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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?

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2 Answers 2

up vote 1 down vote accepted

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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Question: Usually the functional form of F is given. E.g. $F(x,y) = y^2 dx + (2xy-1) dy$. So in this case, can I say the equation is $F(x,y,z) = -1 dx + 0 dy + 1 dz$ (normalized)? –  Legendre Aug 28 '12 at 12:22
    
I wrote a vector and you wrote a covector. There is a natural bijection, essentially given by the inner product of $\mathbb{R}^3$. Yes, you can imagine they are the same thing. –  Siminore Aug 28 '12 at 12:25
    
Sorry, I meant: if we see vector fields as vector valued functions, then the vector field is $F(x,y,z) = \frac{1}{\sqrt{2}}(-1,0,1)$? The function that maps all $x,y,z$ input to a constant vector $\frac{1}{\sqrt{2}}(-1,0,1)$? (sorry, I'm quite bad at this because I didn't take undergrad physics :p) –  Legendre Aug 28 '12 at 12:39
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Yes, it is a constant vector field as you say. –  Siminore Aug 28 '12 at 12:41
    
Thank you for the active feedback. :) –  Legendre Aug 28 '12 at 12:41
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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.

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Thanks for your help! –  Legendre Aug 28 '12 at 12:40
    
you are welcome!! –  Santosh Linkha Aug 28 '12 at 15:31
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