all edges of a cube are expanding at a rate of 3 centimeters per second. How fast is the volume changing when each edge is 10 centimeters?
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Answer:
Volume is changing by [tex]900 \mathrm{cm}^{3} / \mathrm{sec}[/tex]
Solution:
As per the problem, all edges of the cube are expanding at a rate of [tex]3 \mathrm{cm} / \mathrm{sec}[/tex]
So,[tex]\left(\frac{d s}{d t}\right)[/tex] = [tex]3 \mathrm{cm} / \mathrm{sec}[/tex]
We also know that the volume [tex]V=s^{3}[/tex]----- (i)
Differentiating the volume from equation (i) we get,
[tex]\frac{d v}{d t}=3 s^{2} \times\left(\frac{d s}{d t}\right)[/tex]
[tex]=\left(3 s^{2} \times 3\right)[/tex]
[tex]=9 s^{2}[/tex]
As given in the problem each edge = 10 cm.
Hence,[tex]\frac{d v}{d t}=9 \times\left(10^{2}\right)[/tex]
[tex]=(9 \times 100) \mathrm{cm}^{3} / \mathrm{sec}[/tex]
[tex]=900 \mathrm{cm}^{3} / \mathrm{sec}[/tex]