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I was given the following task:

We have an astroid $x=acos^3(t), y=asin^3(t)$ and two points on it $A(a,0), B(0,a)$. We need to find point $M$, which belongs to the arc $AB$, and the length of arc $AM$ should be equal to $\frac{1}{4}$ length of arc $AB$.

As I understood I need to find such $x_0, y_0$ for point $M(x_0, y_0)$ to meet all the conditions. From the task we have $x(t), y(t)$, which means that we need to find the proper $t_0$ for $x(t_0)=x_0, y(t_0)=y_0$. Let the length of $AB$ equals $l$ and $\frac{1}{4}l=l_0$, then as follows from my definite-integrals course:

$$l=\int_0 ^\frac{\pi}{2}\sqrt{[-3acos^2(t)sin(t)]^2+[3asin^2(t)cos(t)]^2}dt=$$$$3a\int_0 ^\frac{\pi}{2}\sqrt{cos^2(t)sin^2(t)[cos^2(t)+sin^2(t)]}dt=$$$$3a\int_0^\frac{\pi}{2}cos(t)sin (t)dt$$

Here I make a substitution $u=sin(t)$, then $du=cos(t)dt, u_1=sin(0)=0, u_2=sin(\frac{\pi}{2})=1$, so now we have:

$$3a\int_0 ^1 udu=\frac{3au^2}{2} \Biggr|_0^1=\frac{3a}{2}$$

First question. Is the integrand solved correctly?

Second question. What should I do next?

P.S. Right now we are dealing with definite-integrals on our math course, so it is supposed that I use definite-integral in this task.

blitzar787
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  • I think you want your upper limit of integration to be $\frac{\pi}{2}$ instead of a in finding the length of AB. – user84413 May 05 '15 at 19:59
  • Why? This is just parametric graph in Cartesian coordinate system, and to find AB, I need to integrate all the way from $x_1$ to $x_2$, isn't it? Could you explain why I need to stop at $\frac{\pi}{2}?$ – blitzar787 May 05 '15 at 20:06
  • @user84413 thankx, I think I understood what is wrong. I take limits of integration from x axis but I integrate dt, so my limits should be angulars. Thats why I need to integrate to $\frac{pi}{2}$. But it is still hard for me to make it so clear that I can crack it. – blitzar787 May 05 '15 at 20:28

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Hint:

Set $\displaystyle 3a\int_0^{t_1}\sin t\cos t\;dt=\frac{1}{4}\cdot 3a\int_0^{\frac{\pi}{2}}\sin t\cos t\;dt$, and then solve for $t_1$.

user84413
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