parametric equation

**ling_c_0202** · August 29th 2006, 01:02 AM

I don't know how to prove it...

(a) for parametric equations x = f(t) and y = g(t), where t is a parameter, show that
$ \frac{d^2y}{dx^2}\:= \frac{\frac{d}{dt}(\frac{dy}{dx})} {\frac{dx}{dt}}$

(b) Given the parametric equations

$ x = \frac{t^2}{2}\ and\ y = 1-t$

find $ \frac{d^2y}{dx^2}\$

**Glaysher** · August 29th 2006, 01:37 AM

(a)

$\frac{d}{dt}(\frac{dy}{dx}) = \frac{d^2y}{dx^2}\frac{dx}{dt}$

Divide by $\frac{dx}{dt}$ and you have shown that the right hand side is the same as the left hand side

(b)

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

Differentiate with respect to t and then divide by $\frac{dx}{dt}$

**ThePerfectHacker** · August 29th 2006, 05:39 AM

Originally Posted by ling_c_0202

(b) Given the parametric equations

$ x = \frac{t^2}{2}\ and\ y = 1-t$

find $ \frac{d^2y}{dx^2}\$

$\frac{dx}{dt}=t$ and $\frac{dy}{dt}=-1$
Thus,
$\frac{dy}{dx}=\frac{-1}{t}$
---
To find the second, take the derivative of this derivative you just found.
$\frac{d}{dt}(-1/t)=\frac{1}{t^2}$
And,
$\frac{dx}{dt}=t$
Thus,
$\frac{d^2y}{dx^2}=\frac{1}{t^3}$

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