Answer:
\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{2}
Explanation:
Hello,
The suitable differential equation for this case is:
[tex]\frac{dV}{dt}=10\frac{cm^3}{s}[/tex]
As we're looking for the change in height with respect to the time, we need a relationship to achieve such as:
[tex]\frac{dh}{dt} = ?*\frac{dV}{dt}[/tex]
Of course, [tex]?=\frac{dh}{dV}[/tex].
Now, since the volume of a cone is [tex]V=\pi r^2h/3[/tex] and the ratio [tex]r/h=15/20=3/4[/tex] or [tex]r=3/4h[/tex], the volume becomes:
[tex]V=\pi (\frac{3}{4} h)^2h/3= \frac{3}{16}\pi h^3[/tex]
We proceed to its differentiation:
[tex]\frac{dV}{dh} =\frac{9}{16} \pi h^2\\\frac{dh}{dV} =\frac{16}{9 \pi h^2}[/tex]
Then, we compute [tex]\frac{dh}{dt}[/tex]
[tex]\frac{dh}{dt} = \frac{16}{9 \pi h^2}*\frac{dV}{dt}\\\frac{dh}{dt} = \frac{16}{9\pi h^2}*10\frac{cm^3}{s} =\frac{160}{9 \pi h^2}[/tex]
Finally, at h=2:
[tex]\frac{dh}{dt}_{h=2cm} =\frac{160}{9\pi 2^2}\\\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{s}[/tex]
Best regards.
