1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I think you want your upper limit of integration to be $\frac{\pi}{2}$ instead of a in finding the length of AB. $\endgroup$ – user84413 May 5 '15 at 19:59
  • $\begingroup$ 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 $\frac{\pi}{2}?$ $\endgroup$ – blitzar787 May 5 '15 at 20:06
  • $\begingroup$ @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. $\endgroup$ – blitzar787 May 5 '15 at 20:28
2
$\begingroup$

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 $t_1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.