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NAME: Due Date: 11/07/2016 Instructor: Jay Taylor Math 223 - Vector Calculus (Fall 2016) Homework 4 R 1. Using the Fundamental Theorem of Calculus for Line Integrals compute the line integral C F · dr where F = y e xy i + xe xy j + (cos z)k and C is the curve line from (0, 0, π) to (1, 1, π) followed by the parabola z = πx 2 in the plane y = 1 to the point (3, 1, 9π). 2. The path C is a line segment of length 10 in the plane starting at (2, 1). For f (x, y ) = 3x + 4y consider Z gradf · dr . C (a) Where should the other end of the line segment C be placed to maximise the value of the integral? (b) What is the maximum value of the integral? 3. The force exerted by gravity on a solid of mass m is F = −mgk; here g is the gravitational constant. (a) Find the work done against this force in moving from the point (1, 0, 0) to the point (1, 0, 2π) along the helix x = cos t, y = sin t, z = t by calculating the line integral. (b) Is F path independent? Explain your answer. Solutions 1. We need to write F as a gradient vector field to apply the Fundamental Theorem of Calculus for Line Integrals (FTOCLI). In other words we need to find a potential function f for F . Assume f (x, y , z) is such a function then F = grad f and we must have ∂f = y e xy ⇒ f = e xy + g(y , z). ∂x Differentiating with respect to y we have xe xy = ∂f ∂g ∂g = xe xy + ⇒ = 0 ⇒ g(y , z) = h(z) ∂y ∂y ∂y is a function of z only. Differentiating with respect to z we have cos(z) = ∂f ∂h = ⇒ h(z) = sin(z) + C. ∂z ∂z Therefore f (x, y , z) = e xy + sin(z) is a potential function for F . The curve starts at (0, 0, π) and ends at (3, 1, 9π) so by the FTOCLI we have Z Z F · dr = (grad f ) · dr = f (3, 1, 9π) − f (0, 0, π) = (e 3 + 0) − (e 0 + 0) = e 3 − 1. C C 2. (a) We have grad f = 3i + 4j . The value of the line integral will be maximised when the line goes in the same direction as grad f . Let r (t) = (2 + 3t)i + (1 + 4t)j be a parameterisation for the line starting at (2, 1) and going in the same direction as grad f . We want the distance from (2, 1) to be 10 so we have 102 = (3t)2 + (4t)2 = 25t 2 ⇒ t 2 = 4 ⇒ t = ±2. Therefore setting the end point to be (8, 9) maximises the value of the integral. (b) By FTOCLI we have Z grad f · dr = f (8, 9) − f (2, 1) = 50 C is the maximum value of the integral. 3. (a) A parameterisation for the curve is given by r (t) = cos ti + sin tj + tk with 0 6 t 6 2π. We therefore have Z Z 2π F · dr = F (r (t)) · r 0 (t) dt C 0 Z 2π = (−mgk) · (− sin ti + cos tj + k) dt 0 Z 2π = −mg dt 0 = −2πmg. The amount of work done against the force is thus 2πmg. (b) We have F = grad f where f (x, y , z) = −mgz so F is a gradient vector field so is path independent.