Problem statement: Show that every tangent line to the curve $x^{2/3} + y^{2/3} = 1$ in the first quadrant has the property that portion of the line in the first quadrant has length 1.
The textbook says to use impicit differentiation. Here's what I've done:
$$ x^{2/3} + y^{2/3} = 1 $$
$$ (x^{2/3})' + (y^{2/3})' = (1)' $$
$$ \frac{2}{3x^{1/3}} + \frac{2}{3y^{1/3}} y'= 0 $$
$$ y'= -(\frac{y}{x})^{1/3} $$
Could someone explain to me the next steps, please? Any help is appreciated.
On
Starting from "the other end", we may also consider that a line which has intercepts $ \ (X \ , \ 0) \ $ and $ \ (0 \ , \ Y) \ \ , \ X , Y \ > \ 0 \ , $ and a segment within the first quadrant of length $ \ 1 \ $ has the slope $ \ -\frac{Y}{X} \ = \ -\frac{\sqrt{1 - X^2}}{X} \ \ , $ since $ \ X^2 + Y^2 \ = \ 1 \ \ . $
As you and Ankit Saha have noted, the slope of the tangent line to the curve at $ \ (x_0 \ , \ y_0) \ $ is $ \ \large{- \frac{y_0^{1/3}}{x_0^{1/3}}} \ $ , which is thus $$ m \ \ = \ \ - \frac{(1 - x_0^{2/3})^{1/2}}{x_0^{1/3}} \ \ , $$
by applying the (unit) hypocycloid equation. The tangent line equation can then be expressed as $$ y \ - \ y_0 \ \ = \ \ - \frac{(1 - x_0^{2/3})^{1/2}}{x_0^{1/3}} · (x \ - \ x_0 ) \ \ . $$ From this, we obtain the relation between its intercepts $$ Y \ - \ {(1 - x_0^{2/3})^{3/2}} \ \ = \ \ - \frac{(1 - x_0^{2/3})^{1/2}}{x_0^{1/3}} · (0 \ - \ x_0 ) \ \ \ , $$ $$ 0 \ - \ {(1 - x_0^{2/3})^{3/2}} \ \ = \ \ - \frac{(1 - x_0^{2/3})^{1/2}}{x_0^{1/3}} · (X \ - \ x_0 ) $$ $$ \Rightarrow \ \ Y \ \ = \ \ (1 - x_0^{2/3})^{3/2} \ + \ (1 - x_0^{2/3})^{1/2} · x_0^{2/3} \ \ = \ \ (1 - x_0^{2/3})^{1/2} · 1 \ \ \ , $$ $$ \frac{(1 - x_0^{2/3})^{1/2}}{x_0^{1/3}} · X \ \ = \ \ (1 - x_0^{2/3})^{3/2} \ + \ (1 - x_0^{2/3})^{1/2} · x_0^{2/3} \ \ = \ \ Y \ \ . $$ Since $$ Y^2 \ = \ 1 - x_0^{2/3} \ \ = \ \ \frac{ 1 - x_0^{2/3}} {x_0^{2/3}} · X^2 \ \ \Rightarrow \ \ X^2 \ \ = \ \ x_0^{2/3} \ \ , $$ we may conclude that $ X^2 + Y^2 \ = \ 1 \ \ , $ and thus that the segment of any tangent line to the hypocycloid (also called an "astroid") within the first quadrant has the required slope and unit length.
[In fact, since the curve equation has "symmetry about the origin", the full curve appears in all four quadrants. With proper handling of signs, we may state the same proposition for any tangent line to the complete curve within any quadrant.
This problem is connected with the "sliding ladder" problem, in which a segment of fixed length is moved so that its endpoints stay on the coordinate axes: the set of all the segments "bounds" a hypocycloid. This application of Plücker "line coordinates" is discussed, for instance, in Chapter 10 of Eli Maor's The Pythagorean Theorem: A 4000-Year History.]
Consider the tangent to the curve at $(x_0, y_0)$. As you have found out, its slope is given by $-\left(\frac{y_0}{x_0}\right)^\frac13$
Thus, the equation of line is $$ y - y_0 = -\left(\frac{y_0}{x_0}\right)^\frac13 (x-x_0)$$
The $x$ and $y$ intercepts can be obtained by substituting $y = 0$ and $x = 0$ respectively $$ \left(x_0 + x_0^\frac13 y_0^\frac23, 0 \right) $$ $$ \left(0, y_0 + x_0^\frac23 y_0^\frac13 \right)$$
The portion of the tangent in the first quadrant is equal to the distance between these two points, whose square is given by $$ \begin{align} &\left(x_0 + x_0^\frac13 y_0^\frac23 \right)^2 + \left(y_0 + x_0^\frac23 y_0^\frac13 \right)^2 \\ &= x_0^2 + 2x_0^\frac43 y_0^\frac23 + x_0^\frac23 y_0^\frac43 + y_0^2 + 2x_0^\frac23 y_0^\frac43 + x_0^\frac43 y_0^\frac23 \\ &= x_0^2 + 3x_0^\frac43 y_0^\frac23 + 3x_0^\frac23 y_0^\frac43 + y_0^2 \\ &= \left(x_0^\frac23 + y_0^\frac23 \right)^2 \\ &= 1 \end{align}$$ because $(x_0, y_0)$ lies on the curve $x^\frac23 + y^\frac23 = 1$