Question

The Altitude of a triangle is increasing at a rate of 8cm/s while its area is increasing at the rate of 12cm^2/s. At what rate is the base of the triangle changing when the altitude is 20 cm and the area is 100 cm^2 ?

Answer 1

Given;

$\\\frac{da}{dt}=8cm/min\\\frac{dA}{dt}={12}cm^2/min$ with a=altitude, A=area, t=time

The question requires us to find $\\\frac{db}{dt}$ when a=20cm and A=100cm².

$A=\frac{1}{2}ba\\\frac{d}{dt}(A)=\frac{d}{dt}(\frac{1}{2}ba)\\\frac{dA}{dt}=\frac{db}{dt}a+\frac{1}{2}b\frac{da}{dt}$

Using the formula for area, we can tell that b=10cm.

Substituting ;

$12=\frac{1}{2}\frac{db}{dt}20+\frac{1}{2}(10)(8)\\\frac{db}{dt}=-2.8cm/min$

The base is changing at a rate of -2.8cm/min