Multivariable Calculus Practice Midterm 2 Solutions Prof. Fedorchuk

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1 Multivariable Calculus Practice Midterm Solutions Prof. Fedorchuk. ( points) Let f(x, y, z) xz + e y x. a. (4 pts) Compute the gradient f. b. ( pts) Find the directional derivative D,, f(,, ). c. ( pts) Find the unit vector pointing in the direction along which f(x, y, z) increases most rapidly at the point (,, ). Solution: (a) We have f z xe y x, e y x, x and so f(,, ),,. (b) Therefore D,, f(,, ) f(,, ),,,,,, /. (c) The direction of the largest increase of f(x, y, z) at (,, ) is f(,, ),, and the unit vector in this direction is,,.. ( points) a. (5 pts) Find an equation of the tangent plane at (, ) to the graph of f(x, y) xy xy + x y. Solution: The equation of the tangent plane at (, ) is: We now compute z f(, ) f x (, )(x ) + f y (, )(y ). f(, ) f x (, ) (y y + 9x y) (,) f y (, ) (xy x + x ) (,) Therefore, the equation of the tangent plane at (, ) is: z 5 (x ) + 9(y ).

2 b. (5 pts) Find an equation of the tangent plane at the point (,, ) to the surface g(x, y, z) ze x + e z+ + xy + y. Solution: We compute g(x, y, z) ze x + y, x +, e x + e z+. The normal vector of the tangent plane at the point (,, ) is so the tangent plane is g(,, ),,, x + (y ) + (z + ).. ( points) The surface contains curves r (t) (t, t, t ) and r (u) (, sin(u), cos(u)). Find the tangent plane to the surface at the point (,, ) and the line perpendicular to the tangent plane and passing through the origin (,, ). Solution: We are going to use the fact that the tangent plane to a surface at a point P contains the tangent lines at P of all curves contained in the surface and passing through P. Here we have two curves contained in the surface and passing through P (,, ). The tangent line of the first curve at P (corresponding to t ) has direction vector v r (),, and the tangent line of the second curve at P (corresponding to u π/4) has direction vector w r (π/4),,. Since the normal vector of the tangent plane to the surface at P is perpendicular to these two vectors, we can take the normal vector of the tangent plane to be v w 5,,. Finally, the equation of the tangent plane at P (,, ) is 5(x ) + (y ) + (z ). The line perpendicular to the tangent plane and passing through the origin (,, ) is x 5t y t z t

3 4. ( points) The function f(x, y) x + y + xy + 9x has critical point. Find it, and identify it as a local minimum, local maximum, or a saddle point. Solution: To find critical points, we have to solve the system { f x (x, y) x + y + 9 f y (x, y) y + x Solving, we obtain y and x 6. Hence (x, y) ( 6, ) is the only critical point. Next we use the nd derivative test: The determinant at ( 6, ) is f xx ( 6, ) f yy ( 6, ) f xy ( 6, ). D( 6, ) f xx ( 6, )f yy ( 6, ) (f xy ( 6, )) 4 >. Since f xx ( 6, ) >, the nd derivative test says that ( 6, ) is a local minimum. 5. ( points) Consider the function f(x, y) x y xy. Find and classify all critical points of f. Solution: To find critical points, we have to solve the system { f x (x, y) x y y y(x y) f y (x, y) x 6xy x(x 6y) The only solution of this system is (x, y) (, ), hence (, ) is the only critical point. Since f xx (, ) f yy (, ) f xy (, ), the nd derivative test is inconclusive. However note that along y x, the function is f(x, x ) x 5 x 5 x 5. Since x 5 has neither maximum nor minimum at x ( x 5 is strictly decreasing because ( x 5 ) x 4 ). We conclude that (, ) is a saddle point of f(x, y), i.e. (, ) is neither local minimum nor local maximum. 6. ( points) Find the maximum and minimum values of f(x, y) x y subject to the constraint x 4 + y. Solution: The method of Lagrange multipliers says that extrema of f(x, y) subject to the constraint are found among the solutions of f(x, y) λ g(x, y), where g(x, y) x 4 +y. We rewrite this condition in the following form: f x λg x λ(x/) f y λg y λ(y) x 4 + y

4 or λx/ λy x 4 + y This gives x /λ and y /λ. Plugging these into the constraint, we obtain: (/λ) 4 + ( /λ) /λ, which gives λ or λ. It follows that the two critical points of f(x, y) subject to the constraint are (, ) and (, ). We have f(, ) 4 and f(, ) 4. Therefore, f(, ) 4 is the maximum value of f(x, y) subject to the constraint x + 4 y ; and f(, ) 4 is the minimum value of f(x, y) subject to the constraint x + 4 y. 7. ( points) Evaluate each of the following iterated integrals. a. (5 pts) Solution: x e x+y dydx x ( ) x ex+y dx (ex e x )dx ( ) e x 6 ex e x+y dydx ( e 6 e e ) (/6 /) 6 e + /. b. (5 pts) Solution: The region 6 4 x sin(y )dydx R {(x, y) x 6, x y 4} is given to us in the form of an y-simple region. Since it is difficult to integrate sin(y ), we are going to rewrite R in the form of an x-simple region. To do this, observe that x y 4 implies that x y and that for any y in the interval [, 4], x has to satisfy only x y. Hence R {(x, y) y 4, x y }. 4

5 Therefore 6 We now compute 4 y 4 sin(y )dxdy x sin(y )dydx y y sin(y )dy sin(u)du sin(y )dxdy. ( cos(64) cos()) ( cos(64)). 8. ( points) Compute the volume of the solid in the first octant bounded by x + y + z 4. Solution: We are integrating the function f(x, y) 4 x y over the region R {(x, y) : x, y, x + y 4} {(x, y) : x, y x}. The volume is thus x (4 x y)dydx (4y xy y ) x dx (x 4x + 4)dx (x / x + 4x) 8/. 5