# Parametric Differentiation

• Feb 16th 2011, 12:24 PM
alibond07
Parametric Differentiation
The diagram (attached) shows a sketch of the curve with parametric equations 'x=t^3', 'y=3t^3 along with the normal at the point with parameter t=2.
Attachment 20830

Find the equation of this normal and the coordinates of this point where it cuts the curve again.

I found dy/dx to be 6t/(3t^2-12), and am stumped on where to go from here. Clearly the normal at t=2 has something to do with this but I can't presently see how. Any ideas guys?

Thanks for any help
• Feb 16th 2011, 12:28 PM
TheEmptySet
Quote:

Originally Posted by alibond07
The diagram (attached) shows a sketch of the curve with parametric equations 'x=t^3', 'y=3t^3 along with the normal at the point with parameter t=2.
Attachment 20830

Find the equation of this normal and the coordinates of this point where it cuts the curve again.

I found dy/dx to be 6t/(3t^2-12), and am stumped on where to go from here. Clearly the normal at t=2 has something to do with this but I can't presently see how. Any ideas guys?

Thanks for any help

The equation you gave us does not match the graph

$x=t^3,y=3t^3 \implies y=3(x)$ which is a line!
• Feb 16th 2011, 12:33 PM
alibond07
Sorry the pair of equations is ' x=t^3-12t' , 'y=3t^2'
• Feb 16th 2011, 12:58 PM
alibond07
Quote:

Originally Posted by TheEmptySet
The equation you gave us does not match the graph

$x=t^3,y=3t^3 \implies y=3(x)$ which is a line!

Sorry the pair of equations is ' x=t^3-12t' , 'y=3t^2'
• Feb 16th 2011, 01:03 PM
TheEmptySet
Quote:

Originally Posted by alibond07
Sorry the pair of equations is ' x=t^3-12t' , 'y=3t^2'

Okay (just FYI your normal is at the wrong point and normal mean perpendicular to the surface).

Your Idea is sound first you need to take the derivative

$\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{2t}{t^2-4}$

So the slope of the normals line is perpendicular to the tangent this gives
$\displaystyle m_{\perp}=\frac{t^2-4}{2t}\bigg|_{t=2}=0$

So the equation of the normal line is horizontal and since $y(2)=3(2)^2=12$ So the equation of the normal line $y_{N}=12$

Now we just need to find another value of $t$ such that $y(t)=12 \implies 3t^2=12$

Can you finish from here?
• Feb 16th 2011, 01:17 PM
alibond07
Quote:

Originally Posted by TheEmptySet
Okay (just FYI your normal is at the wrong point and normal mean perpendicular to the surface).

Your Idea is sound first you need to take the derivative

$\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{2t^2}{t^2-4}$

So the slope of the normals line is perpendicular to the tangent this gives
$\displaystyle m_{\perp}=\frac{t^2-4}{2t^2}\bigg|_{t=2}=0$

So the equation of the normal line is horizontal and since $y(2)=3(2)^2=12$ So the equation of the normal line $y_{N}=12$

Now we just need to find another value of $t$ such that $y(t)=12 \implies 3t^2=12$

Can you finish from here?

Okay thanks, how come out results for $\displaystyle \frac{dy}{dx}$ are different?

I don't think I'm getting you sadly.
• Feb 16th 2011, 01:31 PM
TheEmptySet
Quote:

Originally Posted by alibond07
Okay thanks, how come out results for $\displaystyle \frac{dy}{dx}$ are different?

I don't think I'm getting you sadly.

They are not, you need to simplify your answer. e.g factor out a 3 and reduce. Please be more specific about what you don't understand.
• Feb 16th 2011, 01:40 PM
alibond07
Quote:

Originally Posted by TheEmptySet
They are not, you need to simplify your answer. e.g factor out a 3 and reduce. Please be more specific about what you don't understand.

I'm confused about how you got from (6t)/(3t^2-12) to (2t^2/ t^2 -4).
• Feb 16th 2011, 02:23 PM
TheEmptySet
Quote:

Originally Posted by alibond07
I'm confused about how you got from (6t)/(3t^2-12) to (2t^2/ t^2 -4).

$\displaystyle \frac{6t}{3t^2-12}=\frac{3\cdot 2t}{3(t^2-4)}=\frac{\not{3}\cdot 2t}{\not{3}(t^2-4)}=\frac{ 2t}{t^2-4}$

There is a typo this is correct.