Respuesta :
Answer:
The Volume is decreasing at a rate of [tex]156\pi[/tex] [tex]cm^3[/tex] per minute.
Step-by-step explanation:
Given a cone of radius r and perpendicular height h
Volume of the cone, [tex]V=\frac{1}{3}\pi r^2 h[/tex]
Since the height of the cone is fixed, the rate of change of the volume of the Cone, [tex]\frac{dV}{dt}[/tex] is given as:
[tex]V=\frac{1}{3}\pi r^2 h \\\frac{dV}{dt}=\frac{h\pi}{3} \frac{d}{dt} r^2\\\frac{dV}{dt}=\frac{2rh\pi}{3} \frac{dr}{dt}[/tex]
We are to determine the rate of change of the Volume, V when:
The radius is decreasing at a rate of 2 cm per minute, [tex]\frac{dr}{dt} = -2 cm/min[/tex]
Height, h=9 cm
Radius = 13cm
[tex]\frac{dV}{dt}=\frac{2rh\pi}{3} \frac{dr}{dt}\\=\frac{2*13*9\pi}{3}* ( -2)\\\frac{dV}{dt}=-156\pi \:cm^3/min[/tex]
The Volume is decreasing at a rate of [tex]156\pi[/tex] [tex]cm^3[/tex] per minute.
