# Thread: Parametric Arc Length and Tangent Equations

1. ## Parametric Arc Length and Tangent Equations

Find the exact length of the curve:

x=5+15t^2
y=6+2t^3
0, greater than or = to, t, greater than or = to, 4

The other problem is:

Find equations of tangents to x=3t^2+8, y=2t^3+8, that passes through (11,0).

I had NO idea how to do those. I did a couple steps of each and got to something that I'm unsure of. Any help would be greatly appreciated!

2. Originally Posted by thegame189
Find the exact length of the curve:

x=5+15t^2
y=6+2t^3
0, greater than or = to, t, greater than or = to, 4

The other problem is:

Find equations of tangents to x=3t^2+8, y=2t^3+8, that passes through (11,0).

I had NO idea how to do those. I did a couple steps of each and got to something that I'm unsure of. Any help would be greatly appreciated!
The Formula for arc length is $\displaystyle s=\int_{t_0}^{t_1}\sqrt{\left(\frac{dx}{dt}\right) +\left(\frac{dy}{dt}\right)}dt=\int_0^4\sqrt{(30t) ^2+(6t)^2}dt=6\sqrt{26}\int_0^4tdt=$

$\displaystyle 3\sqrt{26}t^2|_0^4=48\sqrt{26}$

For the second...

$\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{6t}{6t}=1$

So the slope of the function is m=1. You should be able to finish from here.

3. The first problem is y=6+2t^3, not squared.
You end up with (30t)^2 + (6t^2)^2 under the radical. How do I do that?

4. ## Whoops

Okay lets try again.

$\displaystyle s=\int_{t_0}^{t_1}\sqrt{\left(\frac{dx}{dt}\right) +\left(\frac{dy}{dt}\right)}dt=\int_0^4\sqrt{(30t) ^2+(6t^2)^2}dt=6\int_0^4t\sqrt{25+t^2}dt=$

so let $\displaystyle u=25+t^2$ and $\displaystyle du=2tdt$

so we get...

$\displaystyle 3\int_{25}^{41}\sqrt{u}du=2u^{3/2}|_{25}^{41}=2(41\sqrt{41}-125)$