Math 263 Final. (b) The cross product is. i j k c. =< c 1, 1, 1 >

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1 Math 63 Final Problem 1: [ points, 5 points to each part] Given the points P : (1, 1, 1), Q : (1,, ), R : (,, c 1), where c is a parameter, find (a) the vector equation of the line through P and Q. (b) P Q P R. (c) an equation in Cartesian form (i.e. not vector form) for the plane passing through all three points. (d) the angle between the vectors P Q and P R. Is there a value of c for which the vectors are parallel? Perpendicular? olution: (a) A vector in the direction of the line is P Q =<, 1, 1 >, and a vector giving a point on the line is OP =< 1, 1, 1 >. Hence the parameterized vector equation of the line is r =< 1, 1, 1 > +t <, 1, 1 >. (b) The cross product is i j k c =< c 1, 1, 1 > (c) The cross product of P Q and P R gives a vector normal to the plane. Moreover, OP =< 1, 1, 1 > gives a point on the plane. Thus, an equation for the plane is < c 1, 1, 1 > (r < 1, 1, 1 >) = Written out explicitly: (c 1)x y + z = c 3 (d) We have P Q =<, 1, 1 > and P R =< 1, 1, c >. Hence P Q P R = 1 + c. Also, P Q = and P R = + c. Hence the angle between the vectors is given by ( ) c + 1 cos c The vectors are parallel if this angle is zero. i.e. for 1 = c c or c c + 3 =,

2 which has no real solution for c. Hence, the vectors are never parallel. The vectors are perpendicular when c = 1. Problem : [15 points, 5 points to each part] Consider a particle on a path given by r(t) =< t, cos t, sin t >. (a) etermine the speed and acceleration of the particle. (b) Find the unit tangent and unit normal vectors at t =. (c) Calculate the curvature κ at t =. What is the limit of κ when t? olution: (a) By differentiating: r =< t, sin t, cos t >, r =<, cos t, sin t > which gives the velocity and acceleration. The speed is r = 1 + 4t, and r = 5. (b) The unit tangent is T = r r = < t, sin t, cos t > 1 + 4t <,, 1 > at t =. The unit normal is Now, N = T T T = <, cos t, sin t > 1 + 4t 4t < t, sin t, cos t > (1 + 4t ) 3/ <, 1, > at t =. Hence, N =<, 1, > / 5. (c) Finally, the curvature is given by κ(t) = T r 5 at t =. For large t, Hence, T <, cos t, sin t > t r t, T 1 t giving κ.

3 Problem 3: [15 points, 5 points to each part] The surface given by x a + y a + z b = 1 is a spheroid centred at the origin with semi-axes a, a and b along the x, y and z axes. (a) Calculate the outward unit normal vector n at an arbitrary point on the spheroid. ( ) (b) Write out the equation for the tangent plane to the spheroid at point a a,,. (c) Find points on the spheroid where the tangent plane is parallel to the yz plane. olution: (a) The surface of the spheroid is whose unit normal vector is ince we have f(x, y, z) = x a + y a + z b 1 =, f = n = n = f f. x a, y a, z, b x, y, a z b. x + y + a4 z b 4 (b) The normal vector n = a a,, at the given point r = plane is given by n (r r ) =, which can be written as x + y = a. a a,,. The tangent (c) For the tangent plane to be parallel to the yz plane, the normal vector has to be parallel to the x axis. Thus, we require the y and z components of n to vanish: y =, z =. There are two points on the spheroid that satisfy these equations: (±a,, ). Problem 4: [1 points, 5 points to each part] Consider the solid object bounded by the parabloid z = x y on top and the cone z = x + y at the bottom. (a) Calculate the surface area of the solid region.

