Math 261 Solutions to Sample Final Exam Problems

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1 Math 61 Solutions to Sample Final Eam Problems 1 Math 61 Solutions to Sample Final Eam Problems 1. Let F i + ( + ) j, and let G ( + ) i + ( ) j, where C 1 is the curve consisting of the circle of radius, centered at the origin and oriented counterclockwise, and where C is the curve consisting of the line segment from (, ) to (1, 1) followed b the line segment from (1, 1) to (, 1). (a) Calculate the line integral of F over C 1. (b) Calculate the line integral of G over C 1. (c) Calculate the line integral of F over C. (d) Calculate the line integral of G over C. Before we begin, let us check to determine whether F or G are conservative vector fields. Since ( ) 4 ( + ), we see that F is a conservative vector field. To find a potential function f for F, we integrate as follows: f(, ) d + P() f(, ) ( + )d + + Q() Therefore, f(, ) + is a potential function for F. On the other hand, since ( + ) 1 is not equal to ( ) 1, we see that G is not a conservative vector field. (a) Since C 1 is a closed curve and F is a conservative vector field, we know that C 1 F d r without doing an calculations. So our final answer is. (b) Since G is not a conservative vector field, we must do this integral b parameterizing C 1. We can represent C 1 b the parametric curve r(t) cost i + sint j, where t π. Therefore, we have π G d r G( cost, sin t) ( sint i + cost j)dt C 1 so our final answer is 8π. π π π ( sin t + cost) i + ( sin t cost) j) ( sint i + cost j)dt [ 4(sin t + cos t) 4 sintcost + 4 sintcost] dt ( 4) dt, (c) Since F is conservative, we can use the potential function f(, ) + that we calculated above, and the Fundamental Theorem of Calculus for Line Integrals as follows: C F d r f(, 1) f(, ) 1 1 Therefore, our final answer is 1. (d) Since G is not conservative, we must parameterize the two line segments comprising C and calculate this line integral using brute force. The line segment from (, ) to (1, 1) can be parameterized as r 1 (t) t i + t j, where t 1, and the line segment from (1, 1) to (, 1) can be parameterized as r (t) (1 + t) i + j, where t 1. Therefore, we have 1 1 G d r G( r 1 (t)) r 1(t)dt + G( r (t)) r (t)dt C (t i + j) ( i + j)dt + (t 1 + ( + t))dt (6t + 4)dt 7, 1 (( + t) i + t j) idt

2 Math 61 Solutions to Sample Final Eam Problems so our final answer is 7.. Set up but do not evaluate an iterated integral that gives the volume of the solid region that lies below the sphere + + z and above the paraboloid z +. We begin b finding the projection of the curve of intersection of the two surfaces onto the -plane. Specificall, we substitute z + into the equation of the sphere to obtain: z + z z + z (z + )(z 1) z 1. Therefore, the intersection of these two surfaces takes place when z 1; substituting z 1 into the equation of the paraboloid ields the circle + 1. It follows that the projection of this solid onto the -plane is given b the region D {(r, θ) : r 1, θ π} (see diagram to the right), so we have z Volume π 1 r r r dz dr dθ (in clindrical coordinates) dz d d (in rectangular coordinates). Let F i+z sin(z) j+ sin(z) k. Calculate C F d r, where C is the path from A (,, 1) to B (, 1, ) shown in the figure to the right. B z A Calculating, we observe that curlf, so the vector field F is conservative and we can use the Fundamental Theorem of Calculus for Line Integrals, provided that we can find a potential function, which we do below: f(,, z) d + p(, z) f(,, z) z sin(z)d cos(z) + q(, z) f(,, z) sin(z)dz cos(z) + r(, ) Therefore, f(,, z) cos(z) is a potential function for F, so b the Fundamental Theorem of Calculus for Line Integrals, we have F d r f(, 1, ) f(,, 1) (9 cos) ( cos) 1 cos. C 4. Let H i j + z k. Calculate S H d A, where S is the surface of the cube with corners at (,, ), (1,, ), (, 1, ), (1, 1, ), (,, 1), (1,, 1), (, 1, 1), and (1, 1, 1), oriented outward. Note that S is the surface of the solid region given b Therefore, b the Divergence Theorem, we have H da div H dv S E {(,, z) : 1, 1, z 1}. E ( + z)d dz d ( + z)dz d (z z + z ) z1 z d ( )d

