Answer sheet: Final exam for Math 2339, Dec 10, 2010

Transcription

1 Answer sheet: Final exam for Math 9, ec, Problem. Let the surface be z f(x,y) ln(y + cos(πxy) + e ). (a) Find the gradient vector of f f(x,y) y + cos(πxy) + e πy sin(πxy), y πx sin(πxy) (b) Evaluate f(, ). Find the tangent plane to this surface at the point (,,f(, ) ). f(, ) ln( + cos(π) + e ) ln(e ). The tangent plane of z f(x,y) at (x,y,z ) is z f(x,y ) f x (x,y )(x x ) + f y (x,y )(y y ) Plugging in (x,y ) (, ) into f(x,y), we get f x (, ), f y (, ) e. Therefore the tangent plane is z (x ) + e (y ), or y e z e. (c) Find the symmetric equation of the normal line at the point (,, f(, ) )? The normal line is perpendicular to the tangent plane, therefore the direction of the normal line is the same as the normal direction of the tangent plane. From (b) we see the normal direction of the tangent plane at (,, ) is parallel to, e,, Therefore the symmetric equation of the normal line at (,, ) is e (y ) z x. (Each equation represents a plane, and their intersection gives the normal line.)

2 Problem. Let f(x,y) ln(x + e y ) + x y cos(t ) dt. (a) (b) Evaluate f x Evaluate f y f x x (ln(x + e y ) + f y y (ln(x + e y ) x y y x cos(t ) dt) cos(t ) dt) x x + e y + cos(x ) ey x + e y cos(y ) (c) Find f xy and f yx. Are they equal? f xy f yx f x y ( ) x y x + e + y cos(x ) xey (x + e y ) f y x ( ) e y x x + e y cos(y ) xey (x + e y ) Yes they are equal, which should be the case according to the Clairaut s Theorem.

3 Problem. Evaluate the following integrals: } () cos(x +y 4) da, (x,y) 4 x + y 4 + π/, y expressed in polar coordinates is (r,θ) r } 4 + π/, θ π. cos(x + y 4) da 4+π/ cos(r 4) rdrdθ 4 4+π/ dθ 4 cos(r 4)dr 4+π/ π sin(r 4) 4 π(sin(π/) sin()) π () 9 x cos(y ) dydx x The integration region is } (x,y) x, x y 9 (x,y) y 9, x } y. Therefore 9 9 y 9 9 x x cos(y ) dydx x cos(y ) dxdy cos(y )x x x y cos(y )y dy sin(y ) sin(8) 4 dy 9 cos(y ) dy

4 Problem 4. Evaluate the following integrals using appropriate coordinate systems y (a) x + y da, where is the semi-circular region enclosed by (x ) + y 9 in the top half plane. } Use polar coordinates, (r, θ) θ π/, r 6 cos(θ) y π/ x + y da / 6cos(θ) / 6cos(θ) r sin(θ) dr dθ 8 cos (θ) d cos(θ) / / r sin(θ) r r r drdθ r6 cos(θ) r sin(θ)dθ 8 d cos (θ) 8 cos (θ) π/ 6 (b) Evaluate R x da, where R (x,y) } x 4 + y 9, y. Apply change of variables as x r cos(θ), y r sin(θ), this transforms the semi-ellipse into a semicircle } (r,θ) r, θ π. The Jacobian of this transformation is x x J r θ y y 6r(cos (θ) + sin (θ)) 6r. r θ Therefore x da 4r cos (θ) J drdθ R π 6r4 r π 4r cos (θ) drdθ + cos(θ) dθ r 4r dr

