Calculus III: Final Exam with Solutions

CALCULUS 3 FINAL EXAM I have neither given nor received aid on this exam. Name: If you have a question raise your hand and remain seated. In order to receive full credit your answer must be complete, legible and correct. Show all of your work, and give adequate explanations. DO NOT WRITE IN THIS BOX! Problem Points 1 15 pts 2 15 pts 3 15 pts 4 15 pts 5 20 pts 6 15 pts 7 30 pts 8 15 pts 9 15 pts 10 30 pts 11 15 pts TOTAL 200 pts Score 1. (15 pt) Find the area of the region enclosed by the curve 1. (15 pt) Find the area of the region enclosed by the curve x = 1 − t22, y = t(1 − t22), −1 ≤ t ≤ 1 x = 1 − t , y = t(1 − t ), −1 ≤ t ≤ 1 (see the picture below) (see the picture below) 0.5 0 0.5 1 -0.5 I By Green’s Theorem, A = −y dx with C the above curve oriented counter-clockwise. Thus C I Z 1 Z 1 8 2 A= −y dx = −t(1 − t )(−2t)dt = 2 (t2 − t4 )dt = . 15 C −1 −1 2. (15 pt) Find the equation for the plane tangent to the paraboloid z = 2x2 + 3y 2 that is also parallel to the plane 4x − 3y − z = 10 2. (15 pt) Find the equation for the plane tangent to the paraboloid z = 2x2 + 3y 2 that is also parallel to the plane 4x − 3y − z = 10 Write G(x, y, z) = 2x2 + 3y 2 − z, so ∇G(x, y, z) = h4x, 6y, −1i. This tells us that h4x, 6y, −1i is a normal vector for the tangent plane to our surface at the point (x, y, z). For the tangent plane at a point to be parallel to 4x − 3y − z = 10, we need h4x, 6y, −1i to be parallel to h4, −3, −1i. This will certainly happen when x = 1 and y = − 21 . Solving for z in the equation for our surface, we get that the tangent plane at (1, − 12 , 11 ) is parallel to 4x − 3y − z = 10. At 4 this point the equation for the tangent plane will be given by 4(x − 1) − 3(y + 21 ) − (z − 11 ) = 0. 4 3. (15 pt) Find the flux of F(x, y, z) = (x + y)i + (y + z)j + (z + x)k across the portion of the plane x + y + z = 1 in the first octant oriented by unit normals with positive components. We know Φ = RR F • n dS. Let u = x and v = y. Then our surface is given by σ r(u, v) = hu, v, 1 − u − vi , 0 ≤ u ≤ 1, 0 ≤ v ≤ 1 − u. Next, n dS = (ru × rv ) dA = (h1, 0, −1i × h0, 1, −1i) dA = h1, 1, 1i dA . Then ZZ F • n dS Φ= σ Z 1 Z 1−u hu + v, v + (1 − u − v), (1 − u − v) + ui • h1, 1, 1i dv du = 0 Z 0 1 Z 1−u = 2 dv du 0 0 =1 4. (15 pt) Evaluate the integral Z 0 1 Z 1 √ cos (y 3 ) dy dx x by changing the order of integration. Z 0 1 Z 1 √ Z 3 1 Z cos (y ) dy dx = x 0 Z = y2 cos (y 3 ) dx dy 0 1 cos (y 3 )y 2 dy 0 1 = sin(y 3 ) 3 sin(1) = 3 1 0 5. A wire 12 cm long is cut into three or fewer pieces, with each piece bent into a square. (a) (10 pt) What is the minimal total area of the squares? Justify your answer. Let x, y, z denote the lengths of the three segments of the wire after it is
cut, so x+y+z

