Answer:
The rate of change of the radius of the surface of the water when the radius of the surface of the water is 2 feet is approximately 0.637 feet per second.
Step-by-step explanation:
The volume of the cone ([tex]V[/tex]), in cubic feet, is defined by the following equation:
[tex]V = \frac{\pi}{3}\cdot r^{2}\cdot h[/tex] (1)
Where:
[tex]r[/tex] - Radius, in feet.
[tex]h[/tex] - Height, in feet.
And there is the following ratio of the radius to the height is:
[tex]\frac{r}{h} = k[/tex] (2)
By applying (2) in (1):
[tex]h = \frac{r}{k}[/tex]
[tex]V = \frac{\pi}{3\cdot k}\cdot r^{3}[/tex] (3)
And the rate of change of the radius is found by differentiating on (3):
[tex]\dot V = \frac{\pi}{k}\cdot r^{2}\cdot \dot r[/tex] (4)
Where:
[tex]\dot V[/tex] - Rate of change of the volume, in cubic feet per second.
[tex]\dot r[/tex] - Rate of change of the surface of the water, in feet per second.
[tex]\dot r = \frac{k\cdot \dot V}{\pi\cdot r^{2}}[/tex]
If we know that [tex]k = \frac{1}{3}[/tex], [tex]\dot V = 24\,\frac{ft^{3}}{s}[/tex] and [tex]r = 2\,ft[/tex], then the rate of change of the radius of the surface of the water is:
[tex]\dot r = \frac{\left(\frac{1}{3} \right)\cdot \left(24\,\frac{ft^{3}}{s} \right)}{\pi\cdot (2\,ft)^{2}}[/tex]
[tex]\dot r = 0.637\,\frac{ft}{s}[/tex]
The rate of change of the radius of the surface of the water when the radius of the surface of the water is 2 feet is approximately 0.637 feet per second.
