# Calculus of parametric function

• Apr 24th 2013, 05:13 PM
amthomasjr
Calculus of parametric function
My professor gave us a sheet that had the calculus of functions in different formats. The sheet isn't making any sense to me though.

$\displaystyle x=f(t)$

$\displaystyle y=g(t)$

Derivatives: $\displaystyle \frac{dy}{dt}=f'(t)$, $\displaystyle \frac{dx}{dt}=g'(t)$, $\displaystyle \frac{dy}{dx}=\frac{f'(t)}{g'(t)}$

Integrals: $\displaystyle \int_ \alpha ^ \beta g(t)f'(t)dt$

Arc Length: $\displaystyle \int_ \alpha ^ \beta \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy} {dt}\right)^2} dt$

Should the letters correspond?
x goes with f
y goes with g
• Apr 24th 2013, 05:59 PM
christianwos
Re: Calculus of parametric function
You are right. The letters x and y should be inverted. What the sheet says is that the both x and y depend on the 'parameter' t. therefore you can write both x and y as functions of t. Hence the definition of x=f(t) and y=g(t).

For example, you can think of x and y as the coordinates of motion of a particle in two dimensions. As the particle moves in time, the coordinates of the point, that is (x,y), change. I think your prof just mistakenly inverted x and y.
• Apr 24th 2013, 06:18 PM
amthomasjr
Re: Calculus of parametric function
Thanks for clearing that up