f x = 2e xy +y(2x+y)e xy = (2+2xy+y 2 )e xy.

Transcription

1 gri (rg38778) Homework 11 gri (11111) 1 This print-out should have 3 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. Find lim (x,y) (,) 1 1. points 9xy 2 x 2 +y2, if it exists. 1. The limit does not exist correct Find lim (x,y) (,) 2 1. points 2xy 4 x 2 +y8, if it exists. 3. The limit does not exist. correct Find exists points lim (x,y,z) (,,) xy +2yz 2 +6xz 2 x 2 +y 2 +z 4, if it 2. The limit does not exist. correct Find f xy when 4 1. points f(x, y) = (2x+y)e xy. 1. f xy = (1 xy)e xy 2. f xy = (2+xy)e xy 3. f xy = (2+xy)(2x+y)e xy correct 4. f xy = (1+xy)(2x+y)e xy 5. f xy = (2 xy)(2x+y)e xy 6. f xy = (1+xy)e xy Differentiation with respect to x gives f x = 2e xy +y(2x+y)e xy = (2+2xy+y 2 )e xy. So differentiating again, but now with respect to y, we get f xy = (2x+2y)e xy +x(2+2xy +y 2 )e xy. f xy = (4x+2y +2x 2 y +xy 2 )e xy = 2(2x+y)e xy +xy(2x+y)e xy. f xy = (2+xy)(2x+y)e xy points

2 gri (rg38778) Homework 11 gri (11111) 2 Determine f xx +f yy when f(x,y) = (x 2)(y +4)(x+y +5). 1. f xx +f yy = 2(x+y +6) 2. f xx +f yy = 2(x+y +2) correct 3. f xx +f yy = x+y f xx +f yy = x+y f xx +f yy = 2(x+y 6) Using the product rule to differentiate f(x,y) = (x 2)(y +4)(x+y +5) first with respect to x then with respect to y we obtain f x = (y +4)(x+y +5)+(x 2)(y +4) f y = (x 2)(x+y +5)+(x 2)(y +4). Repeating this we next obtain f xx = 2(y +4), f yy = 2(x 2). f xx +f yy = 2(x+y +2). 6 (part 1 of 4) 1. points A function f is defined by f(x,y) = (x+1)(y 6)(x+y 8). (i) Determine f x for the function f. 1. f x (x,y) = (x 6)(x+2y 14) 2. f x (x,y) = (x+1)(x 2y 7) 3. f x (x,y) = (y 6)(2x+y 7) correct 4. f x (x,y) = (y 6)(2x y +9) 5. f x (x,y) = (y +1)(2x+y +9) Using the product rule to differentiate f(x,y) = (x+1)(y 6)(x+y 8) with respect to x we obtain f x = (y 6)(x+y 8)+(x+1)(y 6). f x (x,y) = (y 6)(2x+y 7). 7 (part 2 of 4) 1. points (ii) Determine f y for the function f. 1. f y (x,y) = (y +6)(2x+y 7) 2. f y (x,y) = (x+1)(x+2y 2) 3. f y (x,y) = (x+1)(x+2y 14) correct 4. f y (x,y) = (x 6)(x 2y 14) 5. f y (x,y) = (y 6)(2x+y 14) Using the product rule to differentiate f(x,y) = (x+1)(y 6)(x+y 8) with respect to y we obtain f y (x,y) = (x+1)(x+y 8)+(x+1)(y 6). f y (x,y) = (x+1)(x+2y 14). 8 (part 3 of 4) 1. points (iii) Determine f xx +f yy for the function f.

3 gri (rg38778) Homework 11 gri (11111) 3 1. (f xx +f yy )(x,y) = x+y 5 2. (f xx +f yy )(x,y) = 2(x+y 5) correct 3. (f xx +f yy )(x,y) = 2(x+y +7) Using the product rule to differentiate f x (x,y) = (y 6)(2x+y 7) with respect to y we obtain f xy (x,y) = 2x+2y (f xx +f yy )(x,y) = x+y (f xx +f yy )(x,y) = 2(x+y 7) Differentiating f x (x,y) = (y 6)(2x+y 7) with respect to x we obtain f xx (x,y) = 2(y 6); similarly, differentiating f y (x,y) = (x+1)(x+2y 14) with respect to y we obtain f yy (x,y) = 2(x+1). (f xx +f yy )(x,y) = 2(x+y 5). 9 (part 4 of 4) 1. points (iv) Determine f xy for the function f. 1. f xy (x,y) = 2x+2y 13 correct 2. f xy (x,y) = x+2y f xy (x,y) = 2x+2y f xy (x,y) = 2x+2y f xy (x,y) = 2x+y points Use the Chain Rule to find s when s s s s s s z = x 2 +4xy +y 2, x = 4s 2t, y = st. = 4x+16y +4xs+2ys = 8x 8y +4xt+2yt = 8x+16y +4xs+2ys = 8x+16y +4xt+2yt correct = 4x 8y +4xs+2ys = 4x 8y +4xt+2yt By the Chain Rule for Partial Differentiation, s = s + s. Now while = 2x+4y, s = 4 = 4x+2y, s = t. s = 4(2x+4y)+t(+4x+2y).