5 Problem 5. Let the density function T be T(x,y,z) ln(x + y 4 z ). () Find the gradient vector T(x,y,z) and evaluate T(,, ). T(x,y,z) x + y 4 z x, 4y, z T(,, ), 4,,, () Find the directional derivative v T(x,y,z) where v is a unit vector parallel to,,. Then evaluate v T(,, ). First, normalize,, to get v,,. v T(x,y,z) T(x,y,z) v x + y 4 z x, 4y, z,, ( x + z) x + y 4 z ( x + z) (x + y 4 z ) v T(,, ) ( + ) ( + ) () At the point (,, ), what is the unit direction u that will minimize u T(,, )? What is this minimum? (As a simple verification, check if this value with v T(,, ) from above. It should be smaller.) enote θ the angle between T(x,y,z) and u. Note that u T(x,y,z) T(x,y,z) u T(x,y,z) u cos(θ) T(x,y,z) cos(θ) The minimum u T(x,y,z) is obtained when cos(θ). That is, u is in the opposite direction of T(x,y,z). Hence the unit vector u for fastest decrease at (,, ) should be u T(,, ) T(,, ) The minimum value of u T(,, ) is,, 6. T(,, ) 6.

6 Problem 6. The mapping between spherical coordinates (ρ, θ, φ) and rectangular coordinates (x,y,z) is x ρ sin(φ) cos(θ), y ρ sin(φ) sin(θ), z ρ cos(φ); where ρ, θ π, φ π. (a) Find the equation in the rectangular coordinates for the surface θ π. What is the shape of this surface? When θ π, we see x ρ sin(φ) cos( π ) ρ sin(φ) y ρ sin(φ) sin( π ) ρ sin(φ). Hence y x. Notice that φ π, we see x. Therefore θ π is half of the plane y x, with x. (Another solution: Note that y x ρsin(φ)sin(θ) ρ sin(φ) cos(θ) tan(θ) tan(π ), therefore y x. Again, it is just half of the plane with y.) (b) Find the equation in the rectangular coordinates for the surface φ π. What is the shape of this surface? When φ π, we see x ρ sin( π ) cos(θ) ρ cos(θ) y ρ sin( π ) sin(θ) ρ sin(θ) z ρ cos( π ) ρ Hence z x + y. Notice also that z. Therefore φ π is the cone z x + y. (It is the top half cone of z x + y.) (Another solution: Multiplying cos(φ) cos( π) on both sides by ρ we get ρ cos(φ) ρ. Noticing that ρ cos(φ) z and ρ x + y + z we get z x + y + z. Simplifying this we get z x + y, with z.) (c) Find the equation in the rectangular coordinates for the surface ρ cos(φ) sin(φ) cos(θ). What is the shape and position of this surface? Multiplying ρ on both sides of ρ cos(φ) sin(φ) cos(θ) leads to ρ ρ cos(φ) ρ sin(φ) cos(θ) x + y + z z x (x + ) + y + (z ). The surface is a sphere with radius centered at (,, ).

7 Problem 7. Find the volume of the solid bounded by z 4+(x +y ) and z x y. The intersection of the two circular paraboloids is z 4 + (x + y ) x y. Therefore x + y 4, which means the projection } of the solid onto the x-y plane is (r,θ) r, θ [, π]. The volume can be conveniently calculated using cylindrical coordinates V π x y 4+(x +y ) r dzda 4+r π π( 6) π dzda (6 4r ) r drdθ 6r 4r dr

8 Problem 8. Evaluate E x + y ex +y +z dv, where E is the region bounded by x + y + z 4, y, and z. We change variables into spherical coordinates as x ρ sin φ cos θ, y ρ sin φ sin θ, z ρ cos φ. Note in spherical coordinates, y means θ [,π], z means φ [,π/]. Therefore E (ρ,θ,φ) ρ, θ [,π], φ [, π } ]. x + y ex +y +z dv E / π π dθ eρ / eρ ρ ρ π 4 (e4 e) ρ sin(φ) eρ ρ sin(φ) dρ dφ dθ dφ dρ e ρ ρ dρ

