Thread: parametric equation

1. parametric equation

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

(b) Given the parametric equations

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

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

2. (a)

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

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

(b)

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

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

3. Originally Posted by ling_c_0202
(b) Given the parametric equations

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

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