Math 210, Final Exam, Spring 2010 Problem 1 Solution

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1 Problem Solution. The position vector r (t) t3î+8tĵ+3t ˆk, t describes the motion of a particle. (a) Find the position at time t. (b) Find the velocity at time t. (c) Find the acceleration at time t. (d) Find the length of the path traveled by the particle during the time t. Solution: (a) The position at time t is: r () 3î+8()ĵ+3() ˆk 8î+36ĵ+ 3 ˆk (b) The velocity is the derivative of position. v(t) r (t) 3t î+8ĵ 3t ˆk Therefore, the velocity at time t is: v() 3()î+8ĵ 3() ˆk î+8ĵ 3 4 ˆk (c) The acceleration is the derivative of velocity. a(t) v (t) 6tî+6t 3 ˆk Therefore, the acceleration at time t is: a() 6()î+6() 3 ˆk î+ 3 4 ˆk (d) The length of the path traveled by the particle is: L r (t) dt (3t ) +8 +( 3t ) dt 9t t 4 dt It turns out that a simple antiderivative of the integrand does not exist. There was a typo in the original problem. The ĵ-component of r (t) should have been 8t not 8t.

2 Problem Solution. (a) For f(x,y) e (x+)y find the derivatives: f x, f y, f x, f x y, f y (b) Find the gradient of f at the point (,3). Solution: (a) The first partial derivatives of f(x,y) are The second derivatives are: (b) The gradient of f at (,3) is: f x ye(x+)y f y (x+)e(x+)y f x x ( ) ye (x+)y y e (x+)y x f y y ( ) (x+)e (x+)y (x+) e (x+)y y f x y ( ) (x+)e (x+)y e (x+)y +y(x+)e (x+)y x f(,3) fx (,3),f y (,3) 3e (+)3,(+)e (+)3 3e 9,3e 9

3 Problem 3 Solution 3. (a) Find a potential function for the vector field F(x,y,z) ( z)î+yĵ xˆk (b) Integrate F over the straight line from (,,) to (,,). [You may calculate this directly or you may use a potential function.] Solution: (a) By inspection, a potential function for the vector field F is: ϕ(x,y,z) x xz + y To verify, we calculate the gradient of ϕ: ϕ ϕx î+ϕ y ĵ+ϕ z ˆk ( z)î+yĵ xˆk F (b) Using the Fundamental Theorem of Line Integrals, the value of the line integral is: F d s ϕ(,,) ϕ(,,) C [ ()()+ ] () [ ()()+ ] ()

4 Problem 4 Solution 4. (a) Find the critical points of the function f(x,y) x 3 3x y. (b) Use the second derivative test to classify each critical point as a local maximum, local minimum, or saddle. Solution: (a) By definition, an interior point (a,b) in the domain of f is a critical point of f if either () f x (a,b) f y (a,b), or () one (or both) of f x or f y does not exist at (a,b). The partial derivatives of f(x,y) x 3 3x y are f x 3x 3 and f y y. These derivatives exist for all (x,y) in. Thus, the critical points of f are the solutions to the system of equations: f x 3x 3 () f y y () The two solutions to Equation () are x ±. The only solution to Equation () is y. Thus, the critical points are (,) and (,). (b) We now use the Second Derivative Test to classify the critical points. The second derivatives of f are: f xx 6x, f yy, f xy The discriminant function D(x, y) is then: D(x,y) f xx f yy f xy D(x,y) (6x)( ) () D(x,y) x The values ofd(x,y)atthe critical points andtheconclusions of thesecond Derivative Test are shown in the table below. (a,b) D(a,b) f xx (a,b) Conclusion (, ) 6 Saddle Point (, ) 6 Local Maximum ecall that (a,b) is a saddle point if D(a,b) < and that (a,b) corresponds to a local maximum of f if D(a,b) > and f xx (a,b) <.

