MAT 211 Final Exam. Fall Jennings.

Transcription

1 MAT 211 Final Exam. Fall 218. Jennings. Useful formulas polar coordinates spherical coordinates: SHOW YOUR WORK! x = rcos(θ) y = rsin(θ) da = r dr dθ x = ρcos(θ)cos(φ) y = ρsin(θ)cos(φ) z = ρsin(φ) dv = ρ 2 sin(φ) dρ dφ dθ cylindrical coordinates are like polar coordinates, except with z 1. (1 points) Parametrize the line through the points (1,2,3) and (5,9,14). 4,7,11 = 5 1,9 2,14 3 is the vector that points from (1,2,3) to (5,9,14) so the line is parametrized by r(t) = (1,2,3)+t 4,7,11 = 1+4t,2+7t,3+11t. 2. (1 points) Find the angle between the vectors 3,1,2 and 4,6,3. Let v = 3,1,2 and w = 4,6,3. If θ is the angle between v and w then so v w = v vvw cos(θ) = cos(θ) ( ) 24 θ = arccos (1 points) Find the volume of the parallelipiped (in other words, box ) whose adjacent edges are the vectors 1,2,3, 1,1,2, and 2,1,4. The volume is plus or minus the value of the triple product 1,2,3 ( 1,1,2 2,1,4 ) which is the same as the determinant det = 1(4 2) 2( 4 4)+3( 1 2) =

2 4. (1 points) Find an equation for the plane that passes through the points (,1,1), (1,,1), and (1,1,). (1,,1) (,1,1) = 1, 1, (1,1,) (,1,1) = 1,, 1 so 1, 1, 1,, 1 = 1,1,1 is a normal vector, perpendicular to the plane. Thus 1(x)+1(y 1)+1(z 1) = is an equation for the plane. (Of course there are many others.) 5. (1 points) At time t the location of a moving point is given by r(t) = cos(2t),sin(2t),t. Find its acceleration vector. d 2 r dt 2 = d 2sin(2t),2cos(2t),1 = 4cos(2t), 4sin(2t),. dt xy 6. (1 points) Find the limit, if it exists, or show that it does not exist: lim (x,y) (,) x 2 +y 2 Along the line y = the limit is Along the line y = x the limit is lim (x,) (,) lim (x,x) (,) These don t agree so the limit does not exist. x 2 =. x 2 x 2 +x 2 = (1 points) Let f(x,y) = x 4 +x 2 y and x = s+2t u and y = stu 2. Use the chain rule to find f u. Express your answer as a function of s,t,u; your calculations may involve x and y but there should be no x or y in your final answer. f u = f x x u + = f y y u = (4x 3 +2xy)( 1)+(x 2 )(2stu) = 4(s+2t u) 3 2(s+2t u)(stu 2 )+(s+2t u) 2 (2stu) Page 2

3 8. (1 points) Find the directional derivative of the function at the point (x,y) = ( 6,4) in the direction v = 1 2 f(x,y) = sin(2x+3y) ( 3i j ). f v = f v ) = ( 2cos(2x+3y),3cos(2x+3y) (x,y)=( 6,4) ( 2cos() ) 3 3cos() = , 1 2 = (1 points) In what direction does the function g(s,t) = te st have its maximum rate of change at the point (s,t) = (,2)? (Be careful a direction should be a unit vector.) g increases fastest in the direction of its gradient. At (,2) g g = s, g = t (s,t)=(,2) t 2 e st,e st +ste st (s,t)=(,2) = 4,1 Converting this to a unit vector one obtains the direction 4, (15 points) Use Lagrange multipliers to maximize the objective function: f(x,y,z) = xyz subject to the constraint: g(x,y,z) = x 2 +2y 2 +3z 2 = 6. Be sure to give the point(s) where the maximum occurs and the value of f at that point. f = yz,xz,xy, g = 2x,4y,6z so the Lagrange multiplier condition f = λ g says yz = (λ)(2x) and xz = (λ)(4y) and xy = (λ)(6z) Page 3

