Solutions to old Exam 3 problems Hi students! I am putting this version of my review for the Final exam review here on the web site, place and time to be announced. Enjoy!! Best, Bill Meeks PS. There are probably errors in some of the solutions presented here and for a few problems you need to complete them or simplify the answers; some questions are left to you the student. Also you might need to add more detailed explanations or justifications on the actual similar problems on your exam. I will keep updating these solutions with better corrected/improved versions. The first 6 slides are from Exam practice problems but the material also falls on our Final exam.
Problem 3 - Exam Fall 6 Evaluate the integral y x 3 + 1 dx dy by reversing the order of integration. Note that the region R defined by {(x, y) y x 1, y 1} is equal to the region {(x, y) x 1, y x }. Thus, x 3 + 1 da = x 3 + 1 dx dy = y x = 1 3 R [ x x 3 + 1 dy dx = x 3 + 1 y] dx = x 3 + 1 x dx (x 3 + 1) 1 3x dx = 1 1 3 ( 3 (x 3 + 1) 3 ) = 9 ( 3 1).
Problem 3(3) - From Exam Find the iterated integral, x x (x y)dy dx. = x x (x y) dy dx = [x y y ] x x ( x) ( x) (x 3 x ) dx = x 3 +x +x 1 dx = 1 x 4 + 3 x 3 + x 1 x = 1 + 3 + 1 = 5 6. x dx
Problem 3(4) - From Exam Find the iterated integral, x x 3 sin(y 3 )dy dx. (Hint: Reverse the order of integration) Reverse the order of integration: x 3 sin(y 3 ) dy dx = x [ ] x 4 y = 4 sin(y 3 ) dy = y x 3 sin(y 3 ) dx dy y 4 sin(y 3 ) dy. Let u = y 3 and then du = 3y dy. Making this substitution, y 4 sin(y 3 ) dy = 1 sin (y 3 )3y dy = 1 1 1 cos(y 3 ). Hence, y 4 sin(y 3 ) dy = 1 1 1 cos(y 3 ) = 1 (1 cos(1)). 1
Problem 33() - From Exam Evaluate the following double integral. e y da, R = {(x, y) y 1, x y}. R R e y da = = = 1 ey 1 y e y y dy = 1 e y dx dy = e y (y) dy = 1 (e e ) = 1 (e 1). [ ] y e y x dy
Problem 33(3) - From Exam Evaluate the following double integral. x y x da, R = {(x, y) y 1, x y}. R = 1 R y x y x da = y y x x dx dy (y x ) 1 1 1 ( x) dx dy = 3 (y x ) 3 = 1 y 3 dy = 1 3 3 1 4 y 4 1 = 1 1. y dy
Problem 34() - From Exam Find the volume V of the solid under the surface z = x + y and above the region bounded by curves x y = and x y 3 =. = V = y y 3 (x + y ) dx dy = y 4 + y 4 (y 6 + y y 3 ) dy = [ ] y x + y x y 3 dy y 6 y 5 + y 4 dy = y 7 7 y 6 6 + 1 5 y 5 = 1 7 1 6 + 5.
Problem 1(a) - Fall 8 Consider the points A = (1,, ), B = (, 1, ) and C = (1,, 3). Find the parametric equations for the line L passing through the points A and C. A vector parallel to the line L is: v = AC = 1 1,, 3, =,, 3. A point on the line is A = (1,, ). Therefore parametric equations for the line L are: x = 1 y = t z = 3t.
Problem 1(b) - Fall 8 Consider the points A = (1,, ), B = (, 1, ) and C = (1,, 3). Find an equation of the plane in R 3 which contains the points A, B, C. Since a plane is determined by its normal vector n and a point on it, say the point A, it suffices to find n. Note that: n = AB AC = So the equation of the plane is: i j k 1 1 3 = 3, 3,. 3, 3, x 1, y, z = 3(x 1) 3y + z =.
Problem 1(c) - Fall 8 Consider the points A = (1,, ), B = (, 1, ) and C = (1,, 3). Find the area of the triangle with vertices A, B and C. Consider the points A = (1,, ), B = (, 1, ) and C = (1,, 3). Then the area of the triangle with these vertices can be found by taking the area of the parallelogram spanned by AB and AC and dividing by. Thus: Area( ) = AB AC = 1 = 1 3, 3, = 1 i j k 1 1 3 1 9 + 9 + 4 =.
Problem - Fall 8 Find the volume under the graph of f (x, y) = x + xy and over the bounded region in the first quadrant {(x, y) x, y } bounded by the curve y = 1 x and the x and y-axes. = x (x + xy) dy dx = x(1 x ) + x(1 x ) dx = = ( x 6 6 3x 4 4 + x ) 1 (xy + xy ) 1 x dx x 5 3x 3 + x dx = 1 6 3 4 + 1 = 5 1.
Problem 3 - Fall 8 Let I = 1 Sketch the region of integration. x sin(y ) dy dx. Write the integral I with the order of integration reversed. 3 Evaluate the integral I. Show your work. 1 See the blackboard for a sketch. 3 By Fubini s Theorem, y I = sin(y ) dx dy = = 1 4 y sin(y ) dx dy sin(y )x 1 y sin(y ) y dy = cos(y ) 4 dy = 1 (sin(y ) 1 y) dy = 1 (cos 1 1). 4
Problem 4(a, b, c) - Fall 8 Consider the function F(x, y, z) = x + xy + z. 1 What is the gradient F(x, y, z) of F at the point (1,, 1)? Calculate the directional derivative of F at the point (1,, 1) in the direction 1, 1, 1? 3 What is the maximal rate of change of F at the point (1,, 1)?. f = x + y, xy, 1. So, f (1,, 1) = 6, 4, 1. The unit vector u in the direction 1, 1, 1 is u = 1,1,1 3. D u f (1,, 1) = f (1,, 1) u = 6, 4, 1 1 3 1, 1, 1 = 11 3. The maximum rate of change is the length of the gradient: MRC(f ) = f (1,, 1) = 6, 4, 1 = 53.