3 Math 61 Solutions to Sample Final Eam Problems. Let W be the solid region in the first octant that lies below the plane z 6 and inside the clinder + 4 (see figure to the right). Let F z i + ( + 1) j +(e z ) k. (a) Calculate z F da where S1 is the left surface of the solid region W (that is, the portion of W that lies in the plane ), oriented in the positive direction. (b) Calculate S F d A, where S is the curved portion of W that lies on the clinder, oriented awa from the origin. (a) On the left surface of the region, note that n j, and since on the left surface of the region, we have F z i + ( + 1) j + e z k z i + j. Therefore, F da F n da ((z i + j) j)da da Surface Area of 1. (b) For the clindrical surface of the region, we have R, R cosθ cosθ, and R sin θ sinθ. Therefore, on this portion of the surface, our vector field becomes F z i + ( + 1) j + e z k z i + (4 sinθ cosθ + 1) j + sinθ e z k. In addition, we have da (cos θ i + sin θ j) dθ dz. Putting all of these facts together, we obtain π 6 F da (z i + (4 sin θ cosθ + 1) j + sinθ e z k) (cosθ i + sin θ j)dz dθ S π π π 6 (z cosθ + 8 sin θ cosθ + sinθ)dz dθ ( z cosθ + (8 sin θ cosθ + sin θ)z ) z6 (6 cosθ + 48 sin θ cosθ + 1 sinθ)dθ ( 6 sinθ + 16 sin θ 1 cosθ ) π 64. z dθ 6. A plane passes through the points (1,, ), ( 1,, 4), and (,, ). (a) Find the equation of this plane. (b) At what point does the line r(t) (1 + t) i + ( t) j + ( 1 + t) k intersect this plane? Let P (1,, ), let Q ( 1,, 4), and let R (,, ). (a) Let u PR i j and v PQ i 6 j + 6 k. Then the vector Q( 1,,4) n u v 18 i 6 j 1 k P(1,, ) is a normal vector to the plane. Therefore, since (1,, ) is a point on the plane, u R(,, ) a the equation of the plane is given b 18( 1) 6( ) 1(z + ), or, after simplifing, we can also write the equation of the plane as + +z. (b) Since we can rewrite r(t) parametricall as 1 + t, t, z 1 + t, we can substitute into the equation of our plane from part (a) to obtain (1 + t) + ( t) + ( 1 + t) + 6t + t + t t 1 t 1/. v

4 Math 61 Solutions to Sample Final Eam Problems 4 Substituting t 1/ for t in the equation for r(t), we obtain /, 1/, and z 6/, so the point of intersection is given b (/, 1/, 6/), which is our final answer. 7. If w i j + k, u i + a j + k, and v i + j, find: (a) A unit vector parallel to w. (b) The value of a making w perpendicular to u. (c) A unit vector perpendicular to both w and v. (a) w w i j + k 1 i 1 j + k (b) w u a + 6 a 8. (c) First, note that w v i + j + k, so one such vector is given b w v w v i + j + k 1 i + j + k 1 i + 1 j + 1 k. 8. A clindrical tube of radius cm and length cm contains a gas. Since the tube is spinning around its ais, the densit of the gas increases with its distance from the ais. The densit, D, at a distance of r cm from the ais is D(r) 1 + r grams per cubic centimeter. Find the mass of the gas in the tube. We first note that D(r) has units of grams/cm, and dv has units of cm, so the product D(r) dv has units of grams, which ields a mass. Therefore, the mass we are seeking is given b the integral D(r)dV, where W {(r, θ, z) : r, θ π, z }, so we have W Mass W D(r)dV π (1 + r)r dθ dz dr 6π (r + r )dr 8π grams. 9. The kinetic energ, E, of a moving object having mass m kilograms and speed v meters per second is given b E f(m, v) 1 mv Joules. Evaluate and give a complete sentence interpretation of each of the following quantities. (a) f(, ) (b) f v (, ) (c) f m (, ) (a) We have f(, ) 1 ()() 9 Joules. Thus, f(, ) gives the kinetic energ, in Joules, of an object of mass kilograms moving at a speed of meters per second. (b) We have f v (m, v) mv, so that f v (, ) 6 Joules per meter per second. Thus, the fact that f v (, ) 6 indicates the following: for an object having a mass of kilograms, the kinetic energ increases at a rate of about 6 Joules for each meter per second that the speed of the object is increased beond a speed of meters per second. (c) We have f m (m, v) 1 v. Therefore, f m (, ) 1 (9) 4. Joules per kilogram. So the fact that f m (, ) 4. indicates the following: for an object moving at a speed of meters per second, the kinetic energ increases at a rate of about 4. Joules for each kilogram of mass added to the object beond a mass of kilograms.