4 gri (rg38778) Homework 11 gri (11111) 4 s If z = f(x,y) = 8x+16y +4xt+2yt points f x (2, 4) = 2, f y (2, 4) = 3, keywords: points Use partial differentiation the Chain Rule applied to F(x, y) = to determine /dx when F(x, y) = cos(x 3y) xe 2y =. find at t = 5 when x = g(t), y = h(t) g(5) = 2, g (5) = 5. h(5) = 4, h (5) = dx = sin(x 3y) 2xe2y 2sin(x 3y) 3e 2y dx = sin(x 3y)+e2y 3xe 2y 2sin(x 3y) = 5 = 3 = 1 correct = 7 = 9 By the Chain Rule for Partial Differentiation, = f x (g(t), h(t))g (t) + f y (g(t), h(t))h (t). When t = 5, therefore, = f x (g(5), h(5))g (5) t=5 + f y (g(5), h(5))h (5) = f x (2, 4)g (5) + f y (2, 4)h (5). given the values above, dx = dx = sin(x 3y)+e2y 2xe 2y 3sin(x 3y) sin(x 3y)+e2y correct 3sin(x 3y) 2xe2y dx = sin(x 3y)+2xe2y 3sin(x 3y) e 2y dx = sin(x 3y) 2e2y 2sin(x 3y) 3xe 2y Applying the Chain Rule to both sides of the equation F(x, y) =, we see that When F dx dx + F dx = F + F dx =. dx = F F = F x F y. F(x, y) = cos(x 3y) xe 2y =, therefore, = (2)(5) (3)(3) = 1. dx = sin(x 3y) e2y 3sin(x 3y) 2xe 2y.

5 gri (rg38778) Homework 11 gri (11111) 5 dx = sin(x 3y)+e2y 3sin(x 3y) 2xe 2y points Find the directional derivative, f v, of the function f(x, y) = 5+3x y at the point P(2, 1) in the direction of the vector v = 3, f v = 21 5 correct 2. f v = f v = f v = f v = 4 Now for an arbitrary vector v, ( ) v f v = f, v where we have normalized so that the direction vector has unit length. But when then f(x, y) = 5+3x y, f = (3 y)i+ 3 ( ) x j. 2 y At P(2, 1), therefore, f = 3i+3j. P when v = 3, 4, ( ) v f v (2, 1) = 3, 3 v = keywords: points Find an equation for the tangent plane to the graph of at the point (7,,). z = xe y cosz 7 1. x+7y z = 7 correct 2. x+7y z = 7 3. x 7y z = 7 4. x+7y +z = 7 5. x+7y +z = 7 Note that Let xe y cosz z = 7 F(x) = xe y cosz z. The equation to the tangent plane to the surface at the point P(7,,) is given by F x P (x 7)+F y P (y )+F z P (z ). Since F x = e y cosz, F x P = 1, F y = xe y cosz, F y P = 7, F z = xe y sinz 1, F z P = 1 it follows that the equation of the tangent plane is x+7y z = 7.

6 gri (rg38778) Homework 11 gri (11111) 6 keywords: points Ifr(x) isthe vectorfunction whosegraph is trace of the surface z = f(x, y) = 3x 2 2y 2 2x+3y on the plane y = 2x, determine the tangent vector to r(x) at x = tangent vector = 1, 2, 4 2. tangent vector = 1, 2, 6 correct 3. tangent vector = 2, 1, 4 4. tangent vector = 1,, 6 5. tangent vector = 2, 2, 6 6. tangent vector = 2,, 4 The graph of z = f(x, y) = 3x 2 2y 2 2x+3y is the set of all points (x, y, f(x, y)) as x, y vary in 3-space. So the intersection of the surface with the plane y = 2x is the set of all points But (x, 2x, f(x, 2x)), < x <. f(x, 2x) = 5x 2 +4x. the surface the plane y = 2x intersect in the graph of r(x) = x, 2x, 5x 2 +4x. Now the tangent vector to the graph of r(x) is the derivative r (x) = 1, 2, 1x+4. at x = 1 the graph of r(x) has tangent vector = 1, 2, 6. keywords: points Use Lagrange multipliers to find the maximum minimum values of the function f(x,y) = 5x 2 y, subject to the constraint 5x 2 +2y 2 = f max = 2, f min = 2 2. f max = 5, f min = 5 correct 3. f max = 5, f min = 4. f max = 1, f min = 1 5. f max =, f min = points Use Lagrange multipliers to find the maximum value of the function f(x,y,z) = x+2y subject to the constraints x+y +z = 9, y 2 +z 2 = correct