Problem 4(d) - Fall 8 Consider the function F(x, y, z) = x + xy + z. Find the equation of the tangent plane to the level surface F(x, y, z) = 4 at the point (1,, 1). Recall that the gradient of F(x, y, z) = x + xy + z is normal n to the surface. Calculating, we obtain: F(x, y, z) = x, xy, 1 n = F(1,, 1) = 6, 4, 1. The equation of the tangent plane is: 6, 4, 1 x 1, y, z +1 = 6(x 1)+4(y )+(z +1) =.
Problem 5 - Fall 8 Find the volume V of the solid under the surface z = 1 x y and above the xy-plane. The domain of integration for the function z = 1 x y described in polar coordinates is: D = {(r, θ) r 1, θ π}. In polar coordinates r = x + y and so z = 1 r. Applying Fubini s Theorem, V = π π = (1 r )r dr dθ = [ 1 r 1 ] 1 4 r 4 = 1 4 π dθ = π π dθ = π. (r r 3 ) dr dθ ( 1 1 4 ) dθ
Problem 6(a) - Fall 8 Determine whether the following vector fields are conservative or not. Find a potential function for those which are indeed conservative. 1 F(x, y) = x + e x + xy, xy sin(y). G(x, y) = 3x y + e x + y, x 3 + xy + 3y. On this slide we only consider the function F(x, y). Note that F(x, y) = P(x, y)i + Q(x, y)j, where P(x, y) = x + e x + xy and Q(x, y) = xy sin(y). Since P y (x, y) = x y = Q x (x, y), the vector field F(x, y) is not conservative.
Problem 6(b) - Fall 8 Determine whether the following vector fields are conservative or not. Find a potential function for those which are indeed conservative. 1 F(x, y) = x + e x + xy, xy sin(y). G(x, y) = 3x y + e x + y, x 3 + xy + 3y. On this slide we only consider the function G(x, y). Since y (3x y + e x + y ) = 3x + y = x (x 3 + xy + 3y ), there exists a potential function f(x, y), where f = G. Note that: f y = x 3 + xy + 3y = f(x, y) = x 3 y + xy + y 3 + g(x), where g(x) is some function of x. Since f x = 3x y + e x + y, f x = (x 3 y + xy + y 3 + g(x)) = 3x y + y + g (x) = 3x y + e x + y x = g (x) = e x = g(x) = e x + constant. Hence, f(x, y) = x 3 y + xy + y 3 + e x is a potential function.
Problem 7 - Fall 8 Evaluate the line integral yz dx + xz dy + xy dz, C where C is the curve starting at (,, ), traveling along a line segment to (1,, ) and then traveling along a second line segment to (1,, 1). The parameterizations C 1 (t), C (t) are: C 1 (t) = t,, t 1 C (t) = 1, t, t t 1 So, C 1 (t) = 1,, and C (t) =,, 1. Thus, C 1 yz dx + xy dy + xy dz = [( 1) + (t ) + (t )] dt =. Also, C yz dx + xz dy + xy dz = [(t t ) + (1 t ) + (1 t 1)] dt = 4t 4t dt = =. So, the entire integral equals + =. = 4 1
Problem 8(a) - Fall 8 Use Green s Theorem to show that if D R is the bounded region with boundary a positively oriented simple closed curve C, then the area of D can be calculated by the formula: Area(D) = 1 y dx + x dy Recall Green s Theorem: P dx + Q dy = Hence, C C D ( Q x P y ) da. 1 y dx + x dy = ( x C D x y y ) da = 1 (1 + 1) da = da = Area(D) D D
Problem 8(b) - Fall 8 Consider the ellipse 4x + y = 1. Use the above area formula to calculate the area of the region D R with boundary this ellipse. (Hint: This ellipse can be parametrized by r(t) = 1 cos(t), sin(t) for t π.) The ellipse has parametric equations x = 1 cos t and y = sin t, where t π. Using the formula in the previous theorem, we have: Area(D) = 1 x dy y dx = 1 = 1 C π π ( 1 cos t)(cos t) dt (sin t)( 1 sin t) dt 1 (cos t + sin t) dt = π 4 = π.
Problem 9(a) - Spring 8 For the space curve r(t) = t 1, t, t/, (a) find the velocity, speed, and acceleration of a particle whose position function is r(t) at time t = 4. The velocity v(t) is equal to r (t): v(t) = r (t) = t, t, 1 v(4) = 8, 8, 1. The acceleration a(t) is equal to v (t): a(t) = v (t) =,, a(4) =,,. The speed s(4) is equal to v(4) : 64 + 64 + 14 = s(4) = v(4) = 18 + 1 4.
Problem 9(b) - Spring 8 For the space curve r(t) = t 1, t, t/, (b) find all points where the particle with position vector r(t) intersects the plane x + y z =. Plug the x, y and z-coordinates of r(t) into the equation of the plane and solve for t: (t 1) + t ( t ) = t t 1 = (t + 1)(t 1) = = t = 1 or t = 1. Next evaluate the points on r(t) at these times to obtain the points of intersection: r(1) = 1 1, 1, 1 =, 1, 1 r( 1 ) = ( 1 ) 1, ( 1 ), 1 ( 1 ) = 3 4, 1 4, 1 4.
