ENGG1410-E: Short Test 2 Name: Student ID: Write all your answers on this sheet, and use the back if necessary. Problem 1 (30%). Let C be the curve r(t) = [5t, t2 , 3] from t = 0 to t = 1. Calculate R C t ds. Answer: Z Z 1 s t t ds = 0 C Z dx dt 2 + dy dt 2 + dz dt 2 dt 1q 52 + (2t)2 + 02 dt = 0 Z = 1 p t 4t2 + 25 dt. 0 = = = Z 1 1p 2 4t + 25 d(4t2 + 25). 8 0 1 1 (4t2 + 25)3/2 . 12 0 1 (293/2 − 253/2 ). 12 Problem 2 (30%). Let f (x, y, z) = ex y + 5z. Compute the directional derivative of f (x, y, z) in the direction of [1, 1, 1] at point (1, 2, 3). √ √ √ Answer: The unit vector in the direction of [1, 1, 1] is u = [1/ 3, 1/ 3, 1/ 3]. Thus, the directional derivative equals ∂f ∂f ∂f , , ·u ∇f (x, y, z) · u = ∂x ∂y ∂z ∇f (x, y, z) · u = [ex y, ex , 5] · u √ √ √ = [2e, e, 5] · [1/ 3, 1/ 3, 1/ 3] √ = (3e + 5)/ 3. Problem 3 (40%). Consider the curve C that is the intersection of the following two faces: x2 + y + z 2 = 5 x + y = 1. Give a tangent vector of C at point (0, 1, 2). Answer: From the two equations, we have x2 − x + z 2 = 4 ⇒ ⇒ x2 − x + 1/4 + z 2 = 17/4 2 z x − 1/2 2 √ + √ = 1 17/2 17/2 Hence we can write C in the parametric form [x(t), y(t), z(t)] where √ 1 17 x(t) = + cos t 2 2 √ 1 17 y(t) = 1 − x = − cos t 2 2 √ 17 sin t z(t) = 2 √ √ √ Therefore, a tangent vector is [x0 (t), y 0 (t), z 0 (t)] = [− 217 sin t, 217 sin t, 217 cos t]. The point (0, 1, 2) √ √ is given by t satisfying ( 17/2) cos t = −1/2 and ( 17/2) sin t = 2. Therefore, a tangent vector at this point is [−2, 2, −1/2]. 2

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