Hypocycloid. Oh so fun!

**Columbo** · October 2nd 2008, 07:30 PM

At the risk of asking too many questions on my first day.

I need some more help.

How do I "Show that every tangent line to the curve x^(2/3) +y^(2/3)=1 in the first quadrant has the property that the portion of the line in the first quadrant has length 1" using implicit differentiation?

Thanks bunches for any help.

**Soroban** · October 2nd 2008, 08:06 PM

Hello, Columbo!

Show that every tangent line to the curve $x^{\frac{2}{3}} +y^{\frac{2}{3}} \;=\;1$
in the first quadrant has the property that the portion of the line
in the first quadrant has length 1, using implicit differentiation.

Find the derivative . . .

. . $\frac{2}{3}x^{-\frac{1}{3}} + \frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx} \;=\;1 \quad\Rightarrow\quad \frac{dy}{dx} \;=\;-\frac{y^{\frac{1}{3}}}{x^{\frac{1}{3}}}$

Let $A(p,q)$ be a point on the astroid in Quadrant 1.
. . Note that: . $p^{\frac{2}{3}} + q^{\frac{2}{3}} \:=\:1$

The slope of the tangent at $A$ is: . $m \;=\;-\frac{q^{\frac{1}{3}}}{p^{\frac{1}{3}}}$

The equation of the line through $A$ with slope $m$ is:

. . $y -q \;=\;-\frac{q^{\frac{1}{3}}}{p^{\frac{1}{3}}}(x - p) \quad\Rightarrow\quad y \;=\;-\frac{q^{\frac{1}{3}}}{o^{\frac{1}{3}}}x + p^{\frac{2}{3}}q^{\frac{1}{3}} + q$

. . $y \;=\;-\frac{q^{\frac{1}{3}}}{p^{\frac{1}{3}}}x + q^{\frac{1}{3}}\underbrace{\left(p^{\frac{2}{3}} + q^{\frac{2}{3}}\right)}_{\text{This is 1}}$ . $\Rightarrow\quad\boxed{ y \;=\;-\frac{q^{\frac{1}{3}}}{p^{\frac{1}{3}}}x + q^{\frac{1}{3}}}$

This tangent has: .x-intercept $\left(p^{\frac{1}{3}},\:0\right)$ . . . and y-intercept $\left(0,\:q^{\frac{1}{3}}\right)$

The length of that segment is: . $L \;=\;\sqrt{\left(p^{\frac{1}{3}}-0\right)^2 + \left(0-q^{\frac{1}{3}}\right)^2}$

. . $\text{and we have: }\;L \;=\;\sqrt{\underbrace{p^{\frac{2}{3}} + q^{\frac{2}{3}}}_{\text{This is 1}}} \;=\;\sqrt{1} \;=\;1 \quad\hdots\quad \text{ There!}$

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