# Thread: Derivative of parametric equations?

1. ## Derivative of parametric equations?

A particle is moving in the xy-plane and its position at time t is given by
$\displaystyle x=cos(\frac{\pi }{3}t)$
and $\displaystyle y=2sin(\frac{\pi }{3}t)$. When t=3 the speed of the particle is?

-Ok i know that i have to take the derivative. But i don't remember how to take the derivative of parametric equations...

all help is appreciated..

2. Originally Posted by dandaman
A particle is moving in the xy-plane and its position at time t is given by
$\displaystyle x=cos(\frac{\pi }{3}t)$
and $\displaystyle y=2sin(\frac{\pi }{3}t)$. When t=3 the speed of the particle is?

-Ok i know that i have to take the derivative. But i don't remember how to take the derivative of parametric equations...

all help is appreciated..
Recall the Chain Rule:

$\displaystyle \frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}$

3. Originally Posted by dandaman
A particle is moving in the xy-plane and its position at time t is given by
$\displaystyle x=cos(\frac{\pi }{3}t)$
and $\displaystyle y=2sin(\frac{\pi }{3}t)$. When t=3 the speed of the particle is?

-Ok i know that i have to take the derivative. But i don't remember how to take the derivative of parametric equations...

all help is appreciated..
$\displaystyle \frac{dx}{dt} = -\frac{\pi}{3}\sin\left(\frac{\pi t}{3}\right)$

$\displaystyle \frac{dy}{dt} = \frac{2\pi}{3}\cos\left(\frac{\pi t}{3}\right)$

speed = $\displaystyle \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}$

4. ok so i have
$\displaystyle \frac{dy}{dx}=\frac{dy}{dt} \frac{dt}{dx}$
and
$\displaystyle \frac{dy}{dt}=\frac{2\pi }{3}cos(\frac{\pi }{3}t)$
and
$\displaystyle \frac{dx}{dt}=-\frac{\pi }{3}sin(\frac{\pi }{3}t)$

but i have dx/dt not dt/dx.
do i have to take the reciprocal so i can plug it into the equation?

5. Originally Posted by dandaman
$\displaystyle \frac{dy}{dx}=\frac{dy}{dt} \frac{dt}{dx}$

ignore this ...
you do not need to find $\displaystyle \frac{dy}{dx}$