= y 2 z 2 x 2 12. Bending each segment into a square yields a total area of A = 4 + 4 + 4 . Replacing z with 12 − x − y (and simplifying), we have 1 A(x, y) = (x2 + y 2 + xy − 12x − 12y + 72) . 8 Now, x + y + z = 12 forces x and y to be confined to the region in the xy-plane, R, which is the triangle with vertices (0, 0), (12, 0), (0, 12) (this is just the projection of surface x + y + z = 12, with x, y, z ≥ 0 onto the xy-plane). We now minimize and maximize A over R. Find Interior Critical Points: We simultaneously solve Ax = 81 (2x + y − 12) = 0 and Ay = 18 (2y + x − 12) = 0 to find the one interior critical point (4, 4, 4). Test Critical Points and Boundary Points: On the portion of the boundary of R that lies on the x-axis, we have y = 0 and 0 ≤ x ≤ 12. Substituting into A we get A = 18 (x2 − 12x + 72), 0 ≤ x ≤ 12 . The absolute extrema of 18 (x2 − 12x + 72) on 0 ≤ x ≤ 12 occur when x = 0, 12, so we must consider the points (0, 0, 12), (12, 0, 0) . On the portion of the boundary of R that lies on the y-axis, we have x = 0 and 0 ≤ y ≤ 12. Similar to the previous case, we find that we must consider the points (0, 0, 12), (0, 12, 0) . Finally, on the portion of the boundary of R that lies on the line y = 12 − x, we have y = 12−x and 0 ≤ x ≤ 12. Substituting into A we get A = 81 (x2 −12x+72), 0 ≤ x ≤ 12 . As before, the absolute extrema of 18 (x2 − 12x + 72) on 0 ≤ x ≤ 12 occur when x = 0, 12, so we must consider the points (0, 12, 0), (12, 0, 0) . Testing all the points that we have listed, we find that the minimum total area is 3 cm2 occurring when each segment is of length 4 cm. (b) (10 pt) What is the maximal total area of the squares? Justify your answer. Using the above information, we find that the maximum total area is 9 cm2 occurring when we make no cut and have only one segment of length 12 cm. I 6. (15 pt) Evaluate F • dr, where F(x, y, z) = −3y 2 i + 4zj + 6xk and C is the triangle in the C plane z = 21 y with vertices (2, 0, 0), (0, 2, 1) and (0, 0, 0) with a counterclockwise orinetation looking down the positive z-axis. I ZZ We use Stoke’s Theorem, so F • dr = (curlF) • ndS . Let u = x and v = y. Then our C σ surface, σ, is given by 1 r(u, v) = hu, v, vi , 2 0 ≤ u ≤ 2, 0 ≤ v ≤ 2 − u. Next, 1 1 n dS = (ru × rv ) dA = (h1, 0, 0i × h0, 1, i) dA = h0, − , 1i dA . 2 2 1 Note that we have to choose between ±h0, − 2 , 1i to correctly orient the surface, but the ’+’ is chosen to correspond to the orientation of the curve. Then I ZZ F • dr = (curlF) • ndS C σ Z 2 Z 2−u = Z0 2 Z0 2−u = 1 h−4, −6, 6vi • h0, − , 1i dv du 2 (3 + 6v) dv du 0 Z = 0 2 (−3u2 + 9u − 6) du 0 = −11 7. The helix r1 (t) = cos ti + sin tj + tk intersects the curve r2 (t) = (1 + t)i + t2 j + t3 k at the point P (1, 0, 0). (a) (15 pt) Find the tangent line to each curve at P . Considering when r2 is at P , we see that the intersection occurs when t = 0. Now, r01 = h− sin t, cos t, 1i, so the tangent line to the curve given by r1 at t = 0 will have direction given by h0, 1, 1i. Thus the tangent line is given by h1, 0, 0i+th0, 1, 1i = h1, t, ti. Similarly, we find that the tangent line to the curve given by r1 at t = 0 is given by h1 + t, 0, 0i. (b) (15 pt) Find the angle between the tangent lines at P . The angle between them, θ, can be found by solving cos θ = d1 • d2 ||d1 || ||d2 where d1 and d2 are direction vectors for the lines. Thus cos θ = so θ = π2 . h0, 1, 1i • h1, 0, 0i =0 ||h0, 1, 1i|| ||h1, 0, 0i 8. (15 pt) Evaluate RR sin R 1 (x 2
