Download Math 223 - Vector Calculus (Fall 2016) Homework 4

NAME:
Due Date: 11/07/2016
Instructor: Jay Taylor
Math 223 - Vector Calculus (Fall 2016)
Homework 4
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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.