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Line integral setup

 
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Sep15-12, 06:58 PM   #1
 

Line integral setup


1. The problem statement, all variables and given/known data
A squirrel weighing 1.2 pounds climbed a cylindrical tree by following the helical path

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

How much work did it do?

2. Relevant equations

[itex]\int_{C} \vec{F} \cdot d\vec{r}[/itex]

3. The attempt at a solution
I've defined a curve [itex]C[/itex] by the vector

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

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!
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Sep15-12, 09:48 PM   #2
 
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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.
Sep16-12, 01:50 AM   #3
 
Got it, thanks.

So I should get my answer with the following integral:

[itex]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 ?[/itex]

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.
Sep16-12, 02:32 AM   #4
 

Line integral setup


I'm not still new at line integrals so take this with a grain of salt.
Quote by SithsNGiggles View Post
So I should get my answer with the following integral:

[itex]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 ?[/itex]
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 [itex]E_p=mgh[/itex] so [itex]W=mgΔh[/itex] with m being mass, g gravitational acceleration and h is the heigth.
Sep16-12, 12:38 PM   #5
 
Sorry, I meant I can take it from there. Thanks though!
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