# Parametric Curves

• Oct 19th 2009, 09:46 AM
xpack
Parametric Curves
Find the equations of the tangents to the given curve that pass through the point (12, 10).
x = 6t^2 + 6
y = 4t^3 + 6

I'm following some example but I get confused after

http://www.webassign.net/cgi-bin/sym...-2x%20%2B%2034
and
http://www.webassign.net/cgi-bin/sym...expr=x%20-%202

I understand that m=1 and that makes x-2. But I dont understand is that at one point it looks like m = -2 which is seen in -2x+34 but I dont see how they got m = -2 can someone explain this please.
• Oct 19th 2009, 11:19 AM
Quote:

Originally Posted by xpack
Find the equations of the tangents to the given curve that pass through the point (12, 10).
x = 6t^2 + 6
y = 4t^3 + 6

I'm following some example but I get confused after

http://www.webassign.net/cgi-bin/sym...-2x%20%2B%2034
and
http://www.webassign.net/cgi-bin/sym...expr=x%20-%202

I understand that m=1 and that makes x-2. But I dont understand is that at one point it looks like m = -2 which is seen in -2x+34 but I dont see how they got m = -2 can someone explain this please.

You know how to find the derivative of a parametric curve right?

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

In your situation this is just:

$\frac{dy}{dt}=t$

Now you just solve the parametric equations for t at that point:

$x=12=6t^2+6$
$y=10=4t^3+6$

Find the t, and that will be the slope of the tangent line.
• Oct 19th 2009, 01:12 PM
xpack
Yeah but in both equations it +1 -1 but in order to the get the answer -2x+34 the slope or m would have to be -2 and i dont know how to get that
• Oct 19th 2009, 01:31 PM
Quote:

Originally Posted by xpack
Yeah but in both equations it +1 -1 but in order to the get the answer -2x+34 the slope or m would have to be -2 and i dont know how to get that

The solution to $12=6t^2+6$ is $\pm1$. The solution to $10=4t^3+6$ is $t^3=1$, therefore $t=1$. The only commons solution between them is 1. The value of $(-1)^3$ is -1, so that can't be a solution for the second equation.

EDIT:

Oh, wait, I'm sorry. I didn't realize you had already figured that out. Somehow I missed the last paragraph in your original post.
• Oct 19th 2009, 01:37 PM
xpack
Yeah so just confused on how they got m to equal -2 in first answer
• Oct 19th 2009, 02:00 PM
I'm kind of with you now. It's not obvious to me why there are two tangents at that point.

I have a very similar problem in my calculus book.

- Find equations of the tangents to the curve $x=3t^2+1$ and $y=2t^3+1$ that pass through the point (4,3).

In this problem I can only come up with y=x-1. So I'm confused now too. It makes sense that parametric curves would have two tangents at certain points since they tend to form loops. I just found a resource that might shed some light on how these work:

If you scroll down to example 1, they discuss this.

Pauls Online Notes : Calculus II - Tangents with Parametric Equations
• Oct 19th 2009, 02:14 PM
xpack
Tried following the link you gave me but still coming up with the samething
• Oct 19th 2009, 02:22 PM