Problem 1 - Spring 8 Let D be the region of the xy-plane above the graph of y = x and below the line y = x. (a) Determine an iterated integral expression for the double integral xy da D (b) Find an equivalent iterated integral to the one found in part (a) with the reversed order of integration. (c) Evaluate one of the two iterated integrals in parts (a), (b). Part (a) Part (b) Part (c) = D x xy da = y y x x xy dx dy. xy dy dx. [ ] x 1 xy xy dy dx = x ( 1 x 3 1 ) x 5 dx = 1 8 x 4 1 1 x 6 x 1 dx = 1 8 1 1 = 1 4
Problem 11 - Spring 8 Find the absolute maximum and absolute minimum values of f (x, y) = x + y y in the set D = {(x, y) : x + y 4}. Find critical points of f (x, y): f = x, 4y = = x = and y = 1. Use Lagrange multipliers to study max and min values of f on the circle g(x, y) = x + y = 4: f x, 4y = λ g = λ x, y. We get x = λx = λ = 1 or x =. If λ = 1, then 4y = y = y = 1. Plugging in g(x, y), gives (, 1 ), (, ±) and (± 3, 1). We get f (, 1 ) = 1, f (, ) = 4, f (, ) = 1, f (± 3, 1) = 3. The maximum value of f (x, y) is 1 and its minimum value is 1.
Problem 1(a) - Spring 8 Let D be the region in the first quadrant x, y that lies between the two circles x + y = 4 and x + y = 9. (a) Describe the region D using polar coordinates. (b) Evaluate the double integral 3x + 3y da. D The domain is: D = {(r, θ) r 3, θ π }. Calculate the integral using the substitution: = π = D π 3 x = r cos θ y = r sin θ da = r dr dθ; (3x + 3y)dA = π 3 3r (cos θ + sin θ) dr dθ = (3r cos θ + 3r sin θ)r dr dθ π (7 8)(cos θ + sin θ) dθ = 19(sin θ cos θ) [ ] 3 r 3 (cos θ + sin θ) dθ π = 19 + 19 = 38.
Problem 13(a) - Spring 8 (a) Find z z and at the point (1,, 1) for the implicit x y function z determined by the equation x 3 + y 3 + z 3 3xyz =. Consider the function F(x, y, z) = x 3 + y 3 + z 3 3yxz. Since F(x, y, z) is constant on the surface, the Chain Rule gives: F x x x + F y y x + F z z x = F x + F z z x = F x y y + F y y y + F z z y = F y + F z z y =. Thus, z x = F x F z z y = F y F z = 3x + 3yz 3z 3xy = 3y + 3xz 3z 3xy = x + yz z xy, = y + xz z xy.
Problem 13(b) - Spring 8 (b) Is the tangent plane to the surface x 3 + y 3 + z 3 3xyz = at the point (1,, 1) perpendicular to the plane x + y 3z =? Justify your answer with an appropriate calculation. Since F(x, y, z) = x 3 + y 3 + z 3 3xyz is constant along the surface F(x, y, z) =, F is normal (orthogonal) to the surface. Calculating, we obtain: F = 3x 3yz, 3y 3xz, 3z 3xy F(1,, 1) = 3, 3, 3, which is the normal vector to the tangent plane of the surface. Since the normal to the plane x + y 3z = is, 1, 3 and 3, 3, 3, 1, 3 =, it is perpendicular.
Problem 14(a) - Spring 8 (a) Consider the vector field G(x, y) = 4x 3 + xy, x. Show that G is conservative (i.e. G is a potential or a gradient vector field), and use the fundamental theorem for line integrals to determine the value of G dr, C where C is the contour consisting of the line starting at (, ) and ending at ( 1, 1). Since y (4x 3 + xy) = x = x (x ), there exists a potential function F(x, y), where F = G. Note that: F y = x = F(x, y) = x y + g(x), where g(x) is some function of x. Since F x = 4x 3 + xy, F x = (x y + g(x)) = xy + g (x) = 4x 3 + xy x = g (x) = 4x 3 = g(x) = x 4 + constant. Hence, F(x, y) = x 4 + x y is a potential function. By the fundamental theorem of calculus for line integrals, G dr = F dr = F( 1, 1) F(, ) = 8 = 6. C C
Problem 14(b) - Spring 8 (b) Now let T denote the closed contour consisting of the triangle with vertices at (,),(1,), and (1,1) with the counterclockwise orientation, and let F(x, y) = 1 y y, xy. Compute F dr directly (from the T definition of line integral). The curve T is the union of the segment C 1 from (, ) to (1, ), the segment C from (1, ) to (1, 1) and the segment C 3 from (1, 1) to (, ). Parameterize these segments as follows: C 1 (t) = t, t 1 C (t) = 1, t t 1 C (t) = 1 t, 1 t t 1 F dr = T F dr + C 1 F dr + C F dr C 3 =, 1, dt+ 1 1 t t, t, 1 dt+ (1 t) (1 t), (1 t) 1, 1 dt = + t dt + 3 t + t 1 dt = 3 t + 3t 1 dt = 1 t3 + 3 t 1 1 t = 1 + 3 1 = 1.
Problem 14(c) - Spring 8 Let F(x, y) = 1 y y, xy. (c) Explain how Green s theorem can be used to show that the integral T F dr in (b) must be equal to the area of the region D interior to T. By Green s Theorem, F dr = = D T (xy) x T ( 1 y y) dx + xy dy ( 1 y y) da = y = da. D D (y y + 1) da Since 1 = D da is the area of the triangle D, then the integral 1 in part (b) is equal to this area.
