Length of the curve - parametric

[Poetria]

Homework Statement

[/B]
Find the definite integral formula for the length of the curve for
$$0 \leq t \leq \frac \pi 2$$

$$x = 2*cos^k(t)$$
$$y = 2*sin^k(t)$$

for general $$k \gt 0$$2. The attempt at a solution

I don't understand why this is wrong:

$$\int_0^\frac \pi 2\ \sqrt{{(2*k*cos(t)*(-sin(t))}^2+{(2*k*sin(t)*cos(t)})^2} dt$$

Not in LaTex - sqrt((2*k*cos(t)*(-sin(t)))^2+(2*k*sin(t)*cos(t))^2)

An arc length, parametric form. I have taken derivatives of course. :(

 

[Orodruin]

How did you arrive at that expression? Please show us your attempt, not just the end result!

 

[Poetria]

How did you arrive at that expression? Please show us your attempt, not just the end result!

Of course, but I am very slow with LaTex. Just a moment.

 

[Poetria]

Of course, but I am very slow with LaTex. Just a moment.

I took a formula:

$$ds = \sqrt { (\frac {dx} {dt})^2 + ({\frac {dy} {dt}})^2}$$

and put the derivatives with respect to t. I used the chain rule of course.

 

[Orodruin]

I took a formula:

$$ds = \sqrt { \frac {dx} {dt}^2 + {\frac {dy} {dt}}^2}$$

and put the derivatives with respect to t. I used the chain rule of course.

This still does not show your work. If you do this correctly you will not get what you showed in the original post.

 

[Poetria]

This still does not show your work. If you do this correctly you will not get what you showed in the original post.

Well, I have discovered a mistake (mismatched brackets but it was correct in the version for Wolfram Alpha) but I guess it is not what you mean.

 

[Poetria]

Ah ok I think I got it.

 

[Poetria]

Many thanks. So silly a mistake!