7 gri (rg38778) Homework 11 gri (11111) points Use Lagrange multipliers to find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes one vertex in the plane 1. volume = 3 2. volume = 2 4x+2y +z = volume = 1 correct in which case 4x = 2y = z. From the fact that P lies on the plane, therefore, it thus follows that F has a critical point at x = 1, y = 1, z = 2. 2 At this critical point V will be maximized. the box has maximum volume = 1. The solid shown in points 4. volume = volume = 1 2 The rectangular box will have one corner at the origin. if the corner diagonally opposite is P(x, y, z), then the box will have side-lengths x, y z, hence have volume V = xyz. But P lies on the plane 4x+2y +z = 6, so we have to maximize V subject to this last restriction on x, y, z. To use Lagrange multipliers, set F(x, y, z, λ) = xyz λ(4x+2y +z 6). Then F will have a critical point when yz = 4λ, zx = 2λ, xy = λ, 4x+2y +z = 6. From the first set of equations we see that xyz = 4λx, yzx = 2λy, zxy = λz, lies inside the sphere x 2 +y 2 +z 2 = 25 outside the cylinder x 2 +y 2 = 16. Find the volume of the part of this solid lying above the xy-plane. 1. volume = volume = 18π correct 3. volume = 9π 4. volume = 27π 5. volume = volume = 9

8 gri (rg38778) Homework 11 gri (11111) 8 From directly overhead the solid is similar to y r by converting to polar coordinates. 1. I = π(e 4 1) 2. I = 4π(e 16 1) 3. I = 4π(e 4 1) θ 4 x 5 4. I = π(e 16 1) correct 5. I = 2π(e 4 1) In Cartesian coordinates this is the annulus { } R = (x, y) : 16 x 2 +y the volume of the solid above the xyplane is given by the integral V = (25 x 2 y 2 ) 1/2 dx. R To evaluate V we change to polar coordinates. Now { } R = (r, θ) : 4 r 5, θ 2π, so that after changing coordinates the integral becomes V = 5 2π 4 = 2π 5 4 = π 25 r 2 rdrdθ r 25 r 2 dr [ 2 3 (25 u)3/2] 25 16, using the substitution u = r 2. volume = V = 18π points Evaluate the iterated integral 6. I = 2π(e 16 1) In Cartesian coordinates the region of integration is { (x, y) : y } 16 x 2, x 4, which is the shaded region in y θ r The graphic shows that in polar coordinates the region of integration is { } (r, θ) : r 4, θ π/2. in polar coordinates I = 4 π/2 = 2π 4 4e r2 rdθdr 4 6 re r2 dr = π setting u = r 2. x e u du, I = 4 16 x 2 4e x2 +y 2 dx I = π(e 16 1).

9 gri (rg38778) Homework 11 gri (11111) points Evaluate the iterated integral I = 2 4 y 2 4 y 2 2x 2 y 2 dx by converting to polar coordinates. 1. I = 2π 2. I = 7 3 π 3. I = 8 3 π correct 4. I = 1 3 π 5. I = 3π In Cartesian coordinates the region of integration is the set of all points (x, y) such that 4 x 2 x 4 x 2, y 2, which is the shaded region in y r Now cos 2 θsin 2 θ = 1 4 (1+cos2θ)(1 cos2θ) I = 1 4 = 1 4 (1 cos2 2θ) = 1 4 sin2 2θ = 1 8 (1 cos4θ). 2 π = 1 4 π 2 r 5 (1 cos4θ)dθdr r 5 dr = 1 4 π [ 1 6 r6 ] 2 I = 8 3 π points Use cylindrical coordinates to evaluate the integral I = ydv when W is the solid lying above the xy-plane between the cylinders W x 2 +y 2 = 2, x 2 +y 2 = 4, below the plane z = x I = 112. θ 2 x 2. I = 112π The graphic shows that in polar coordinates the region of integration is { } (r, θ) : r 2, θ π. in polar coordinates I = 2 π 2r 5 cos 2 θsin 2 θdθdr 3. I = I = 224π 5. I = correct points

