Derivative of parametric equations?

**dandaman** · September 13th 2009, 11:00 AM

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

**e^(i*pi)** · September 13th 2009, 11:04 AM

Originally Posted by dandaman

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

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

**skeeter** · September 13th 2009, 11:10 AM

Originally Posted by dandaman

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

$\frac{dx}{dt} = -\frac{\pi}{3}\sin\left(\frac{\pi t}{3}\right)$

$\frac{dy}{dt} = \frac{2\pi}{3}\cos\left(\frac{\pi t}{3}\right)<br />$

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

**dandaman** · September 13th 2009, 03:52 PM

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

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

**skeeter** · September 13th 2009, 04:13 PM

Originally Posted by dandaman

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

ignore this ...

you do not need to find $\frac{dy}{dx}$

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