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A particle moves along the path $(t,t^2,t^3)$, $0 \leq t \leq 1$, and experiences a constant unit force parallel to vector $(-1,0,1)$. What is the work done by the force?

Attempt:

I am studying for the GRE Mathematics Subject Test, and I don't know any physics. While preparing for the test, I learned how to integrate line integrals over a vector field.

i.e. $ F(x,y) = M dx + N dy$ is a vector field. $r(t) = (x(t),y(t))$ is a parameterized curve where $a \leq t \leq b$. Then, the line integral over vector field F is

$\int_a^b \! F(x(t),y(t)) \cdot (x'(t),y'(t)) \, \mathrm{d} t$

So, I know that the integral to solve for my question is:

$\int_0^1 \! F(t,t^2,t^3) \cdot (1,2t,3t^3) \, \mathrm{d} t$

Question:

How do I find the equation for the vector field, $F(x,y,z)$? Or is my approach wrong?

2 Answers 2

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You are looking for a vector field $F \colon \mathbb{R}^3 \to \mathbb{R}^3$ such that

  1. $|F|=1$ (constant unit force)
  2. $F$ is parallel to $(-1,0,1)$.

Then $F(x,y,z)=\frac{1}{\sqrt{2}}(-1,0,1)$.

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    Thank you for the active feedback. :)2012-08-28
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Since the force is unit and is parallel to $(-1, 0, 1)$, Force should be divided by $\sqrt{(-1)^2+1^2} =\sqrt 2 $ to make it unit force. The work done would be $\int_0^1 {1 \over \sqrt 2}(-1, 0,1)\cdot(1, 2t, 3t^2)dt$ As it does not vary upon the path it takes.