We are given that a pile of sand has the shape of a cone. The volume is given by:
[tex]V=\frac{1}{3}\pi r^2h[/tex]
Where:
[tex]\begin{gathered} r=\text{ radius} \\ h=\text{ height} \end{gathered}[/tex]
We are given that the height is 1/4 of the radius. Therefore, we have:
[tex]h=\frac{1}{4}r[/tex]
Now, we substitute in the formula for the volume:
[tex]V=\frac{1}{3}\pi r^2(\frac{1}{4}r)[/tex]
Now, we solve the products:
[tex]V=\frac{1}{12}\pi r^3[/tex]
Now, we substitute the value of the radius:
[tex]V=\frac{1}{12}\pi(12m)^3[/tex]
Now, we solve the operations:
[tex]V=452.39m^3[/tex]
Therefore, the volume is 452.39 cubic meters.
To determine ther rate of change of volume is determined by determining the derivative on both sides of the formula for the volume:
[tex]\frac{dV}{dt}=\frac{1}{12}\pi\frac{d}{dt}(r^3)[/tex]
Now, we determine the derivative using the following formula:
[tex]\frac{d}{dx}(f(x))^n=n(f(x))^{n-1}f^{\prime}(x)[/tex]
Applying the rule we get:
[tex]\frac{dV}{dt}=\frac{1}{12}\pi3r^2\frac{dr}{dt}[/tex]
Simplifyiong we get:
[tex]\frac{dV}{dt}=\frac{1}{4}\pi r^2\frac{dr}{dt}[/tex]
Now, we substitute the values:
[tex]\frac{dV}{dt}=\frac{1}{4}\pi(12m)^2(4\text{ m/h\rparen}[/tex]
Solving the operations:
[tex]\frac{dV}{dt}=452.39\frac{m^3}{h}[/tex]
Therefore, the rate of change is 452.39 cubic meters per second.
