Solving Squirrel Work Problem with Line Integral Setup

[SithsNGiggles]

Homework Statement

A squirrel weighing 1.2 pounds climbed a cylindrical tree by following the helical path

x = \cos{t}, y = \sin{t}, z = 4t, 0 \leq t \leq 8 \pi
(distance measured in feet)

How much work did it do?

Homework Equations

\int_{C} \vec{F} \cdot d\vec{r}

The Attempt at a Solution

I've defined a curve C by the vector

\vec{r}(t) = \cos{t} \vec{i} + \sin{t} \vec{j} + 4t \vec{k},
0 \leq t \leq 8 \pi

I'm not sure where to go from here. Specifically, I don't know how to use the weight of the squirrel. Every other problem I've worked on explicitly gave me a vector field to work with.

I know the bounds of the integral will be from 0 to 8π, and that r'(t) will be used.

Thanks in advance!

 

[LCKurtz]

The squirrel's weight points straight down. Try $\vec F = \langle 0,0,-1.2\rangle$. And remember the line integral gives the work done by the force. You should be able to check your answer by comparing the change in potential energy.

 

[SithsNGiggles]

Got it, thanks.

So I should get my answer with the following integral:

W = \int^{8\pi}_{0} (0\vec{i} + 0\vec{j} - 1.2\vec{k}) \cdot (-\sin{t}\vec{i} + \cos{t}\vec{j} + 4\vec{k}) dt ?

This isn't for a physics course, and we haven't learned anything about potential energy. If the integral's setup is right, though, I can't take it from there.

 

[Gullik]

I'm not still new at line integrals so take this with a grain of salt.

So I should get my answer with the following integral:

W = \int^{8\pi}_{0} (0\vec{i} + 0\vec{j} - 1.2\vec{k}) \cdot (-\sin{t}\vec{i} + \cos{t}\vec{j} + 4\vec{k}) dt ?

Take the dot product inside the integral, and integrate the answer.

This isn't for a physics course, and we haven't learned anything about potential energy. If the integral's setup is right, though, I can't take it from there.

Potential energy close to the Earth is E_p=mgh so W=mgΔh with m being mass, g gravitational acceleration and h is the heigth.

 

[SithsNGiggles]

Sorry, I meant I can take it from there. Thanks though!