# Thread: Expressing a Curve in r=r(t) form on an interval I

1. ## Expressing a Curve in r=r(t) form on an interval I

The curve is given by y=the integral from 1 to x sqrt[sqrt(t) -1 dt, 1 <=x<=16
(basically the sqrt sign is over t-1 and the t also has a sqrt over it)
It asks to express the curve in the form r=r(t) on an Interval I and to find the length of the curve and is the curve smooth
*I don't understand by which the curve is given...I'm totally lost

2. Originally Posted by latavee
The curve is given by y=the integral from 1 to x sqrt[sqrt(t) -1 dt, 1 <=x<=16
(basically the sqrt sign is over t-1 and the t also has a sqrt over it)
It asks to express the curve in the form r=r(t) on an Interval I and to find the length of the curve and is the curve smooth
*I don't understand by which the curve is given...I'm totally lost
is this what you mean?

$\displaystyle y = \int_1^x \sqrt{\sqrt{t} - 1} \, dt$ ; $\displaystyle 1 \le x \le 16$

3. ## Yes!

Yes that's exactly it! Thanks! I was just trying to figure out how to use the symbols

4. integrate using the substitution $\displaystyle u = \sqrt{t} - 1$ to find $\displaystyle r(t)$

you know how to find arc length?

5. I know the formula to find the arc length sqrt [1+f'(x)^2]- so do I find the derivative of the function? and where does the [1 to x] factor in?

6. Originally Posted by latavee
I know the formula to find the arc length sqrt [1+f'(x)^2]- so do I find the derivative of the function? and where does the [1 to x] factor in?
familiar with the Fundamental theorem of Calculus ?

$\displaystyle \frac{d}{dx} \left[\int_a^x f(t) \, dt \right] = f(x)$

7. Ok, so the derivative would be:

y'=sqrt[sqrt(x)-1 corect?

8. Originally Posted by latavee
Ok, so the derivative would be:

y'=sqrt[sqrt(x)-1 corect?
$\displaystyle y' = \sqrt{\sqrt{x}-1}$ ... correct.

9. So how would I express that in r(t) form?

10. as stated in my first response ...

Originally Posted by skeeter
integrate using the substitution $\displaystyle u = \sqrt{t} - 1$ to find $\displaystyle r(t)$

11. Ok Thanks!