Question

Calculate the length of the path over the given interval. x(t)=8t^(2), y(t)=6t^(2)-1. 0 less or equal t more or equal than 4

Answer 1

parametric equation of the path is x(t) = 8t^2; y(t) = 6t^2 -1 ; 0<= t<= 4

Answer 2

It's a webwork question. I'm not sure what you mean. Yeah that is the parametric equation

Answer 3

I think it's arc length.

Answer 4

oh, I see, it's arc length from cal2

Answer 5

yeah with integration.

Answer 6

@zepdrix

Answer 7

grr i can't remember D: Lemme think about it.

Answer 8

hahahaa.... relax....

Answer 9

So the length of the path, arc length, will be,\[\large L=\int\limits ds=\int\limits \sqrt{1+\left(\frac{dy}{dx}\right)^2}dx\]

Oh but we're dealing with parametric equations, so the formula we want to use is,\[\large L=\int\limits\limits \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dx}{dt}\right)^2} dt\]

Answer 10

The limits on our integral will be the interval of t,\[\large L=\int\limits_0^4 \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\]

Answer 11

So this shouldn't be too bad, we just need to find a couple of derivatives, then integrate.
Can you do the first one?

\(\large x=8t^2\)

\(\large \dfrac{dx}{dt}=?\)

Answer 12

is that 16t?

Answer 13

\[\large L=\int\limits\limits\limits_0^4 \sqrt{\left(16t\right)^2+\left(\frac{dy}{dt}\right)^2}dt\]Ok good.

Answer 14

Understand what I did? I skipped a step there.
I plugged the 16t in for our dx/dt.

Answer 15

yeah

Answer 16

cool, what do you get for the other derivative? c:

Answer 17

12t

Answer 18

\[\large L=\int\limits\limits\limits\limits_0^4 \sqrt{\left(16t\right)^2+\left(12t\right)^2}dt\]

Ok good. So we've successfully setup our integral using the derivative pieces that we needed.

Can you solve it from here? Just requires a little simplification, and an easy integral.

Answer 19

Yeah. I will try. I will post if I have any problem. Thanks for the help.