Differentiation Problem

**fdylan** · March 28th 2010, 11:18 AM

So here's the problem:

A curve is represented parametrically by x = (t²-1)², y = t³. Find dy/dx in terms of t and show that
$ \frac {d^2 y}{dx^2} = \frac {-3(t^2 +1)}{16t(t^2- 1)^3} $

I tried to work it out, and I got

dy/dx = $ \frac{3t^2}{4t^3-4t} $ (not sure if it's correct). And about the second derivative, it got too complicated.

Thanks for any help.

**e^(i*pi)** · March 28th 2010, 11:58 AM

Originally Posted by fdylan

First of all I'm not sure whether it's in the correct Section.

So here's the problem:

A curve is represented parametrically by x = (t²-1)², y = t³. Find dy/dx in terms of t and show that
$ \frac {d^2 y}{dx^2} = \frac {-3(t^2 +1)}{16t(t^2- 1)^3} $

I tried to work it out, and I got

dy/dx = $ \frac{3t^2}{4t^3-4t} $ (not sure if it's correct). And about the second derivative, it got too complicated.

Thanks for any help.

$x = (t^2-1)^2 \: \rightarrow \: \frac{dx}{dt} = 4t(t^2-1) = 4t^3-4t$

$\frac{dt}{dx} = \frac{1}{\frac{dx}{dt}} = \frac{1}{4t(t^2-1)}$ (this will be used later in the chain rule)

$y = t^3 \: \rightarrow \: \frac{dy}{dt} = 3t^2$

The chain rule says that $\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}$

$\frac{dy}{dx} = \frac{3t^2}{2t(t^2-1)} = \frac{3t}{4t^2-4}$ I have expanded the brackets to make the next step easier

No problem with your original answer although you can cancel out a t.

For the second part use the quotient rule.

$u = 3t \: \rightarrow u' = 3$

$v = 4t^2-4 \: \rightarrow v' = 8t$

**fdylan** · March 28th 2010, 12:07 PM

Many Thanks. Cutting out 't' makes it a lot easier I guess.

**e^(i*pi)** · March 28th 2010, 12:11 PM

Originally Posted by fdylan

for the x part, I worked it out different.
$ x = (t^2 -1)^2 x = (t^2 -1)(t^2 -1) x = t^4 -2t^2 +1 \frac{dx}{dt} = 4t^3 -4t \frac{dy}{dx} = \frac{3t^2}{4t^3 -4t} $

It still works, I just used the chain rule.

$\frac{3t^2}{4t^3-4t} = \frac{3t^2}{4t(t^2-1)} = \frac{3t}{4(t^2-1)}$

t cancels

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