5 Math 61 Solutions to Sample Final Eam Problems 1. Given below are level curve diagrams for two functions, f and g. (a) Which of the two functions above is linear? Wh? Find a formula for the linear function. (b) Estimate g (1, ) and g (1, ). 1 1 z f(, ) z g(, ) (a) The function z f(, ) is linear because its level curves are equall spaced straight lines whose z-values decrease at a constant rate as ou move from one side of the diagram to the other. To find an equation for f, we first use the diagram to deduce that (,, ), (1,, ), and (,, 4) are all points on the graph of f. Therefore, we have Similarl, m slope of f in -direction z 1 1. n slope of f in -direction z 4. Thus, the equation for our plane is given b z z + m( ) + n( ) 1 ( ) + ( ), which ields a final answer of f(, ) 1 +. (b) Using the level curve diagram, we see that (1,, 4) and (,, ) are points on the graph of g. Therefore, we have g (1, ) z 4 1 1, so our approimation for g (1, ) is 1/. Similarl, since the point (1, 1, 6) also lies on the graph of g, we have g (1, ) z 6 4 1, so our approimation of g (1, ) is /. 11. Suppose that the temperature at a point (, ) is given b the formula T(, ) where and are measured in meters. 6 + degrees Celsius, + 1 (a) If ou move awa from the point (1, 1) in the direction of the point (, ), is the temperature increasing or decreasing? At what rate? (b) In what direction should ou move awa from the point (1, 1) for the temperature to remain constant? Before doing parts (a) and (b), we calculate a formula for the gradient function since we know that we will need this for both parts of the problem. We have T(, ) 6( + + 1) 1, so that T(, ) 6( + + 1) i 6( + + 1) 4 j 6( + + 1) ( i + 4 j). Therefore, T(1, 1) 6( ) ( i + 4 j) i 64 j. (a) We first note that we will be moving from (1, 1) to (, ), which means we are moving in the direction of the vector i + j. Therefore, the vector u (1/ ) i + (/ ) j describes a unit vector in the direction of our motion. Therefore, we have T u (1, 1) T(1, 1) u ( i 64 j) ( 1 i + j)

6 Math 61 Solutions to Sample Final Eam Problems 6 and we therefore see that the temperature is decreasing at a rate of about 71.6 degrees Celsius per meter. (b) Here, we are looking for a unit vector u u 1 i + u j such that T u (1, 1). Thus, we have T u (1, 1) T(1, 1) u ( i 64 j) (u 1 i + u j) u 1 64u u 1 u. Since we also know that u must be a unit vector, we also know that u 1, so u 1 + u 1 ( u ) + u 1 u 1 u 1 and u 1 u Therefore, we must move in the direction of ( / ) i + (1/ ) j in order for the temperature to remain constant. 1. Find the equation of the line containing the point (1,, ) that is parallel to both of the planes + z 4 and + z 1. Note that n 1 i + j k and n i j + k are normal vectors to the respective planes. Thus, since the desired line is parallel to both planes, it is perpendicular to both normal vectors. Therefore, n 1 n i j k is parallel to the line, so the line is given b the vector equation We could also write the answer in parametric form as r(t) (1 + t) i + ( t) j + ( t) k. 1 + t, t, z t. 1. Let W be the solid object consisting of two solid clinders meeting at right angles at the origin. One clinder is centered on the -ais, between and with radius and the other is centered on the -ais between and with radius (see figure below). Let S be the whole surface of W ecept for the circular end of the clinder centered at (,, ). The boundar of S is a circle, C, and the surface S is oriented outward. Let F i+ j+z k, and suppose ou are also told that F curl(z i+z j+ k). (a) Suppose that ou want to calculate S F d A. Write down two other integrals that have the same value as S F d A. (b) Calculate S F d A using whatever method is easiest. z Before we do part (a) and part (b), let s make some preliminar observations. B Stokes Theorem, we have F da S C (z i + z j + k) d r, where C is oriented as shown in the diagram to the right. Furthermore, letting S be the portion of the plane that lies inside C, we see that S has the same boundar curve as S, so that b Stokes Theorem, we have F da (z i + z j + k) d r F da, S C S where S is oriented in the negative direction. C n

7 Math 61 Solutions to Sample Final Eam Problems 7 (a) B the observations above, we see that S F d A is equal to the line integral C (z i + z j + k) d r, where C is oriented as shown in the diagram above. Similarl, we see that S F d A is also equal to S F d A, where S is the disk of radius on the interior of C, oriented in the negative direction. (b) Since the surface S has equation, and since n j is orthogonal to S at ever point on S with the orientation described above, we have F da F n da ( i + j + z k) ( j)da ( )da S S S S ( )(Area of S ) 1π.