Answer:
The rate of change of the volume of the cylinder at that instant = [tex]791.28\ mm^3/hr[/tex]
Step-by-step explanation:
Given:
Rate of increase of base of radius of base of cylinder = 7 mm/hr
Height of cylinder = 1.5 mm
Radius at a certain instant = 12 mm
To find rate of change of volume of cylinder at that instant.
Solution:
Let [tex]r[/tex] represent radius of base of cylinder at any instant.
Rate of increase of base of radius of base of cylinder can be given as:
[tex]\frac{dr}{dt}=7\ mm/hr[/tex]
Volume of cylinder is given by:
[tex]V=\pi\ r^2h[/tex]
Finding derivative of the Volume with respect to time.
[tex]\frac{dV}{dt}=\pi\ h\ 2r\frac{dr}{dt} [/tex]
Plugging in the values given:
[tex]\frac{dV}{dt}=\pi\ (1.5)\ 2(12)(7) [/tex]
[tex]\frac{dV}{dt}=252\pi [/tex]
Using [tex]\pi=3.14[/tex]
[tex]\frac{dV}{dt}=252(3.14) [/tex]
[tex]\frac{dV}{dt}=791.28\ mm^3/hr[/tex] (Answer)
Thus rate of change of the volume of the cylinder at that instant = [tex]791.28\ mm^3/hr[/tex]
