Question

The altitude of a triangle is increasing at a rate of 1 cm/min while the area of the triangle is increasing at a rate of 2cm2 /min. at what rate is the base of the triangle changing when the altitude is 10 cm and the area is 100 cm2 ?

Answer 1

We have the area of the triangle $A(t)= \frac{bh}{2}$.

$b=\frac{2A}{h}$

Differentiating with respect to time t,

$\frac{db}{dt}=\frac{d}{dt}(\frac{2A}{h} )$

$\frac{db}{dt}=(\frac{2}{h} )\frac{dA}{dt}-(\frac{2A}{h^2} )\frac{dh}{dt}$

Substituting numerical values, the rate of change of base length is

$\frac{db}{dt}=(\frac{2}{10} )2-(\frac{200}{10^2} )1\\ \frac{db}{dt}=-1.6 cm/min$