4 (b) Calculate the volume of the solid region. olution: (a) It will be helpful to sketch the parabloid and the cone first. The intersection between the two surfaces is a circle x + y = 1, z = 1. We will use the formula for the surface area of curved surfaces: ( ) ( ) A() = da, x y where is the circular region x + y = 1. We first calculate the partial derivatives for the top and the bottom part of the surface: Top (parabloid): x Bottom (cone): x = = x, y = y, x x, = +y y y x +y, 1 + ( x Thus, the total area is ( ) A() = x + 4y da = π+ ) + ( y) = 1 + 4x + 4y ; 1 + ( x π dθ ) + ( y) = r rdr = π+ π 6 (53/ 1). (b) The solid volume is given be the following double integral: V = [ ( x y ) x + y ] da = π dθ 1 ( r r)rdr = 5π 6. Problem 5: [15 points, 5 points to each part] Let f(x, y, z) = e x sin(y)z. (a) Compute F = f. Is F conservative? Justify. (b) how that f is a solution of Laplace s equation ( f) =. (c) Use the divergence theorem to show that the flux of F across any simple closed surface is. olution: (a) F = e x sin(y)z, e x cos(y)z, e x sin(y). As F = f, it is conservative by definition. (b) Note that ( f) = f xx + f yy + f zz. In our case: f xx = e x sin(y)z, f yy = e x sin(y)z, and f zz =.

5 Therefore, ( f) = f xx + f yy + f zz =. (c) As F is a vector field whose components have continuous partial derivatives everywhere, we can use divergence theorem to get F d = div F dv, where = E. On the other hand, o, we conclude F d =. E div F = ( f) =. Problem 6: [5 points, 5 points to each of (a-c) and 1 points to part (d)] Consider the vector field F(x, y, z) = (sin x + y )i + (cos y + z )j + (e z + x )k. (a) Compute curl F and div F. (b) Is F conservative? Justify your answer. (c) Can F be written as the curl of another vector field G? Justify your answer. (d) Let C be the curve of intersection of the cylinder x +y = x and plane z = x oriented counterclockwise as viewed from above. enote by the part of the plane z = x that is bounded by C. (i) Parametrize C. (ii) Parametrize. (iii) Use tokes Theorem to calculate C F dr [Hint: You may need cos (θ) = 1 (1 + cos(θ)).] olution: (a). curl F = z, x, y, and div F = cos(x) sin(y) e z. (b) No. As curl F, F is not conservative. (c) No. As div F, there is no vector field G which satisfies curl G = F. (d) (i) Note that if (x, y, z) is on the surface of the cylinder, after setting x = r cos(θ) and y = r sin(θ), we observe that r (cos (θ) + sin (θ)) = r cos(θ), which implies r = cos(θ). To have r we also require θ [ π/, π/]. This means that the surface of the cylinder can be parametrized by x = cos (θ), y = cos(θ) sin(θ) and

6 z = z where θ [ π/, π/]. Using the fact that C is the intersection of this cylinder and z = x, we obtain a parametrization of C: x = cos (θ), y = cos(θ) sin(θ), z = cos (θ), θ [ π/, π/]. Note: ome students may write the cylinder as (x 1) + y = 1, and then parametrize C as x = 1 + cos θ, y = sin θ, z = 1 + cos θ, θ π. (ii) is the graph of the function z = g(x, y) = x with (x, y) where is the disk on the xy-plane with radius 1, centered at (1, ). Therefore, a parametrization of is given by Alternatively, passing to polar coordinates, x = x, y = y, z = x, (x, y). x = r cos(θ), y = r sin(θ), z = r cos(θ), where (r, θ) = {(r, θ) : π/ θ π/, and r cos(θ)} Note: ome students may parametrize as x = 1 + r cos θ, y = r sin θ, z = 1 + r cos θ, r 1, θ π. (iii) Using tokes Theorem, we have F dr = C curl F d. As is the graph of g(x, y) = x, we can use Equation from the book, which yields curl F d = ( z) g x da = = = 16 3 = 16 3 = 4 3 = π π/ cos(θ) π/ π/ cos(θ) π/ π/ π/ π/ π/ x da cos 4 (θ)dθ ( r cos θ drdθ + cos(θ) + 1 ) cos(4θ) dθ [ 3 θ + sin(θ) sin(4θ) ] π/ π/

7 Note: Alternatives solutions that we may see, depending on which formula to use and on the parametrization: (1) Normal vector to being n =< 1/,, 1/ >, curl F d = (z y)d = which can then be calculated using one of the parametric forms. () Using Formula from the book, curl F d = Using the alternative polar parametrization: π xda = dθ 1 xda. (x y)dxdy, (1 + r cos θ) r dr = π.