+ y) cos 1 (x 2
− y) dA, where R is the triangular region whose vertices are (0, 0), (2, 0) and (1, 1). We start with a change of coordinates. Set u = 12 (x + y) and v = 21 (x − y), so that x = u + v and y = u − v. The region R transforms to the uv-region S which is a triangle with vertices ∂(x,y) = −2. Thus, (0, 0), (1, 0) and (1, 1). Calculating the Jacobian, we find that ∂(u,v)  ZZ sin R    ZZ ∂(x, y) 1 1 (x + y) cos (x − y) dA = sin u cos v dA 2 2 ∂(u, v) S Z 1Z u = 2 sin u cos v dv du 0 0 Z 1 2 sin2 u du = Z0 1 (1 − cos(2u)) du = 0 1 = − sin 2 + 1 2 R (3,2) 9. (15 pt) Consider (0,0) (2xey + 1)dx + x2 ey dy. Show that the integral is independent of path and evaluate the integral using the Fundamental Theorem of Line Integrals. To check that the vector field appearing in this integral is conservative it is enough to check the equality of the partial derivatives, since it is defined in the whole plane: ∂ ∂ 2 y (x e ) = 2xey = (2xey + 1) ∂x ∂y R y Let φ be the potential. Then ∂φ = 2xe + 1, φ = (2xey + 1)dx = x2 ey + x + C(y). ∂φ = x2 ey . ∂x ∂y Hence C 0 (y) = 0, C is a constant, and we can choose C = 0. Then φ = x2 ey + x and Z (3,2) (0,0) (2xey + 1)dx + x2 ey dy = φ(3, 2) − φ(0, 0) = 9e2 + 3 10. Find (a) (15 pt) the equation of the plane through P (1, −1, 2), Q(2, 1, 3), and R(−1, 2, −1). −→ −→ We have: P Q = h1, 2, 1i, P R = h−2, 3, −3i, i j k −→ −→ P Q × P R = 1 2 1 = −9i + j + 7k −2 3 −3 Therefore the equation of the plane is −9(x − 1) + (y + 1) + 7(z − 2) or −9x + y + 7z − 4 = 0 (b) (15 pt) parametric equations of the line of intersection of the planes x + 2y + z = 1 and x − y + 2z = −8. Normal vectors of the planes are h1, 2, 1i and h1, −1, 2i. The direction vector of the intersection line is given by the cross-product of the normal vectors, i.e. by i j k 1 2 1 = 5i − j − 3k 1 −1 2 Now look for a point in the intersection, e.g. the one which has z = 0. We then have x+2y = 1, x − y = −8. From this x = −5, y = 3, i.e. the point (−5, 3, 0) is in the intersection of the planes. From this, the parametric equations of the line are x = −5 + 5t y =3−t z = −3t 11. (a) (1 pt) Write a formula expressing the directional derivative Dn f of a function f along a unit vector n in terms of ∇f and n. Dn f = ∇f • n (b) (14 pt) Let σ be the sphere x2 + y 2 + z 2 = 1, let n be an outward unit normal, and 2 x 2 2 let RR Dn f be the directional derivative of f (x, y, z) = x + e cos y + y + z . Evaluate Dn f dS. σ ZZ ZZ ZZZ ∇f • ndS = Dn f dS = σ σ 2 div∇f dV G 2 2 where G is the interior of the sphere x + y + z = 1. Now ∇f = h2x + ex cos y, −ex sin y + 2y, 2zi, and div∇f = 6. Therefore   ZZ ZZZ 4π Dn f dS div∇f dV = 6V olume(G) = 6 = 8π 3 σ G