# Expanding isosceles triangle The legs of an isosceles right tri- angle increase in length at a rate of 2 m/s. a. At what rate is the area of the trian

Right triangles and trigonometry
Expanding isosceles triangle The legs of an isosceles right tri- angle increase in length at a rate of 2 m/s.
a. At what rate is the area of the triangle changing when the legs are 2 m long?
b. At what rate is the area of the triangle changing when the hypot- enuse is 1 m long?
c. At what rate is the length of the hypotenuse changing?

2020-12-14
Consider a isosceles right triangle ABC,

Consider AB = x is the length of a leg of a isosceles right triangle an AC = y is a hypotenuse. Given that, $$\displaystyle\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}={2}$$ m/s
a) We know that area of a right triangle $$\displaystyle=\frac{{1}}{{2}}\times$$ base $$\displaystyle\times$$ height
$$\displaystyle{A}=\frac{{1}}{{2}}{\left({x}\right)}{\left({x}\right)}$$
$$\displaystyle{A}=\frac{{1}}{{2}}{x}^{{2}}$$ (1)
Differentiate A with respect to t,
$$\displaystyle\frac{{{d}{A}}}{{{\left.{d}{t}\right.}}}=\frac{{d}}{{{\left.{d}{t}\right.}}}{\left(\frac{{1}}{{2}}{x}^{{2}}\right)}$$
$$\displaystyle\frac{{{d}{A}}}{{{\left.{d}{t}\right.}}}=\frac{{1}}{{2}}{\left({2}{x}\right)}\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}$$
$$\displaystyle\frac{{{d}{A}}}{{{\left.{d}{t}\right.}}}={x}\frac{{{\left.{d}{x}\right.}}}{{\left.{d}{t}\right.}}{\left({2}\right)}$$
$$\displaystyle\frac{{{d}{A}}}{{{\left.{d}{t}\right.}}}={\left({2}\right)}{\left({2}\right)}$$ [Given]
$$\displaystyle\frac{{{d}{A}}}{{{\left.{d}{t}\right.}}}={4}$$
Hence area of the triangle changing at $$\displaystyle{4}\frac{{m}^{{2}}}{{s}}$$.
b) In isosceles right triangle ABC,
$$\displaystyle{A}{C}^{{2}}={A}{B}^{{2}}+{B}{C}^{{2}}$$ [Using pythoras theorem]
$$\displaystyle{y}^{{2}}={x}^{{2}}+{x}^{{2}}$$
$$\displaystyle{y}^{{2}}={2}{x}^{{2}}$$
$$\displaystyle{1}={2}{x}^{{2}}$$ [Given that hypotenuse y=1m]
$$\displaystyle{x}^{{2}}=\frac{{1}}{{2}}$$
$$\displaystyle{x}=\frac{{1}}{\sqrt{{2}}}$$
Substtute $$\displaystyle{x}=\frac{{1}}{\sqrt{{2}}}{\quad\text{and}\quad}\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}={2}$$ in (2)
$$\displaystyle\frac{{{d}{A}}}{{{\left.{d}{t}\right.}}}={\left(\frac{{1}}{\sqrt{{2}}}\right)}{\left({2}\right)}$$
$$\displaystyle\frac{{{d}{A}}}{{{\left.{d}{t}\right.}}}=\sqrt{{2}}$$
Hence area of the triangle changing at sqrt2 m^2/sZSK.
c) In isosceles right triangle ABC,
$$\displaystyle{A}{C}^{{2}}={A}{B}^{{2}}+{B}{C}^{{2}}$$ [Using pythoras theorem]
$$\displaystyle{y}^{{2}}={x}^{{2}}+{x}^{{2}}$$
$$\displaystyle{y}^{{2}}={2}{x}^{{2}}$$
$$\displaystyle{y}=\sqrt{{2}}{x}$$
Differentiate y with respect to t,
$$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}=\frac{{d}}{{{\left.{d}{t}\right.}}}{\left(\sqrt{{2}}{x}\right)}$$
$$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}=\sqrt{{2}}\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}$$
$$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}=\sqrt{{2}}$$ (2)
$$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}={2}\sqrt{{2}}$$
Hence hypotenuse change at $$\displaystyle{2}\sqrt{{2}}$$ m/s