10 gri (rg38778) Homework 11 gri (11111) 1 Use cylindrical coordinates to evaluate the integral I = (2y 2 +3z 2 )dv W when W is the solid shown in in cylindrical coordinates W consists of all points (r, θ, z) with r 1, θ 2π, z r. So I can be written as the repeated integral = ( 2π( r = ( 2π ( 2π ) ) (2r 2 sin 2 θ +3z 2 ) dθ rdr r 3( 2sin 2 θ+1 ) ) dθ rdr ) r 4 (2 cos2θ)dθ dr. I = 4π r 4 dr = 4 5 π. that lies above the xy-plane, below the cone z 2 = x 2 +y 2, within the cylinder x 2 +y 2 = I = points The solid W consists of all points enclosed by the cylinder the sphere shown in x 2 +y 2 = 1 x 2 +y 2 +z 2 = 4 2. I = 1π 3. I = 6 5 π 4. I = 4 5 π correct 5. I = 2π Since x = rcosθ, y = rsinθ, z = z, in cylindrical coordinates, the cylinder becomes r = 1 while the cone becomes z = r. Use cylindrical coordinates to find the volume of W.

11 gri (rg38778) Homework 11 gri (11111) volume = 2π (8+3 3/2) 2. volume = 4π (8+3 3/2) 3. volume = 2π 3 4. volume = 4π 3 5. volume = 2π 3 (8 3 3/2) ( 8 3 3/2) correct (8+3 3/2) 6. volume = 4π (8 3 3/2) As a triple integral volume(w) = W 1dV. But in rectangular coordinates, W consists of all points (x, y, z) such that x 2 +y x 2 y 2 z 4 x 2 y 2. while in cylindrical polar coordinates, W consists of all points (r, θ, z) such that r 1, θ 2π, 4 r 2 z W has volume given by I = = 4π ( 2π( 4 r 2 4 r 2 4 r 2. r 4 r 2 dr. ) ) dθ rdr To evaluate this last integral we use the substitution u 2 = 4 r 2. For then W has volume = 4π 3 (8 33/2 ) points The surface S shown in consists of the portion of the sphere where x 2 +y 2 +z 2 = 16 x 2 +y 2 4. Use spherical polar coordinates (ρ, θ, φ) to describe S. 1. S = all points P(ρ, θ, φ)} with ρ = 4, θ 2π, π 6 φ π. 2. S = all points P(ρ, θ, φ)} with ρ = 2, θ 2π, π 3 φ 2π 3. so 3 I = 4π 2 2udu = 2rdr, u 2 du = [ 4π 3 3 u3] S = all points P(ρ, θ, φ)} with ρ = 2, θ 2π, π 6 5π 6.

12 gri (rg38778) Homework 11 gri (11111) S = all points P(ρ, θ, φ)} with ρ = 4, θ 2π, π 3 φ 2π S = all points P(ρ, θ, φ) with correct ρ = 4, θ 2π, π 6 φ 5π S = all points P(ρ, θ, φ)} with ρ = 2, θ 2π, π 3 φ π. In spherical polar coordinates (ρ, θ, φ), x = ρsinφcosθ, y = ρsinφsinθ, z = ρcosφ, with θ 2π ψ π. We need to find further restrictions on ρ, θ, φ so that x 2 +y 2 +z 2 = 16, x 2 +y 2 4. Now ρ 2 = x 2 +y 2 +z 2 = 16, i.e., ρ = 4. But then, z 2 = 16cos 2 φ = 16 x 2 y S consists of all points P with ρ = 4 θ 2π, π 6 φ 5π points Use spherical coordinates to evaluate the integral I = 4zdV W. when W is bounded by the hemisphere the xy-plane. 1. I = π correct 2. I = 7 4 π 3. I = 3 2 π 4. I = 3 4 π 5. I = 5 4 π z = 1 x 2 y 2, In spherical polar coordinates (ρ, θ, φ), x = ρcosθsinφ, y = ρsinθsinφ, z = ρcosφ, while the Jacobian of the mapping is (ρ, θ, φ) (x, y, z) (x, y, z) (ρ, θ, φ) = ρ2 sinφ. On the other h, in spherical polar coordinates, W consists of all points (ρ, θ, φ) such that But ρ 1, θ 2π, φ π 2. I = π/2 2π π/2 cosφsinφdφ = 1 2 4ρ 3 cosφsinφdφdθdρ. [ cos 2 φ ] π/2 = 1 2.

