# 1st and 2nd derivatives of parametric equations

• Dec 8th 2011, 05:22 AM
Stuck Man
1st and 2nd derivatives of parametric equations
I have found dy/dx correctly. I know I have not used the easiest method of obtaining the second derivative. It is correct except that the final numerator should be to the power 3. I cannot see where I have gone wrong. I followed the formula correctly at the start. It is the same as the quotient rule except that the denominator is cubed not squared.
• Dec 8th 2011, 06:46 AM
skeeter
Re: 1st and 2nd derivatives of parametric equations
$\displaystyle \frac{d}{dx} \left(\frac{dy}{dx} = \frac{2t}{t^2+1}\right)$

$\displaystyle \frac{d^2y}{dx^2} = \frac{(t^2+1)(2 \frac{dt}{dx}) - (2t)(2t \cdot \frac{dt}{dx})}{(t^2+1)^2} =\frac{(t^2+1)(2) - (2t)(2t)}{\frac{dx}{dt} (t^2+1)^2}$

sub in what you got for $\displaystyle \frac{dx}{dt}$ in the denominator and clean up the algebra.
• Dec 8th 2011, 08:46 AM
Stuck Man
Re: 1st and 2nd derivatives of parametric equations
Thats the easier way. I want to know why I have got the wrong result as I have done it.
• Dec 8th 2011, 09:36 AM
skeeter
Re: 1st and 2nd derivatives of parametric equations
Quote:

Originally Posted by Stuck Man
Thats the easier way. I want to know why I have got the wrong result as I have done it.

well, I hope someone has the patience and fortitude to find that error for you.
• Dec 8th 2011, 11:07 AM
Stuck Man
Re: 1st and 2nd derivatives of parametric equations
The mistake is in the second line. I tried to make a common denominator. I should have multiplied instead. The denominator should have a power of 6 as I did it.