dx dy. We have y 1 and x 2 2y. The integral to be computed is The inner integral is 2 2y The outer integral is 1 2 2y x + y dx dy. x + y dx = x2 2 + (2 xy x=2 2y 2y)2 x= = + (2 2y)y 2 = 2(1 y) 2 + 2(1 y)y = 2(1 2y + y 2 ) + 2(y y 2 ) = 2(1 y). 1 2(1 y) dy = 2 ) (y y2 y=1 y= 2 = 1.