Problem 15(a,b,c) - Fall 7 Let I = 4 (a) Sketch the region of integration y e x3 dx dy. (b) Write the integral I with the order of integration reversed. (c) Evaluate the integral I. Show your work. (a) (b) Make your sketch and from the sketch, we see: 4 x I = e x3 dx dy = e x3 dy dx. y (c) We next evaluate the integral using part (b). x [ ] x I = e x3 dy dx = e x3 y = 1 3 dx = e x3 x dx e x3 (3x ) dx = 1 3 ex3 = 1 3 (e8 e ) = 1 3 (e8 1).
Problem 16 - Fall 7 Find the distance from the point (3,, 7) to the line L x = 1 + t, y = t, z = 1 + 3t. Note that the plane T passing through P = (3,, 7) with normal vector the direction n = 1, 1, 3 of the line L must intersect L in the point Q closest to P. We now find Q. The equation of the plane T is: (x 3) (y ) + 3(z + 7) =. Substitute in this equation the parametric values of L and solve for t: = (1 + t) 3 [( t) ] + 3[(1 + 3t) + 7] = 11t +. Hence, t = and Q = 1, 4, 5. The distance from P and Q is d = 4 + + = 4, and thus the distance from P to L.
Problem 17(a) - Fall 7 Find the velocity and acceleration of a particle moving along the curve r(t) = t, t, t 3 at the point (, 4, 8). Recall that the velocity v(t) = r (t) and the acceleration a(t) = r (t): v(t) = r (t) = 1, t, 3t a(t) = v (t) =,, 6t. As the point (, 4, 8) corresponds to t = on r(t), v() = 1, 4, 1 a() =,, 1.
Problem 17(b) - Fall 7 Find all points where the curve in part (a) intersects the surface z = 3x 3 + xy x. Plug the x, y and z-coordinates into the equation of the surface and solve for t. t 3 = 3t 3 + t 3 t = 3t 3 t = t(3t 1) = = t = or t = ± 1. 3 Next plug these t values into r(t) to get the 3 points of intersection: r() =,,, r( 1 3 ) = 1 3, 1 3, 1 3 3 r( 1 3 ) = 1 3, 1 3, 1 3 3.
Problem 18 - Fall 7 Find the volume V of the solid which lies below the sphere x + y + z = 4, above the xy-plane, and inside the cylinder x + y = 3. We first describe in polar coordinates the domain D R for the integral. D = {(r, θ) r 3, θ π}. The graphing function is z = 4 x y = 4 r. This gives the volume: = 1 V = π 3 π 3 = 1 3 π 4 r da = π 3 4 r r dr dθ (4 r ) 1 1 π [ ( r) dr dθ = 3 (4 r ) 3 π 7 (1 8) dθ = 3 dθ = 7 π 3 θ = 14π 3. ] 3 dθ
Problem 19(a) - Fall 7 Consider the line integral 1 + x dx + xy dy, C where C is the triangular path starting from (, ), to (, ), to (, 3), and back to (, ). (a) Evaluate this line integral directly, without using Green s Theorem. The curve C is the union of the segment C 1 from (, ) to (, ), the segment C from (, ) to (, 3) and the segment C 3 from (, 3) to (, ). Parameterize these segments: C 1 (t) = t, t 1 C (t) =, 3t t 1 C 3 (t) = t, 3 3t t 1. Thus, C 1 + x dx + xy dy = 1 + x dx + xy dy + 1 + x dx + xy dy + 1 + x dx + xy dy C 1 C C 3 1 = 1 + t() dt+ 4 3t(3) dt+ 1 + ( t)( )+( t)(3 3t)( 3)dt ( = 1 + t + 36t + ) 3 t 36t + 7t 36 dt. This long straightforward integral is left to you the student to do.
Problem 19(b) - Fall 7 Consider the line integral 1 + x dx + xy dy, C where C is the triangular path starting from (, ), to (, ), to (, 3), and back to (, ). Evaluate this line integral using Green s theorem. Let D denote the -dimensional triangle bounded by C. Green s Theorem gives: ( (xy) 1 + x dx + xy dy = ( ) 1 + x) da x y C = D y da = = 3 x D y dy dx = 9 4 x dx = 3 4 x 3 = 6. [ y ] 3 x dx
Problem (a) - Fall 7 Consider the vector field F = (y /x )i (y/x)j. Find a function f such that f = F. Suppose f exists. Then: f x = y y = f (x, y) = x where g(y) is a function of y. Then: Hence, dx = y x x + g(y), f y = y ( y + g(y)) = y x x + g (y) = y x. g (y) = = g(y) is constant. Taking the constant g(y) to be zero, we obtain: f (x, y) = y x.
Problem (b) - Fall 7 Consider the vector field F = (y /x )i (y/x)j. Let C be the curve given by r(t) = t 3, sin t for π t π. Evaluate the line integral C F dr. We will apply the potential function f (x, y) = y x in part (a) and the fundamental theorem of calculus for line integrals. We get: C F dr = f (r(π)) f (r( π )) = f (π3, ) f ( π3 8, 1) = 8 π 3
Problem 1 - Fall 6 Find parametric equations for the line L in which the planes x y + z = 1 and x + y + z = 1 intersect. The direction vector v of the line L is parallel to both planes. Hence, v = n 1 n where n 1 = 1,, 1 and n =, 1, 1 are the normal vectors of the respective planes: i j k v = n 1 n = 1 1 = 3, 1, 5. 1 1 Next find the intersection point of L with the xy-plane by setting z = : x y = 1 x + y = 1 = ( 3 5, 1 5, ) is the intersection point. Parametric equations are: x = 3 5 3t y = 1 5 + t z = 5t.
