Given:
The edge length of a cube is changing at a rate of 10 in/sec.
To find:
The rate by which cube's volume changing when the edge length is 3 inches.
Solution:
We have,
[tex]\dfrac{da}{dt}=10\text{ in/sec}[/tex]
We know that, volume of cube is
[tex]V=a^3[/tex]
Differentiate with respect to t.
[tex]\dfrac{dV}{dt}=3a^2\dfrac{da}{dt}[/tex]
Substituting [tex]\dfrac{da}{dt}=10[/tex] and a=3, we get
[tex]\dfrac{dV}{dt}_{a=3}=3(3)^2(10)[/tex]
[tex]\dfrac{dV}{dt}_{a=3}=3(9)(10)[/tex]
[tex]\dfrac{dV}{dt}_{a=3}=270[/tex]
Therefore, the volume increased by 270 cubic inches per sec.
