1. (a) The volume of a piece of cake, with radius r, height h and angle θ, is given by the formula: [Yes! It s a piece of cake.]

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1 1. (a The volume of a piece of cake, with radius r, height h and angle θ, is given by the formula: / 3 V (r,h,θ = 1 r θh. Calculate V r, V h and V θ. [Yes! It s a piece of cake.] V r = 1 r θh = rθh V h = 1 r θ V θ = 1 r h. (b Consider the following function of two variables: / 7 Find f xy at the point (x,y = (π,1. e f (x,y = sinx x x cos(xy. By the Mixed Partials Theorem, f xy = f yx. The latter is much easier to compute since the monster term depends only on x. f xy = f yx = (f y x = (0 + (cos(xy y x = ( sin(xy (xy y x (single-variable chain-rule = (x sin(xy x = sin(xy xy cos(xy f xy (π,1 = sin(π 1 π 1cos(π 1 = π. 1

2 . Consider the two surfaces defined by equations: Surface 1: x y z = 4 Surface : z = (x siny + (a Find the equation of the tangent plane to Surface 1 at the point (x,y,z = (3,,1. Your final answer / 6 must be of the form Ax + By + Cz = D. Let g(x,y,z = x y z, then Surface 1 is the level surface g = 4. A normal vector to the tangent plane at (3,,1 is n = g(3,,1: g g = x, g y, g z = x, y, z n = g(3,,1 = 6, 4,. The equation of the tangent plane with n = 6, 4, and P 0 (3,,1 is: After simplification: 3x y z = 4. 6x 4y z = 6(3 4( (1. (b Show that the tangent plane to Surface 1 at the point (x,y,z = (3,,1 is parallel to the tangent / 4 plane to Surface at the point (x,y,z = (0,0,3. Surface is given as a graph of a two-variable function f (x,y = (x siny +. A normal vector to the tangent plane to Surface at (0,0,3 is given by: f f N = (0,0, x y (0,0, 1 = 3(x + 1, cosy, 1 (x,y=(0,0 = 3,, 1. Therefore, the normal vector n to Surface 1 and the normal vector N to Surface at the given points are parallel (since n = N. The two tangent planes are parallel to each other.

3 3. Let / 10 r(t = e t i + t 3 j + (t lnt tk. At which point(s on the curve is the tangent line horizontal (meaning parallel to the z = 0 plane? Give the (x,y,z coordinates of each such point. The direction of the tangent line is parallel to the velocity vector: ( r (t = e t i + 3t j + lnt + t 1 t 1 k = e t i + 3t j + (lntk. In order for the tangent line to be parallel to the z = 0, we require the k-component of r (t to be 0, i.e. lnt = 0. The only t for it to be true is that t = 1. In (x,y,z-coordinates of the point on the curve when t = 1 is given by: (x,y,z = (e 1,1 3,(1 ln(1 1 = (e,1, 1. 3

4 4. Define the curve r(t = x(ti + y(tj + z(tk by: x(t =t, y(t =3t cost 3sint, where t 0 z(t =3t sint + 3cost. (a Calculate the arc-length (i.e. distance of the curve segment r(t from t = 0 to t = 1. / 5 x (t = 4t y (t = 3cost 3t sint 3cost = 3t sint z (t = 3sint + 3t cost 3sint = 3t cost r (t = x (ti + y (tj + z (tk = 4ti 3t(sintj + 3t(costk r (t = (4t + ( 3t sint + (3t cost = 16t + 9t (sin t + cos t = 16t + 9t = 5t = 5t. The arc-length travelled by r(t from t = 0 to t = 1 is given by: r (t dt = 5tdt = 5t 0 t=1 t=0 = 5. (b Find a positive value k such that the arc-length of the curve segment r(t from t = 0 to t = k is / 5 equal to s, where s is a positive number. Your answer for k should be in terms of s. First set-up the equation for k and s: k s = r (t dt = 0 k 0 = 5k. 5tdt (from part (a Since k has to be positive, by solving the above equation, we have k = s 5. 4

