# parametric derivative

• Feb 4th 2008, 01:25 PM
parametric derivative
If x=t^3, y=2t^2 for t>0, then what does d^2y/dx^2 equal?
• Feb 4th 2008, 02:23 PM
bobak
$\displaystyle \frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}$
$\displaystyle \frac{d^2y}{dx^2} = \frac{d}{dt} \left ( \frac{dy}{dx} \right ) \cdot \frac{dt}{dx}$
• Feb 4th 2008, 02:23 PM
Plato
This is what the question is asking.
$\displaystyle \frac{{d^2 y}}{{dx^2 }} = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right) = \frac{{\frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right)}}{{\frac{{dx}}{{dt}}}}$
Step 1. Express $\displaystyle y' = \frac{{dy}}{{dx}}$ in terms of t.

Step 2. find $\displaystyle \frac{{dy'}}{{dt}}$, i.e. the derivative of step 1 with respect to t.

Step 3. divide $\displaystyle \frac{{dy'}}{{dt}}$ by $\displaystyle \frac{{dx}}{{dt}}$.
• Feb 4th 2008, 03:08 PM
So is this what I should do?
$\displaystyle y=\frac{{2x}}{{t}}$
Step 1:$\displaystyle y'=\frac{{2}}{{t}}$
Step 2: $\displaystyle \frac{{dy'}}{{dt}}=\frac{{-2}}{{t^2}}$
Step 3: $\displaystyle \frac{{dx}}{{dt}}=3t^2$
so the answer would be $\displaystyle \frac{{d^2y}}{{dx^2}}=\frac{{-2}}{{3t^4}}$.
Is that right?
• Feb 4th 2008, 03:31 PM
bobak
Quote:

So is this what I should do?
$\displaystyle y=\frac{{2x}}{{t}}$
Step 1:$\displaystyle y'=\frac{{2}}{{t}}$ No, this is incorrect, you cannot treat t as a constant when you know it is a variable
Step 2: $\displaystyle \frac{{dy'}}{{dt}}=\frac{{-2}}{{t^2}}$
Step 3: $\displaystyle \frac{{dx}}{{dt}}=3t^2$
so the answer would be $\displaystyle \frac{{d^2y}}{{dx^2}}=\frac{{-2}}{{3t^4}}$.
Is that right?

first find $\displaystyle \frac{dx}{dt}$ and $\displaystyle \frac{dy}{dt}$

you should get $\displaystyle \frac{dx}{dt} = 3t^2$ and $\displaystyle \frac{dy}{dt} = 4t$

then you use $\displaystyle \frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}$

giving $\displaystyle \frac{dy}{dx} = 4t \cdot \frac{1}{3t^2}$

want to finish this off ?
• Feb 4th 2008, 03:51 PM
Okay, so $\displaystyle \frac{dy}{dx} = \frac{4}{3t}$
Step 2: $\displaystyle \frac{{dy'}}{{dt}}=\frac{{-4}}{{3t^2}}$
Step 3: $\displaystyle \frac{{dx}}{{dt}}=3t^2$
so the answer would be $\displaystyle \frac{{d^2y}}{{dx^2}}=\frac{{-4}}{{9t^4}}$.
Is that right?
• Feb 5th 2008, 12:24 AM
bobak
Yeah that looks good to me.
• Feb 5th 2008, 12:36 AM
mr fantastic
Quote:

Okay, so $\displaystyle \frac{dy}{dx} = \frac{4}{3t}$
Step 2: $\displaystyle \frac{{dy'}}{{dt}}=\frac{{-4}}{{3t^2}}$
Step 3: $\displaystyle \frac{{dx}}{{dt}}=3t^2$
so the answer would be $\displaystyle \frac{{d^2y}}{{dx^2}}=\frac{{-4}}{{9t^4}}$.