Practice Final Solutions

Transcription

1 Practice Final Solutions Math 1, Fall 17 Problem 1. Find a parameterization for the given curve, including bounds on the parameter t. Part a) The ellipse in R whose major axis has endpoints, ) and 6, ) and whose minor axis has endpoints, 1) and, 5). The ellipse is centered at, ) and has a radius of in the x-direction and in the y-direction, so the parameterization is x, y) = cos t, sin t), t π. Part b) The line segment in R with endpoints 1,,, ) and,,, 1). The parameterization is x 1, x, x, x ) = 1 t)1,,, ) t,,, 1), t 1 which simplifies to x, y) = 1 t, t, t, t), t 1. Part c) The circle of radius 1 in R that is centered on the point,, ) and lies in the plane y = x. The vectors v = 1 1, 1, ) and k,, 1) are orthogonal and lie in the plane y = x, so the circle is This simplifies to x, y, z) = cos t)v sin t)k, t π. x, y, z) = cos t, cos t ), sin t, t π. 1

2 Problem. Evaluate xy ds, where C is the portion of the ellipse x 9y = 6 that lies in the first C quadrant of R. Note that the ellipse goes through the points ±, ) and, ±). Thus we can parameterize C as x, y) = cos t, sin t), t π. Then dx ) ds = dt ) dy dt = sin t) dt cos t) dt = 9 sin t cos t dt so xy ds = C = 6 π/ π/ cos t) sin t) 9 sin t cos t dt cos t sin t 9 sin t cos t dt Substituting u = 9 sin t cos t gives du = 1 cos t sin t dt, so the integral becomes 5 9 [ ] 9 u du = 5 u/ = 8 5.

3 Problem. A circle C of radius 1 is rolling around the outside of the circle x y = 1, as shown in the figures below. A point P on C has initial coordinates 1, ), and moves with C as it rolls. t = t = π/ t = π/ Part a) Find the coordinates of P at t = π/ and t = π/. At t = π/, the circles are tangent at the point It follows that P is at, 1 )., ), so the center of C is at, ). At t = π/, the circles are tangent at the point, 1), so the center of C is at, ). Then P is at 1, ). Partb) Find a parameterization for the curve traced out by P. At time t, the circles are tangent at cos t, sin t), so the center of C is at cos t, sin t). Let vt) be the vector from the center of C to the point P. We know that v) = 1, ), vπ/) =, 1), and vπ/) = 1, ), so it s not hard to guess that vt) = cos t, sin t). Then the parameterization is x, y) = cos t cos t, sin t sin t).

4 Problem. Use geometric reasoning to evaluate the given integral. Parta) x da, where R is the region in R defined by the inequalities x ) y and y. R The inequality x ) y defined a disk of radius centered at, ), and R is the upper half of this disk. Then the area of R is π, and the average value of x on R is, so the integral is 6π. Part b) x y z ) da, where S is the sphere of radius in R centered at the origin. S Note that x y z = 9 on this sphere. The sphere has an area of πr = 6π, so the integral is π. Part c) z da, where T is the torus r ) z = 1 in R. T This torus is symmetric about the plane z =, so the average value of z on T is equal to zero. Thus the integral is. Part d) x 1 x ) ds, where L is line segment in R with endpoints,,, ) and,,, ). L Note that x 1 x = 6 on this line segment. The line segment has length =, so the integral is 1. Problem 5. Evaluate R and z x y. π π/ π/ [ sin φ = π z x y z ) dv, where R is the region in R defined by x y z ρ sin φ ρ cos φ ) dρ dφ dθ = ρ 6 [ 1 ] 1 = π) ρ ] π/ π/ π ) 1 8 dθ π/ π/ ) = π 16 dρ sin φ cos φ dφ ρ

5 Problem 6. Find a parameterization for the given surface, including bounds on the parameters u and v. Part a) The surface r = 1 for 1 z 1. 1 z We use u = θ and v = z, which means that r = 1. Then the parameterization is 1 v x, y, z) = ) cos u 1 v, sin u 1 v, v, π, 1 v 1. Part b) The ellipsoid ) x ) y ) z = 1. 5 Note that this ellipsoid goes through the points ±,, ),, ±, ),,, ±5). We take the parameterization of the unit sphere and stretch it by factors of in the x-direction, in the y-direction, and 5 in the z-direction: x, y, z) = cos u cos v, sin u cos v, 5 sin v). Part c) The paraboloid x = y z for x 1. Since the plane x = 1 cuts the paraboloid along a circle, we should use something like cylindrical coordinates. x, y, z) = u, u cos v, u sin v), u 1, v π. Part d) The torus r 5) z 1) =. We can think of the given equation as a circle in the rz-plane, which can be parameterized by r = 5 cos u, z = 1 sin u, u π Finally we let θ = v, which gives us x, y, z) = 5 cos u) cos v, 5 cos u) sin v, 1 sin u ) 5

