Answer to Question #269645 in Calculus for Miza

Question

given 1/u+1/v=1/f with f as a constant. if f=10 cm and u decrease with the rate of 2 cm/second, find the rate of v when u=40 cm

Answer 1

Given

$$\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}$$

Differentiate both sides with respect to $t$

$$(\dfrac{1}{u}+\dfrac{1}{v})'=(\dfrac{1}{f})'$$

$$-\dfrac{u'}{u^2}-\dfrac{v'}{v^2}=0$$

Solve for $v'$

$$v'=-\dfrac{v^2}{u^2}u'$$

$$v=\dfrac{uf}{u-f}$$

Substitute

$$v'=-\dfrac{f^2}{(u-f)^2}u'$$

Given $f=10\ cm, u'=-2\ cm/s, u=40\ cm$

$$v'=-\dfrac{(10\ cm)^2}{(40\ cm-10\ cm)^2}(-2\ cm/s)=\dfrac{2}{9}\ cm/ s$$

$v$ increase with the rate of $\dfrac{2}{9}$ cm/second.