Apps of differentiation

Hello, jarny!

Quote:

A right triangle is formed in the first quadrant by the x and y axes
and a line through the point (1,2).
a) Write the length L of the hypotenuse as a function of x.

Code:

| (0,b)* | * | * (1,2) | o | * | * - - + - - - - - * - - | (a,0)

The line with intercepts $(a,0),\:(0,b)$ has the equation: . $\frac{x}{a} + \frac{y}{b} \:=\:1$

Since $(1,\,2)$ is on this line: . $\frac{1}{a} + \frac{2}{b} \:=\:1\quad\Rightarrow\quad b \:=\:\frac{2a}{a-1}\;\;{\color{blue}[1]}$

The length of the hypotenuse is given by: . $L^2 \:=\:a^2+b^2\;\;{\color{blue}[2]}$

Substitute [1] into [2]: . $L^2 \;=\;a^2 + \left(\frac{2a}{a-1}\right)^2 \;=\;\frac{a^4 - 2a^3 + 5a^2}{(a-1)^2}$

Therefore: . $L \;=\;\frac{\sqrt{x^4-2x^2 + 5x^2}}{x-1}$