6 Part e) The square in R with vertices 1,, 1, ),, 1,, 1), 1,, 1, ), and, 1,, 1). Note that 1,, 1, ) is opposite from 1,, 1, ). The two vectors along the sides of the square are, 1,, 1) 1,, 1, ) = 1, 1, 1, 1) and, 1,, 1) 1,, 1, ) = 1, 1, 1, 1). So the parameterization is x 1, x, x, x ) = 1,, 1, ) u 1, 1, 1, 1) v1, 1, 1, 1), u 1, v 1 which simplifies to x 1, x, x, x ) = 1 u v, 1 u v, 1 u v, 1 u v), u 1, v 1 Problem 7. Let S be the surface in R defined by x 1, x, x, x ) = u, v, u v, u v) for u 1 and v 1. Evaluate x x da. The Jacobian matrix is We have Then da = 7 du dv = du dv, so S x x da = = [ u v uv S = 1) ) ) 1) 1) ) = 7 vu v) du dv = ] 1 dv = u= v v ) uv v ) du dv dv = [ ] v 1 v = 11 6

7 Problem 8. A line segment in R has initial endpoints,, ) and,, 1). The line segment moves units in the y-direction, while simultaneously rotating 18 around the y-axis, as shown in the following figure. Part a) Find a parameterization for the surface traced out by the segment, including bounds on the two parameters. Let s have the motion occur between t = and t = π. Then the line segment has rotated by an angle t at time t, and has moved in the y-direction by t/π units. Then the endpoints of the line segment are, t ) π, so the parameterization is x, y, z) = 1 u) which simplifies to Part b) x, y, z) = and sin t, t ) π, cos t, t ) π, u sin t, t ) π, cos t, t π, u 1 u sin t, t ) π, u cos t, t π, u 1. Write a double integral for the surface area of this surface. You do not need to evaluate the integral. The Jacobian matrix is u cos t sin t /π u sin t cos t The two columns are orthogonal, with the first having magnitude u /π) and the second being a unit vector. So the surface area is 1 π u 16 dt du. π 7

8 Problem 9. Let M be the -manifold in R defined by x 1 x = x x and 1 x 1 x. Part a) Find a parameterization for M, including bounds on the three parameters u, v, w. Let u = x 1 x = x x. Then 1 u. Moreover, x 1, x ) and x, x ) are points in R that lie on the circle of radius u centered at the origin, so the parameterization is x 1, x, x, x ) = u cos v, u sin v, u cos w, u sin w), 1 u, v π, w π. Part b) Compute x1 x ) dv. M The Jacobian matrix is cos v u sin v sin v u cos v cos w u sin w sin w u cos w The three columns are orthogonal with magnitudes, u, and u respectively, so dv = u du dv dw. Then the integral is π π 1 u dw dv du = 1 u du π dv π dw = 1 π 5 8

9 Problem 1. Let S be the surface in R defined by x y =, y, and z. Find the average value of yz on S. Using cylindrical coordinates, we can parameterize this surface by The Jacobian matrix is x, y, z) = cos u, sin u, v), u π, v. sin u cos u. 1 The two columns are orthogonal with magnitudes and 1, so da = du dv. Then π π ) 9 yz da = sin u)v)) dv du = sin u du v dv = ) = 6. S It is clear geometrically that the surface area of S is 6π, so the average value of yz is 6 6π = 6 π. Problem 11. Let R be the region in R defined by r e z and z 1. Use the coordinates u, v, w defined by x = ue w cos v, y = ue w sin v, z = w to evaluate x y ) dv R Note that the given region corresponds to the ranges u 1, v π, and w 1. The Jacobian matrix is e w cos v ue w sin v ue w cos v e sin v ue w cos v ue w sin v 1 Taking the determinant of this matrix yields ue w, so dv = ue w du dv dw. Then the integral is 1 π 1 u e w) ue w) 1 π 1 dw dv du = u du dv e w dw = πe 1) 8 9