4 If x = or y = or z = then f(x,y,z) = xyz =. Surely the maximum is larger than that, so assume x and y and z. Then the above equations say that λ = yz 2x = xz 4y = xy 6z. Solving also yz 2x = xz 4y y 2x = x 4y 2y 2 = x 2 yz 2x = xy 6z z 2x = x 6z 3z 2 = x 2. Thus the constraint equation g(x,y,z) = x 2 +2y 2 +3z 2 = 6 becomes x 2 +x 2 +x 2 = 6 so x = ± 2, y = ±1, and z = ± 2/3. Thus the maximum of f(x,y,z) = xyz is max f = ( 2)(1)( 2/3) and it occurs at four points: ( 2,1, 2/3),( 2, 1, 2/3),( 2,1, 2/3),( 2, 1, 2/3). 11. (1 points) Evaluate the integral by reversing the order of integration 1 3 3y e x2 dx dy. We re integrating over the triangle y 1, 3y x 3, which is the same triangle as x 3, y x/3. Page 4

5 Thus 1 3 3y e x2 dx dy = = = 3 x/3 3 3 e x2 dy dx ye x2 y=x/3 y= (x/3)e x2 dx Substitute u = x 2, du = 2x dx so du/2 = x dx into the integral to obtain e u (1/2)du = 1 6 ( e 9 1 ). dx 12. (1 points) Set up an integral, in polar coordinates or cylindrical coordinates, that computes the volume of the solid that lies below the paraboloid z = 18 2x 2 2y 2 and above the x,y-plane. Just set it up, you don t need to integrate it. In polar coordinates the top of the solid is z = 18 2r 2 and the top meets the base, where z =, when = 18 2r 2 so r = 3. Thus the volume is 2π 3 (18 2r 2 ) r dr dθ. 13. (1 points) Use spherical coordinates to find the volume of a sphere of radius 1. (Go ahead and integrate this one, it s easy!) 2π π 1 = 1 3 2π π ρ 2 sin(φ) dρ dφ dθ = 1sin(φ)dφ dθ = 1 3 = 1 3 2π π 2π π 2π π ρ 3 3 ρ=1 ρ= (1)(2) dθ = 2 3 (2π) sin(φ) dφ dθ 1( cos(φ)) φ=π φ= dθ Page 5

6 14. (1 points) Let F be the vector field F = (x+y)i+(y z)j+z 2 k and let C be the curve parametrized by r(t) = t 2 i+t 3 j+t 2 k, t 1. Evaluate the line integral C F dr. Using the parametrization, x = t 2 dx = 2t dty = t 3 dy = 3t 2 dtz = t 2 dz = 2t dt so = 1 F dr = (x+y) dx+(y z) dy +z 2 dz C C (t 2 +t 3 )(2t) dt+(t 3 t 2 )(3t 2 ) dt+(t 4 )(2t) dt = 1 2t 3 t 4 +5t 5 dt = Let G = (x+y)i+(y +z)j+(x+z)k. Find (a) (5 points) div G (x+y) x + (y +z) y + (x+y) z = = 3 (b) (5 points) curl G. ( (x+z) i y ) ( (y +z) (x+z) j (x+y) ) ( (y +z) +k (x+y) ) = i j k z x z x y (c) (5 points) Is G the gradient of some potential function? Why or why not? No, because its curl is not zero. 16. (1 points) Determine whether or not the vector field F = (3x 2 +2y 2 )i+(4xy +3)j is the gradient of some potential function f. If it is, find f. Page 6

7 (3x2 +2y 2 ) y so it is a gradient of some function f. + (4xy +3) x = 4y+4y = f(x,y) = 3x 2 +2y 2 = f x 3x 2 +2y 2 dx = x 3 +2xy 2 +C(y) where C(y) is some function of y. 4xy +3 = f y = ( x 3 +2xy 2 +C(y) ) y (1) 4xy +3 = 4xy +C (y) (2) 3y +K = C(y) (3) Therefore where K is some constant. f(x,y) = x 3 +2xy 2 +3y +K 17. (1 points) Use Green s theorem to evaluate the line integral (y +e x ) dx+(2x+cos(y 2 )) dy C where C is the boundary of the region enclosed by the parabolas y = x 2 and x = y 2. (Assume C is oriented in the counterclockwise direction.) Let R be the region enclosed by the parabolas. Green s theorem says C (y +e x ) dx+(2x+cos(y 2 )) dy = (y+e x ) R y = ( 1+2) da = da R R + (2x+cos(y2 )) x da Page 7

8 is the area enclosed by the parabolas. The parabolas intersect at (,) and (1,1) and y = x on the top parabola so the area enclosed by them is R da = 1 x x 2 dx = = 1 3. Page 8