Problem - Fall 6 Consider the surface x + y z = and the point P(1, 1, 1) which lies on the surface. (i) Find the equation of the tangent plane to the surface at P. (ii) Find the equation of the normal line to the surface at P. Recall that the gradient of F(x, y, z) = x + y z is normal (orthogonal) to the surface. Calculating, we obtain: F(x, y, z) = x, y, 4z F(1, 1, 1) =,, 4. The equation of the tangent plane is:,, 4 x 1, y 1, z 1 = (x 1)+(y 1) 4(z 1) = The vector equation of the normal line is: L(t) = 1, 1, 1, + t,, 4 = 1 + t, 1 + t, 1 4t.
Problem 3 - Fall 6 Find the maximum and minimum values of the function on the disc x + y 4. f (x, y) = x + y x We first find the critical points. f = x, y = = x = 1 and y =. Next use Lagrange multipliers to study max and min of f on the boundary circle g(x, y) = x + y = 4: f = x, y = λ g = λ x, y. y = λy = y = or λ = 1. y = = x = ±. λ = 1 = x = x, which is impossible. Now check the values of f at 3 points: f (1, ) = 1, f (, ) =, f (, ) = 8. The maximum value is 8 and the minimum value is 1.
Problem 4 - Fall 6 Evaluate the iterated integral 1 x x + y dy dx. The domain of integration for the function described in polar coordinates is: D = {(r, θ) r 1, θ π }. Since da = dy dx in polar coordinates is: r dr dθ and r = x + y, 1 x π [ r 3 3 x + y dy dx = ] 1 dθ = 1 3 π π dθ = θ 3 π r dr dθ = π 6.
Problem 5(b) - Fall 6 Find the volume V of the solid under the surface z = 4 x y and above the region in the xy-plane between the circles x + y = 1 and x + y = 4. The domain of integration for the function z = 4 x y described in polar coordinates is: D = {(r, θ) 1 r, θ π}. In polar coordinates r = x + y and so z = 4 r. Applying Fubini s Theorem, V = = π π 1 (4 r )r dr dθ = [ r 1 4 r 4 ] 1 dθ = = 9 4 π π π dθ = 9π. 1 (4r r 3 ) dr dθ (8 4) ( 1 4 )dθ
Problem 6(a) - Fall 6 Determine whether the following vector fields are conservative or not. Find a potential function for those which are indeed conservative. (a) F(x, y) = (x + xy)i + (xy y )j. Note that F(x, y) = P(x, y)i + Q(x, y)j, where P(x, y) = x + xy and Q(x, y) = xy y. Since P y (x, y) = x y = Q x (x, y), the vector field F(x, y) is not conservative.
Problem 6(b) - Fall 6 Determine whether the following vector fields are conservative or not. Find a potential function for those which are indeed conservative. (b) F(x, y) = (3x y + y )i + (x 3 + xy + 3y )j. Note that F(x, y) = P(x, y)i + Q(x, y)j, where P(x, y) = 3x y + y and Q(x, y) = x 3 + xy + 3y. Since P y (x, y) = 3x + y = Q x (x, y) and P(x, y) and Q(x, y) are infinitely differentiable on the entire plane, F(x, y) has a potential function f (x, y). Thus, f x = 3x y + y = f (x, y) = Since f y = Q, then 3x y + y dx = x 3 y + y x + g(y). f y = y (x 3 y + y x + g(y)) = x 3 + xy + g (y) = x 3 + xy + 3y. Hence, g (y) = 3y = g(y) = y 3 + C, for some constant C. Taking C =, gives: f (x, y) = x 3 y + y x + y 3.
Problem 7 - Fall 6 Evaluate the line integral C (x + y)dx + (xy + 1)dy, where C is the curve starting at (, ), traveling along a line segment to (1, ) and then traveling along a second line segment to (, 3). Let C 1 be the segment from (, ) to (1, ) and let C be the segment from (1, ) to (, 3). Parameterizations for these segments are: C 1 (t) = t, t t 1 C (t) = 1 t, + t t 1. Now calculate: (x + y) dx + (xy + 1) dy = C (x + y) dx + (xy + 1) dy C 1 C = (t +t) dt+(t +1) dt+ [(1 t) ++t]( 1)+(1 t)(+t)+1 dt = 3t + t dt = t 3 + t + t 1 = 1 + 1 + = 4.
Problem 8 - Fall 6 Use Green s Theorem to evaluate the line integral F dr where C F = y 3 + sin x, xy + cos y and C is the unit circle x + y = 1 which is oriented counterclockwise. First rewrite F dr in standard form: C F dr = (y 3 + sin x) dx + (xy + cos y) dy. C Recall and apply Green s Theorem: P dx + Q dy = C where D is the disk with boundary C: (y 3 +sin x) dx+(xy +cos y) dy = C D C D Q x P y da, D y 3y da = y da. D Next evaluate the integral using polar coordinates: π π [ y da = (r sin θ) r 4 r dr dθ = 4 sin θ = 1 4 π sin θ dθ. ] 1 dθ
Problem 9 (a) Express the double integral R x y x da as an iterated integral and evaluate it, where R is the first quadrant region enclosed by the curves y =, y = x and y = x. First rewrite the integral as an iterated integral. = R (x y x) da = = [ x 3 y 3 [ ( y) 3 y 3 y x y ] y y ] ( y) (x y x) dx dy dy [ y 7 3 y The remaining straightforward integral is left to you the student to do. ] dy
Problem 9 (b) Find an equivalent iterated integral expression for the double integral in (a), where the order of integration is reversed from the order used in part (a). (Do not evaluate this integral. ) = x R (x y x) dy dx + (x y x) da x 1 (x y x) dy dx.
