# Parametrizing y=x^2 in physics

1. Nov 29, 2003

### philipc

How would one might parametrize y = x^2

Here is the problem

"an object moves along the parabola y = x^2 from (0,0) to (2,4). One of the forces acting on the object is F(x,y) = (x+2y)i + (2x+y)j. Calculate the work done by F."

the soluction manual has r(t) = ui + u2j -- [0,2]
I forgot how to parametrize this type of function(y = x^2), and how to get the limits of integration of [0,2]?
Thanks
Philip

2. Nov 29, 2003

### Hurkyl

Staff Emeritus
No it didn't; it probably had something like $\mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j}\; -- \; [0, 2]$ though. What are the $x$ and $y$ components of $r$ as a function of $t$? How does the range of $t$ compare to the ranges of $x$ and $y$? Does that give you any ideas how one might, in general, parametrize the curve $y=f(x) \; \mbox{from} \; (a, f(a)) \; \mbox{to} \; (b, f(b))$?

Last edited: Nov 29, 2003
3. Nov 29, 2003

### philipc

Hurkly,
Sorry I forgot the "^", yes you were right. You would have to forgive me, being an "elder" student, I don't remember a thing from the past, so my foundation in math is a bit weak. Can you give me what level of math that material may have been covered? That way I can do some reading on it.
Thanks for your help,
Philip

4. Nov 30, 2003

### Jacob Chestnut

I first saw simple parameterizations, functions and circles, in my university's Calculus1 course. They appeared again, fairly extensively, in my Multivariable Calculus course.

When you have a function defined as y in terms of x, the simple parameterization is to let x=t, then y=f(t). This gives
r(t)=<t, f(t)> as the position vector.

You will then obtain the limits, in terms of t, from the beginning and ending values of x, since x = t.

I'm sure someone will give you a more rigorous explanation, but I hope this helps.

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