5 5. Let r(t = cost, sint, cost. (a Find the corresponding velocity v(t as a function of t. / v(t = r (t = sint, cost, sint. (b Calculate the cross product r(t v(t of the position with the velocity as a function of time. / 5 i j k r(t v(t = cost sint cost sint cost sint = ( sin t cos ti (sint cost sint costj + ( cos t sin tk = (sin t + cos ti 0j (sin t + cos tk = i k. (c Use the result of your calculation from (b to establish that the curve traced out by r is contained / 3 in a single plane. Write an equation for this plane. By the definition of cross products, r(t v(t is at all times orthogonal to r(t (and v(t as well. Therefore we have (r(t v(t r(t = 0 for any t. From (b, we know r(t v(t is a constant vector given by 1,0,1. Denote the components of r(t by: r(t = x(ti + y(tj + z(tk. We have: 1,0,1 x(t,y(t,z(t = 0 = x(t + z(t = 0. Therefore, the curve traced out by r(t is contained in a single plane, given by an equation: x + z = 0. 5

6 6. When studying the surface of a sphere, rather than the usual (x,y,z-coordinates for R 3, it can often be more convenient to use spherical coordinates: x = cosu cosv y = sinu cosv z = sinv with π u π, π/ v π/. In these coordinates, increasing v corresponds to going north; and increasing u corresponds to going east. Let f = x + yz + z. (a Compute f f and. Express your answers in terms of u and v. / 10 u v By the chain rule, we have f u = f x x u + f y y u + f z z u = (x ( sinu cosv + z (cosu cosv + (y + 1 = (cosu cosv( sinu cosv + sinv cosu cosv = cosu sinu cos v + cosu sinv cosv. f v = f x x v + f y y v + f z z v = (x ( cosu sinv + z ( sinu sinv + = cos u cosv sinv sinu sin v + (y cosv (sinu cosv + 1 cosv [If your answers are correct, you should have f u = 1 f and v = 1 when (u,v = (0,π/4. If you got anything else, you better double-check your calculations in (a.] 6

7 ( (b Let P be the point 1,0, 1 in the (x,y,z-coordinates, which corresponds to (u,v = ( 0, π / 5 4 in the spherical (u, v-coordinates. Regarding f (u, v as a two-variable function of u and v, compute the directional derivative of f heading south-west from P. [Reminder: increasing v corresponds to going north; and increasing u corresponds to going east.] By substituting (u,v = (0,π/4 into the results of (a, we have f v = 1 1 f u = ( = = = 1. f (u,v = f u i + f v j f (0,π/4 = 1 i 1 j. The south-west direction is given by the vector v = i j, and after unitizing the vector becomes u = i j. The directional derivative of f at P in this direction is given by: 1 D u f (0,π/4 = f (0,π/4 u =, 1 1, 1 = 0. (c i. In which direction from P is f increasing most rapidly? Your answer should be one of the / 3 following directions: north, north-east, east, south-east, south, south-west, west, north-west. Do justify your answer. f increases most rapidly at the direction of f (0,π/4 since D w f (0,π/4 = f (0,π/4 cosθ where θ is the angle between f (0,π/4 and the unit direction vector w. It is at the maximum when θ = 0 or equivalently, f (0,π/4 is parallel to the direction vector w. From (b, we have f (0,π/4 = 1 i 1 j, which is the south-east direction. ii. What is the directional derivative of f at P in the direction found in part (c(i? / 1 Let w be the unit vector at the south-east direction, then: w =, D w f (0,π/4 = f (0,π/4 w =, 1, 1 = 1. 7

8 7. Let r(t = x(ti+y(tj+z(tk be the position vector of a particle at time t. Suppose the path of the particle obeys the following law: r (t = ρ(x(t,y(t,z(t where ρ(x,y,z is a three-variable function (that represents the potential energy of the particle. The total (kinetic + potential energy E(t of the particle at time t is defined by: E(t = 1 r (t + ρ(x(t,y(t,z(t. (a Show that the kinetic energy 1 r (t satisfies the following rate of change: / 4 ( d 1 dt r (t = ρ r (t. [For the sake of simplicity, we abbreviate ρ(x(t,y(t,z(t by ρ.] ( d 1 dt r (t = d ( 1 dt r (t r (t = 1 (r (t r (t + r (t r (t (product rule = r (t r (t = ρ r (t. (b Show that the total energy is conserved by proving that: de = 0. [Hint: use part (a] / 6 de dt = d ( 1 dt r (t = ρ r (t + ( ρ = = + d dt ρ(x(t,y(t,z(t ( ρ dx x dt + ρ dy y dt + ρ z dz dt x i + ρ y j + ρ ( dx z k dt i + dy dt j + dz dt k + ( ρ dx x dt + ρ dy y dt + ρ dz + z dt dt ( ρ dx x ( ρ dx x dt + ρ dy y dt + ρ z dt + ρ y dz dt dy dt + ρ z = 0 dz dt 8

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