Problem 3 Calculate the line integral C F dr, where F(x, y) = y xi + xyj, and C is the path starting at (1, ), moving along a line segment to (3, ) and then moving along a second line segment to (, 1). Let C 1 be the segment from (1, ) to (3, ) and let C be the segment from (3, ) to (, 1). Parameterizations for these curves are: C 1 (t) = 1 + t, t t 1 C (t) = 3 3t, t t 1 Next calculate: y x dx + xy dy = C y x dx + xy dy + C 1 y x dx + xy dy = C ( t) (1+t) dt+(1+t)( t)( ) dt+ t (3 3t)( 3) dt+(3 3t)t dt. I leave the remaining long but straightforward calculation to you the student.
Problem 31 Evaluate the integral R y x + y da with R the region {(x, y): 1 x + y, y x.} First describe the domain R in polar coordinates: R = {1 r, θ π 4 }. Rewrite the integral in polar coordinates: y π x + y 4 da = R Now evaluating the integral: π 4 = 3 4 1 π 4 sin θ r 3 dr dθ = 1 π 4 sin θ dθ = 3 4 ( cos θ) π 4 (r sin θ r)r dr dθ. [ ] 1 4 sin θ r 4 = 3 4 1 dθ ( 1 1 ).
Problem 3 (a) Show that the vector field F(x, y) = 1 y + x, x + 1 is y conservative by finding a potential function f (x, y). Suppose f (x, y) is the desired potential function. Then: f x (x, y) = 1 y + x = f (x, y) = x y + x + g(y), where g(y) is a function of y. Then: where K is a constant. f y (x, y) = x y + g (y) = x y + 1 = g (y) = 1 = g(y) = y + K, Taking K =, we obtain: f (x, y) = x y + x + y.
Problem 3 (b) Let C be the path described by the parametric curve r(t) = 1 + t, 1 + t for (a) to determine the value of the line integral C F dr. By the Fundamental Theorem of Calculus for line integrals, F dr = f (x, y ) f (x 1, y 1 ), C where (x 1, y 1 ) is the beginning point for r(t) and (x, y ) is the ending point for r(t). Note that the person who made this problem forgot to give the beginning and ending times, so this is all we can do.
Problem 33 (a) Find the equation of the tangent plane at the point P = (1, 1 1) in the level surface f (x, y, z) = 3x + xyz + z = 1. Recall that the gradient field f (x, y, z) is orthogonal to the level set surfaces of f (x, y). Hence n = f (1, 1, 1) is a normal vector for the tangent plane. Calculating: f = 6x + yz, xz, xy + z f (1, 1, 1) = 6 1, 1, 1 = 5, 1, 1. The equation of the tangent plane is: 5, 1, 1 x 1, y 1, z+1 = 5(x 1) (y 1) (z+1) =.
Problem 33 (b) Find the directional derivative of the function f (x, y, z) at P = (1, 1, 1) in the direction of the tangent vector to the space curve r(t) = t t, t, t t 3 at t = 1. First find the tangent vector v to r(t) at t = 1: r (t) = 4t 1, t 3, t 6t v = r (1) = 3,, 4. The unit vector u in the direction of v is: u = v 3,, 4 =. v 9 By part (a), f at (1, 1, 1) is: f (1, 1, 1) = 5, 1, 1. The directional derivative is: D u f (1, 1, 1) = f (1, 1, 1) 1 9 3,, 4 = 1 9 ( 5, 1, 1 3,, 4 = 1 9.
Problem 34 Find the absolute maxima and minima of the function f (x, y) = x xy + y y. in the region bounded by the lines x =, y = and x + y = 7. This problem is left to you the student to do.
Problem 35 Consider the function f (x, y) = xe xy. Let P be the point (1, ). (a) Find the rate of change of the function f at the point P in the direction of the point Q = (3, ). The unit vector u in the direction of the point (3, ) is: u = Calculate f (1, ): PQ PQ = The directional derivative is:, = 1 1, 1. 4 + 4 f = e xy + xye xy, x e xy f (1, ) = 1, 1. D u f (, 1) = f (1, ) u = 1, 1 1 1, 1 =.
Problem 35 Consider the function f (x, y) = xe xy. Let P be the point (1, ). (b) Give a direction in terms of a unit vector (there are two possibilities) for which the rate of change of f at P in that direction is zero. Let v = a, b be a vector with a + b = 1 and suppose D v f (1, ) =. Calculating, we get: D v f (1, ) = f (, 1) a, b = 1, 1 a, b = a + b. Setting D v f (1, ) = means a + b = = a = b. Then a + b = ( b) + b = b = 1 = b = ± 1. Hence, the two possibilities for v are: 1, 1 and 1, 1.
Problem 36 (a) Find the work done by the vector field F(x, y) = x y, x over the circle r(t) = cos t, sin t, t π. The work W done is: W = F dr = (x y) dx + x dy. C Note r(t) = cos t, sin t parameterizes C and r (t) = sin t, cos t. So, the work done is: W = = π π (cos t sin t)( sin t) dt + C π sin(t) cos(t) dt + = sin (t) π π cos t cos t dt sin (t) + cos (t) dt + π = π.
