Answer:
The volume of the cone is increasing at the rate oof 3848.45 cm³/s when r=15 cm and h=10 cm.
Step-by-step explanation:
The volume of the cone is given by the following formula:
[tex]V = \frac{r^{2}h\pi}{3}[/tex]
In which V is measured in cm³ while r and h are measured in cm.
Suppose that both the radius r and height h of a circular cone are increasing at a rate of 7 cm/s.
This means that [tex]\frac{dh}{dt} = \frac{dr}{dt} = 7[/tex]
How fast is the volume of the cone increasing when r=15 cm and h=10 cm?
This is [tex]\frac{dV}{dt}[/tex] when [tex]r = 15, h = 10[/tex].
[tex]V = \frac{r^{2}h\pi}{3}[/tex]
Applying implicit differentiation:
We have three variables, V, r and h. So
[tex]\frac{dV}{dt} = \frac{2r\frac{dr}{dt}h \pi + r^{2}\pi \frac{dh}{dt}}{3}[/tex]
[tex]\frac{dV}{dt} = \frac{2*15*7*10\pi + 225*\pi*7}{3}[/tex]
[tex]\frac{dV}{dt} = 3848.45[/tex]
The volume of the cone is increasing at the rate oof 3848.45 cm³/s when r=15 cm and h=10 cm.
