# Thread: how to find 2nd derivative of parametric equation?

1. ## how to find 2nd derivative of parametric equation?

here's the parametric equation:
x= tcos(t)
y= tsin(t)

for first derivative i get
dy/dx= $\displaystyle \frac {tcos(t)+sin(t)}{-tsin(t)+cos(t)}$

using:
$\displaystyle d^2y/dx^2$= $\displaystyle \frac{(d/dt)(dy/dt)}{dx/dt}$

this is what i get for the second derivative:

$\displaystyle d^2y/dx^2$= $\displaystyle \frac{-tsin(t)+2cos(t)}{-tsin(t)+cos(t)}$

but the answer in the back of the book is:
$\displaystyle d^2y/dx^2$= $\displaystyle \frac{t^2+2}{(cos(t)-tsint)^3}$

what did i do incorrectly or what am i doing incorrectly?

2. Originally Posted by yoman360
here's the parametric equation:
x= tcos(t)
y= tsin(t)

for first derivative i get
dy/dx= $\displaystyle \frac {tcos(t)+sin(t)}{-tsin(t)+cos(t)}$

using:
$\displaystyle d^2y/dx^2$= $\displaystyle \frac{(d/dt)(dy/dt)}{dx/dt}$

this is what i get for the second derivative:

$\displaystyle d^2y/dx^2$= $\displaystyle \frac{-tsin(t)+2cos(t)}{-tsin(t)+cos(t)}$

but the answer in the back of the book is:
$\displaystyle d^2y/dx^2$= $\displaystyle \frac{t^2+2}{(cos(t)-tsint)^3}$

what did i do incorrectly or what am i doing incorrectly?
If $\displaystyle \frac{\,dy}{\,dx}=\frac{\,dy/\,dt}{\,dx/\,dt}$, then

$\displaystyle \frac{\,d^2y}{\,dx^2}=\frac{\,d}{\,dx}\left(\frac{ \,dy}{\,dx}\right)=\frac{\,d}{\,dx}\left(\frac{t\c os t+\sin t}{\cos t-t\sin t}\right)$

Now observe that $\displaystyle \frac{\,d}{\,dx}=\frac{\,d}{\,dt}\frac{\,dt}{\,dx}$.

Therefore, $\displaystyle \frac{\,d}{\,dx}\left(\frac{t\cos t+\sin t}{\cos t-t\sin t}\right)=\frac{\,d}{\,dt}\left(\frac{t\cos t+\sin t}{\cos t-t\sin t}\right)\cdot\frac{\,dt}{\,dx}$.

Now apply quotient rule, and keep in mind that $\displaystyle \frac{\,dt}{\,dx}=\frac{1}{\,dx/\,dt}=\frac{1}{\cos t-t\sin t}$; you will then get the desired result.

Can you take it from here?