Problem 36 (b) Use Green s Theorem to calculate the line integral C ( y ) dx + xy dy, over the positively (counterclockwise) oriented closed curve C defined by x = 1, y = 1 and the coordinate axes. Recall that Green s Theorem is: ( Q P dx + Q dy = C D x P y where D is the square bounded by C. C Applying Green s Theorem, we obtain: ( y ) dx+(xy) dy = = 3 [ y ] 1 (y+y) dy dx = 3 ) da, 1 dx = 3 dx = 3. y dy dx
Problem 37 (a) Show that the vector field F(x, y) = x y, 1 3 x 3 is conservative and find a function f such that F = f. Since y (x y) = x = x (1 3 x 3 ), the vector field is conservative. Suppose f (x, y) is a potential function for F(x, y). Then: f x (x, y) = x y = f (x, y) = x y dx = x 3 y 3 + g(y), where g(y) is a function of y. Then: f y (x, y) = 3 y (x y 3 + g(y)) = 1 3 x 3 + g (y) = 1 3 x 3 = g (y) = = g(y) = C, for some constant C. Setting C =, we get: f (x, y) = x 3 y 3.
Problem 37 (b) Using the result in part (a), calculate the line integral C F dr, along the curve C which is the arc of y = x 4 from (, ) to (, 16). By the Fundamental Theorem of Calculus, F dr = f (, 16) f (, ) = 8 16 = 18. 3 C
Problem 38 Consider the surface x + y 1 4 z = and the point P(1,, 5) which lies on the surface. (a) Find the equation of the tangent plane to the surface at the point P. (b) Find the equation of the normal line to the surface at the point P. Let F(x, y, z) = x + y 1 4 z and note that F is normal to the level set surface x + y 1 4 z =. Calculating F at (1,, 5), we obtain: F = x, y, 1 z F(1,, 5) =, 4, 5. The equation of the tangent plane is:, 4, 5 x 1, y, z+ 5 = (x 1)+4(y )+ 5(z+ 5) =. The vector equation of the normal line is: L(t) = 1,, 5 +t, 4, 5 = 1+t, +4t, 5+ 5t.
Problem 39 A flat circular plate has the shape of the region x + y 1. The plate (including the boundary x + y = 1) is heated so that the temperature at any point (x, y) on the plate is given by T(x, y) = x + y x. Find the temperatures at the hottest and the coldest points on the plate, including the boundary x + y = 1. First find the critical points: T = x 1, 4y =, = x = 1 and y =. Apply Lagrange multipliers with the constraint function g(x, y) = x + y = 1: T = x 1, 4y = λ g = λ x, y, = 4y = λy = λ = or y =. y = = x = ±1. λ = = x 1 = (x) = 4x = x = 1 and y = ± 3. Checking values: f ( 1, ) = 1 4, f (1, ) =, f ( 1, ) =, f ( 1, ± 3 ) = 9 4. The maximum value is 9 4 and the minimum value is 1 4.
Problem 4 The acceleration of a particle at any time t is given by a(t) = 3 cos t, 3 sin t,, while its initial velocity is v() =, 3,. At what times, if any are the velocity and the acceleration of the particle orthogonal? First find the velocity v(t) by integrating a(t) and using the initial value v() =, 3, : v(t) = a(t) dt = 3 cos t, 3 sin t, dt = 3 sin t + x, 3 cos t + y, t + z. Since v() =, 3,, we get 3 sin() + x, 3 cos() + y, + z = x, 3 + y, z =, 3,, = x = y = z =. Hence, v(t) = 3 sin t, 3 cos t, t. Take dot products and solve for t: a(t) v(t) = 9 cos t sin t 9 sin t cos t + 4t = 4t = = t =.
Problem 41(a) - Spring 9 Consider the two lines r 1 (t) = 1, 1, + t 1, 1, and r (s) =,, + s 1, 1,. Find the point at which they intersect. Solution. To find the point of intersection, set r 1 (t) = r (s). That is, 1 + t = s; 1 t = s; t =. Solve these equations to obtain t = 1, s =. This gives the point of intersection (,, ).
Problem 41(b) - Spring 9 Consider the two lines r 1 (t) = 1, 1, + t 1, 1, and r (s) =,, + s 1, 1,. Find a normal vector of the plane which contains these two lines. Solution. Note that the vectors parallel to the two lines are v 1 = 1, 1, and v = 1, 1,, respectively. So a normal vector of the plane which contains these two lines is i j k n = v 1 v = 1 1 =,, 1 1
Problem 41(c) - Spring 9 Consider the two lines r 1 (t) = 1, 1, + t 1, 1, and r (s) =,, + s 1, 1,. Find the cosine of the angle between the lines r 1 (t) and r (s). Solution. The vectors parallel to the two lines are v 1 = 1, 1, and v = 1, 1,, respectively. Denote by θ the angle between v 1 and v. So cos θ = v 1 v 1, 1, 1, 1, = v 1 v 1, 1, 1, 1, = 1 3 Since the angle between two lines is either acute or right, the 1 cosine of the angle between the lines r 1 (t) and r (s) is 3.
Problem 4(a) - Spring 9 Consider the surface S defined by the equation xy + xz + z = 7. Find the equation of the tangent plane to the surface S at the point (1, 1, ). Solution. Let f (x, y, z) = xy + xz + z. Compute the gradient: f (x, y, z) = y + z, xy, x + z. So a normal vector of the tangent plane to the surface S at the point (1, 1, ) is: n = f (1, 1, ) = 3,, 5. Thus an equation of the tangent plane to the surface S at the point (1, 1, ) is: 3(x 1) + (y 1) + 5(z ) =.
Problem 4(b) - Spring 9 Find the directional derivative of f (x, y, z) = x + xy + z at the point (1,, 3) in the direction v = 1,,. Solution. Compute the gradient of f at (1,, 3): f (x, y, z) = x + y, xy, 1 f (1,, 3) = 6, 4, 1 Normalize the direction: u = 1,, 1,, = 1 3 1,,. Evaluate: D u f (1,, 3) = f (1,, 3) u = 6, 4, 1 1 3 1,, = 16 3.