13 gri (rg38778) Homework 11 gri (11111) 13 I = 2 2π ρ 3 dθdρ = π. I = 2π π/2 2ρ 4 cos 2 φsinφdφdθdρ points Use spherical coordinates to evaluate the integral I = 2z 2 dv when W is bounded by the hemisphere the xy-plane. W z = 1 x 2 y 2, 1. I = 4 15 π correct 2. I = 2 5 π 3. I = 1 3 π 4. I = 2 15 π 5. I = 1 5 π In spherical polar coordinates (ρ, θ, φ), x = ρcosθsinφ, y = ρsinθsinφ, z = ρcosφ, while the Jacobian of the mapping is (ρ, θ, φ) (x, y, z) (x, y, z) (ρ, θ, φ) = ρ2 sinφ. On the other h, in spherical polar coordinates, W consists of all points (ρ, θ, φ) such that ρ 1, θ 2π, φ π 2. But π/2 cos 2 φsinφdφ = 1 3 I = 2 3 2π [ cos 3 φ ] π/2 = 1 3. ρ 4 dθdρ = 4 15 π points Find the Jacobian of the transformation when T : (u, v) (x(u, v), y(u, v)) x = e 4u 2v, y = e 3u+v. (u, v) = 1e7u+3v (u, v) = 1e7u v correct (u, v) = 2e7u+3v (u, v) = 2e7u v (u, v) = 2e7u+3v (u, v) = 2e7u v For general functions the Jacobian of the transformation T : (u, v) (x(u, v), y(u, v))

14 gri (rg38778) Homework 11 gri (11111) 14 is defined by But when then while (u, v) = u u v. v x = e 4u 2v, y = e 3u+v, u = 4e4u 2v, In this case, u = 3e3u+v, (u, v) = e7u v v = 2e4u 2v, v = e3u+v, (u, v) = 1e7u v. 2. I = I = I = I = I = 27 When we see that while x = 3u+2v, x+y = u+5v, (u, v) = y = 3v 2u = 13. by the change of variables formula for double integrals, if T maps the rectangle D = [a, b] [c, d] keywords: points Use the transformation T : (u, v) (x(u, v), y(u, v)) with in the uv-plane onto D as shown in y v d T D D c u a b x x = 3u+2v, y = 3v 2u to evaluate the integral I = (x+y)dx when D is the square with vertices (, ), (2, 3), (5, 1) (3, 2). 1. I = 39 correct D then D = (x+y)dx D = 13 (u+5v) (u, v) dudv d c ( b a ) (u+5v)du dv.

15 gri (rg38778) Homework 11 gri (11111) 15 Now we compute a, b, c, d using the fact that T maps vertices of D to vertices of D. Since T : (u, v) (x, y) = (3u+2v, 3v 2u), it follows that (i) T(u, v) = (3u+2v, 3v 2u) = (, ) when (u, v) = (, ), (ii) T(u, v) = (3u+2v, 3v 2u) = (2, 3) when (u, v) = (, 1), (iii) T(u, v) = (3u+2v, 3v 2u) = (5, 1) when (u, v) = (1, 1), (iv) T(u, v) = (3u+2v, 3v 2u) = (3, 2) when (u, v) = (1, ). D is the square having corners (, ), (, 1), (1, 1), (1, ), in which case But I = 13 ( ) (u+5v)du dv. [ 1 1 (u+5v)du = +5uv] 2 u2 = v. I = 13 ( v ) dv = points Find the Jacobian of the transformation when 1. T : (r, θ) (x(r, θ), y(r, θ)) x = 2e r cosθ, (r, θ) = 6cos2θ y = 3e r sinθ (r, θ) = 5e2r cos2θ (r, θ) = 5e2r (r, θ) (r, θ) = 6 = 6cos2θ correct (r, θ) = 5e2r cos2θ For general functions the Jacobian of the transformation is defined by But when T : (u, v) (x(u, v), y(u, v)) (u, v) = x = 2e r cosθ, u u then by the Product Rule, v. v y = 3e r sinθ. r = 2er cosθ, θ = 2er sinθ, while r = 3e r sinθ, θ = 3e r cosθ. In this case, (r, θ) = 2e r cosθ 2e r sinθ 3e r sinθ 3e r cosθ because = 6e r e r (cos 2 θ sin 2 θ). (r, θ) = 6cos2θ e r e r = 1, cos2θ = cos 2 θ sin 2 θ. keywords: