# Work along a line integral

## Homework Statement

A particle is attracted towards the origin by a force proportional to the cube of its distance from the origin. How much work is done in moving the particle from the origin to the point (2,4) along the path y = x^2 assuming a coefficient of friction $$\mu$$ between the particle and the path?

## The Attempt at a Solution

For the path, set x = t. Thus, y = t^2

$$\vec{r}(t) = t\hat{i} + t^2\hat{j}$$
$$\vec{r}'(t) = \hat{i} + 2t\hat{j}$$

Because we're going from the origin to (2,4), t goes from 0 to 2.

For the force, we know that:

$$\left\|F\right\| = {k(x^2 + y^2)}^{3/2}$$

I want this ideally in vector form, but the best I could come up with:
$${\left\|F\right\|}^{1/3} = kx\hat{i} + ky\hat{j}$$

My main problem is I can't really use $${\left\|F\right\|}^{1/3}$$ to calculate $$\int_{C} {F\cdot dr}$$. If I could just find a way to express F in vector form, I'm pretty much off to the races. (Because I can then calculate the work done by friction by finding the normal to $$\vec{r}(t)$$, multiplying it by $$\mu$$, and dotting it with $$\vec{r}'(t)$$ from $$0\leq t \leq 2$$, etc. etc.)

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Dick
Homework Helper
The vector (x*i+y*j)/sqrt(x^2+y^2) points away from the origin in the direction of (x,y) and has length 1. Just take that vector and adjust it's length and direction.

How would I adjust for the "cube of the distance" though ? Looking at the numerator of that unit vector, there isn't much, I think, that one can do with $$(x\hat{i} + y\hat{j})^3$$

Dick
Homework Helper
The cube of the distance is r^3=(x^2+y^2)^(3/2), like you said. If u=(x*i+y*j)/sqrt(x^2+y^2) has length 1, shouldn't r^3*u have length proportional to r^3? What does that vector simplify to? (x*i+y*j)^3 doesn't even mean anything.

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(facepalm of realization)

I'd probably multiply r^3 and u together, wouldn't I.

Then slap on the k at the front of the product, and equate it to F.

Dick