Problem 43(a) - Spring 9 Find the volume V of the solid under the graph of z = 9 x y, inside the cylinder x + y = 1, and above the xy-plane. Solution. The volume V = D 9 x y da where D = {(x, y) x + y 1}. To compute this double integral, we use polar coordinates: π 9 x y da = (9 r ) r dr dθ = D π = π (9r r 3 ) dr dθ = 17 17 dθ = 4 So the volume V = 17π. 4 θ π π ( 9r r 4 4 ) r=1 dθ r= = 17 4 (π ) = 17 π.
Problem 43(b) - Spring 9 Evaluate the iterated integral x ey dy dx. Solution. x The iterated integral integral D ey da, where ey dy dx is equal to the double D = {(x, y) x 1, x y 1}. D can be rewritten as a type II region So, = D = {(x, y) y 1, x y}. x xe y x=y x= e y dy dx = dy = y e y dx dy ye y dy = 1 1 ey = e 1.
Problem 44(a) - Spring 9 Determine whether or not the vector field F(x, y) = 3e y + xy, xe y + x + y is conservative. If so, find a potential function. A vector field F = P, Q is conservative on a simply connected open domain if and only if P y = Q x everywhere on that domain and the component functions of the vector field are defined and have continuous partial derivatives. Since R is a simply connected open set, it is enough to check that P, Q satisfy P y = Q x. For the vector field F = 3e y + xy, xe y + x + y, we have P y = 6ey + x e y + x = Q x So we conclude F is NOT conservative.
Problem 44(b) - Spring 9 Determine whether or not the vector field G(x, y) = 3x + y, 4xy + 3 is conservative. If so, find a potential function. A vector field G = P, Q is conservative on a simply connected open domain if and only if P y = Q x everywhere on that domain and the component functions of the vector field are defined and have continuous partial derivatives. Since R is simply connected open set, it is enough to check that P, Q satisfy P y = Q x. For the vector field G = (3x + y )i + (4xy + 3)j, we have P y = 4y = Q x So we conclude G is conservative.
Continuation of Solution to 44(b): A potential function for G is a function g such that G = g. To compute g we solve, G = 3x + y, 4xy + 3 = g x, g y = g. Anti-differentiate the component equations with respect to the appropriate variable: g = 3x + y dx = x 3 + xy + h 1 (y) = 4xy + 3 dy = xy + 3y + h (x) So we have h 1 (y) = 3y and h (x) = x 3. Thus g = x 3 + xy + 3y. Be sure to check your answer! Compute g = 3x + y, 4xy + 3 = G
Problem 45(a) - Spring 9 Find parametric equations for the line segment C from the point A = ( 1, 5, ) to the point B = (1, 6, 4). To get parametric equations for L you need a point through which the line passes and a vector parallel to the line. For example, take the point to be A and the vector to be AB. The vector equation of C is C(t) = OA + t AB = 1, 5, + t, 1, 4 = 1 + t, 5 + t, 4t, where O is the origin. So the parametric equations are: x = 1 + t y = 5 + t, t [, 1]. z = 4t
Problem 45(b) - Spring 9 Evaluate the line integral C xz dy + y dz, where C is the line segment in part (a). Solution. By part (a), the parametric equations of C are: x = 1 + t y = 5 + t, t [, 1]. z = 4t So, xz dy + y dz = ( 1 + t)(4t) dt + (5 + t) 4dt C = 4 4t + 8t 3 + 5 + t dt = ( 4 4 3 t3 + t 4 + 5t + 1 ) ( 1 t = 4 4 3 + + 5 + 1 ) = 74 3.
Problem 46 - Spring 9 Let D be the region on the plane bounded by the curves y = x x and the x-axis, and C be the positively oriented boundary curve of D. Use Green s Theorem to evaluate the line integral C (xy + cos(ex )) dx + (x + e cos y ) dy. Green s Theorem states C P dx + Q dy = D ( Q x P y ) da. In this problem, P = xy + cos(e x ) and Q = x + e cos y. So, C (xy + cos(ex )) dx + (x + e cos y ) dy = D ( (x +e cos y ) x (xy+cos(ex )) y ) da = D (x x)da = x x x dydx = xy x x dx = x(x x ) dx = x x 3 dx = 3 x 3 1 4 x 4 = 3 (8) 1 4 (16) = 16 3 16 4 = 4 3.
Solutions to the remaining problems are similar to worked out solutions here and in the other reviews and are left to you the students to do on your own. 47. Find parametric equations for the line in which the planes 3x 6y z = 15 and x + y z = 5 intersect. 48. Find the equation of the plane containing the points P(1, 3, ), Q(, 1, ) and R(,, 1). 49. Find all points of intersection of the parametric curve r(t) = t, t, 1 t t and the plane x + y + z = 3. 5. Find the absolute maximum and minimum of the function f (x, y) = x + y y on the closed disc x + y 5 of radius 5. 51. Evaluate R xy da, where R is the region in the first quadrant bounded by the line y = x and the parabola y = x. 5. Consider the vector field F(x, y) = xy + sin y, x + x cos y + 1. (a) Show that F(x, y) = xy + sin y, x + x cos y + 1 is conservative by finding a potential function f (x, y) for F(x, y). (b) Use your answer to part a to evaluate the line integral C F dr, where C is the arc of the parabola y = x going from (, ) to (, 4). 53. Evaluate the line integral C F dr where F = y + sin x, xy and C is the unit circle oriented counterclockwise. 54. Evaluate the line integral C F dr, where F = y, xy + x and C is the curve starting at (, ), traveling along a line segment to (, 1) and then traveling along a second line segment to (, 3).