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The equation of astroid is $x^{2/3} + y^{2/3} = a^{2/3}$. Find the points where the slope of the tangent to the astroid is equal to $-1$.

I got the derivative to be $-y^{1/3}/x^{1/3}$ and so I set $-y^{1/3}=x^{1/3}$ and therefore $x^{2/3} - x^{2/3} = a^{2/3}$ but now what?

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Let $x=a\cos^3t,y=a\sin^3t$

$$\frac{dy}{dx}=\frac{dy}{dt}/\frac{dx}{dt}=-3\sin t\cos^2 t/3\sin^2 t\cos t=-\cot t$$

We need $$\cot t=-1$$

Method $\#1:$

$$\frac{\cos t}1=\frac{\sin t}{-1}=\pm\sqrt{\frac{\cos^2t+\sin^2t}{1^2+(-1)^2}}=\pm\frac1{\sqrt2}$$

Method $\#2:$

$$\cot t=-1\iff\tan t=-1=-\tan\frac\pi4=\tan\left(\pi-\frac\pi4\right)$$

$$\implies t=n\pi+\left(\pi-\frac\pi4\right)$$ where $n$ is any integer

$$\implies t=\pi+\left(\pi-\frac\pi4\right), \pi-\frac\pi4\pmod{2\pi}$$

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  • $\begingroup$ Why choose x=acos3t,y=asin3t? $\endgroup$ – user192422 Dec 2 '14 at 5:01
  • $\begingroup$ Because that allows you to use the property that $\cos ^2 + \sin ^2 =1$ $\endgroup$ – Alan Dec 2 '14 at 5:05
  • $\begingroup$ @user192422, This is the parametric form $\endgroup$ – lab bhattacharjee Dec 2 '14 at 5:06
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You made two sign errors that screwed you up: $dy/dx=-y^{1/3}/x^{1/3}=\bf{-1}$, so $x^{1/3}=y^{1/3}$ (thus $x=y$ in the reals). It should be simple from here, but watch your signs when you continue!

The other sign error you made, btw, was after your first one: if $x^{1/3}=-y^{1/3}$ (as if say the slope was 1) (so $x=-y$ in $\mathbb{R}$), then what is $x^{2/3}$?

lab bhattacharjee's solution is overcomplicated unless you want to do further work with equations of that form.

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