9 Problem 9. Evaluate the line integrals (a) xe x dy + y dz + 5z dx, C } where C (x,y,z) x t,y et,z t, t. xe x dy + y dz + 5z dx C te t de t + e t dt + 5t 4 dt te t e t dt + te t dt + 5t 4 dt (t + te t + 5t 4 )dt (the nd term is by integration-by-parts) (t + te t et + t 5 ) + e (e e ) + (5 + e ) (b) F d r, where F (x,y,z) ( C z ) i + e y j + y k, z and C is part of a helix curve given by the vector function π r (t) sin(t) i + cos(t) j + t k, t π. Note that on the helix, x sin(t), y cos(t), z t. π F d r F r (t)dt C π π π π t, ecos(t), cos(t) t cos(t) cos(t), sin(t), dt t + e cos(t) ( sin(t)) + cos(t) dt t e cos(t) ( sin(t)) dt π e cos(t) π e e e π e cos(t) d cos(t)

10 Problem. Evaluate z x + y dv, where E is the solid enclosed by a conic region E z x + y and an ellipsoid region (x + y ) + z 4. Note that on E, x + y z 4 (x + y ). We integrate w.r.t. z first. First we project E onto the x-y plane. To determine the largest radius of the projection, we need the intersection of the two surfaces: z 4 x + y (x + y ), So (x + y ) x + y. Hence the projection of E on the x-y plane is a disk with radius : (x,y) x + y }. We can then switch to polar coordinates to evaluate the double integral. Essentially we are using cylindrical coordinates on E (x,y,z) (x,y), x + y z } (x + y ). E z x + y dv π (x +y ) x +y x + y π(8 8 ) 6 π z dz da x + y z (x +y ) z da z x +y r (8 8r ) r drdθ 8 8r dr

11 Problem. (Bonus points) If the length of the diagonal of a rectangular box must be fixed as L, find the dimensions of the box such that its volume is the largest. What is this maximum volume? Let the dimensions of the rectangular box be x, y, z. The diagonal of the box is then x + y + z L. The problem can be modelled as min V (x,y,z) xyz x,y,z> s.t. g(x,y,z) x + y + z L By Lagrange multiplier, We get V λ g x + y + z L yz λx xz λy xy λz Multiply respectively by x,y,z on both sides of the above three equations, we get xyz λx λy λz. Since x, y, z >, we have λ >. Therefore, x y z. Since x + y + z L, we have the optimal dimensions x y z L. By simple geometrical argument, we see these are the dimensions corresponding to the largest volume V max L.

12 Problem. (Bonus points) The least squares method has many significant applications in a vast array of areas, including data fitting and statistical regression analysis. The legendary Gauss is credited for developing the method in 795 (at age 8). He applied the method to determine the orbit of the then newly discovered asteroid Ceres, using the mere amount of data obtained by only 4 days of observation. This allowed him (then 4 years old) to correctly predict the position of the asteroid, a feat no one else could do at that time. The prediction was said to have caused great sensation in the early nineteenth century. Here we look at the simplest least squares method: To determine a straight line y mx+b that best approximates n observed data in the x-y-plane. enote the coordinates of the n observed data as (x,y ), (x,y ),, (x n,y n ), the vertical deviation of the point (x i,y i ) from the line y mx + b is d i y i (mx i + b). The method of least squares for this simple case is to find the m and b such that is minimized. That is, the best fit line y mx + b minimizes the sum of the squares of the vertical deviations from the line. Show that the m and b of the best fit line can be obtained from solving the following linear equations ( ) m x i + bn m ( i x i i ) ( ) + b x i i i y i x i y i. (Hint: This problem is not as hard as it sounds, you can try applying the first order necessary condition to a suitable function with variables.) The objective function to minimize is the following function that depends only on m and b, (m,b) d i (y i (mx i + b)). i i Applying the first order necessary condition to (m, b) (using chain rule for each square term), it follows that i b (m,b) (y i (mx i + b)) i m (m,b) x i (y i (mx i + b)) Simplifying the above we get the desired equations to solve for m and b : ( ) (y i (mx i + b)) y i m x i + bn i x i (y i (mx i + b)) i which proves the result. i i i ( x i y i m i i x i ) ( ) + b x i, i i d i