5 Problem 5 Solution 5. Find the maximum and minimum of the function f(x,y) (x ) +y subject to the constraint: ( x ) ( y ) g(x,y) + 3 Solution: We find the minimum and maximum using the method of Lagrange Multipliers. First, we recognize that ( x 3 ) +( y ) is compact which guarantees the existence of absolute extrema of f. We look for solutions to the following system of equations: f x λg x, f y λg y, g(x,y) which, when applied to our functions f and g, give us: ( ) x (x ) λ () 9 ( y y λ () ) ( x ) ( y ) + (3) 3 From Equation () we observe that: If y then Equation (3) gives us: ( y y λ ) 4y λy 4y λy y(4 λ) y, or λ 4 ( x 3 ) + ( ) x 9 x 9 x ±3

6 If λ 4 then Equation () gives us: ( ) x (x ) λ 9 ( ) x (x ) 4 9 x 4x 9 5x 9 x 9 5 which, when plugged into Equation (3), gives us: ( x 3 ) + ( y ) ( ) 9/5 + y y 4 y y 64 5 y ± 8 5 Thus, the points of interest are (3,), ( 3,), ( 9 5, 8 5 ), and (9 5, 8 5 ). We now evaluate f(x,y) (x ) +y at each point of interest. f(3,) (3 ) + 4 f( 3,) ( 3 ) + 6 f( 9 5, 8 5 ) (9 5 ) +( 8 5 ) 6 5 f( 9 5, 8 5 ) (9 5 ) +( 8 5 ) 6 5 Fromthe values above we observe that f attains an absolute maximum of 6 and an absolute minimum of 6 5.

7 Problem 6 Solution 6. Compute the integral xydxdy over the quarter circle {(x,y) : x, y, x + y }. [You may use polar or Cartesian coordinates.] Solution: y x y D x From the figure we see that the region D is bounded on the left by x and on the right by x y. The projection of D onto the y-axis is the interval y. Using the order of integration dxdy we have:

8 xydxdy y xydxdy [ ] y x y dy ( ) ydy y ( ) y ydy ( ) y y 3 dy [ y ] 4 y4 [ () ] 4 ()4 8

9 Problem 7 Solution 7. Compute the integral dx dy dz over the tetrahedron {(x, y, z) : x, y, z, x/3 + y/5 + z/7 }. Solution: The region of integration is shown below.

10 The volume of the tetrahedron is V dx dy dz 3 5 5x/3 7 7x/3 7y/ x/ dz dy dx (7 73 x 75 ) y dy dx [7y 73 xy 7 ] 5 5x/3 y dx [ 7 (5 53 ) x 73 (5 x 53 ) x 7 (5 53 ) ] x dx ( x 35 3 x x x + 35 ) 8 x dx ( x + 35 ) 8 x dx [ 35 x 35 6 x x3 ] (3) 6 (3) (3)3 35

11 Problem 8 Solution 8. Find an equation for the tangent plane to the surface defined by xy + z at the point (,,). Solution: We use the following formula for the equation for the tangent plane: f x (a,b,c)(x a)+f y (a,b,c)(y b)+f z (a,b,c)(z c) because the equation for the surface is given in implicit form. Note that n f(a,b,c) f x (a,b,c),f y (a,b,c),f z (a,b,c) is a vector normal to the surface f(x,y,z) C and, thus, to the tangent plane at the point (a,b,c) on the surface. The partial derivatives of f(x,y,z) xy +z are: f x y f y xy f z 4z Evaluating these derivatives at (,, ) we get: Thus, the tangent plane equation is: f x (,,) 4 f y (,,) ()() 4 f z (,,) 4() 8 4(x )+4(y )+8(z )

12 Problem 9 Solution 9. Compute the integral (3x +y)dx+(x +y 3 )dy C over the counterclockwise boundary of the rectangle using Green s theorem or otherwise. {(x,y) : x 3, y } Solution: Green s Theorem states that F d s C ( g x f ) da y where is the region enclosed by C. The integrand of the double integral is: g x f y ( x +y 3) x y x Thus, the value of the integral is: F d s C ( 3x +y ) ( g x f ) da y (x )da (x )dydx [ xy y ] dx ( ) 4x dx [ ] 3 x x (3) (3)

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