1. ## 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

and

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.

2. 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

and

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.

3. 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

4. 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.

5. Yeah so just confused on how they got m to equal -2 in first answer

6. 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

7. Tried following the link you gave me but still coming up with the samething

8. Originally Posted by xpack
Tried following the link you gave me but still coming up with the samething
me too

I'm going to post the problem in my book, which is almost identical to yours. The example that is on the page show that there is more than one solution for t. This isn't the case in our problems though. So I will post mine in a new thread.

9. I understand now, thank you very much you've been great help