parametric derivative

**chaddy** · February 4th 2008, 01:25 PM

If x=t^3, y=2t^2 for t>0, then what does d^2y/dx^2 equal?

**bobak** · February 4th 2008, 02:23 PM

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

**Plato** · February 4th 2008, 02:23 PM

This is what the question is asking.
$\frac{{d^2 y}}{{dx^2 }} = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right) = \frac{{\frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right)}}{{\frac{{dx}}{{dt}}}}$
Step 1. Express $y' = \frac{{dy}}{{dx}}$ in terms of t.

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

Step 3. divide $\frac{{dy'}}{{dt}}$ by $\frac{{dx}}{{dt}}$ .

**chaddy** · February 4th 2008, 03:08 PM

So is this what I should do?
$y=\frac{{2x}}{{t}} $
Step 1: $ y'=\frac{{2}}{{t}} $
Step 2: $ \frac{{dy'}}{{dt}}=\frac{{-2}}{{t^2}} $
Step 3: $ \frac{{dx}}{{dt}}=3t^2 $
so the answer would be $\frac{{d^2y}}{{dx^2}}=\frac{{-2}}{{3t^4}} $ .
Is that right?

**bobak** · February 4th 2008, 03:31 PM

Originally Posted by chaddy

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

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

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

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

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

want to finish this off ?

**chaddy** · February 4th 2008, 03:51 PM

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

**bobak** · February 5th 2008, 12:24 AM

Yeah that looks good to me.

**mr fantastic** · February 5th 2008, 12:36 AM

Originally Posted by chaddy

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

By the way .... since y = 2t^2 you can express the result as a function of y ......

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