Respuesta :
Answer:
The rate of change in volume is [tex]3136 \ \pi \ cm^3/min[/tex].
Step-by-step explanation:
Given:
radius [tex]r= 14\ cm[/tex]
[tex]\frac{d}{dt}r=4 \ cm/min[/tex]
We need to find the rate of change of Volume [tex]\frac{d}{dt}(V)[/tex]
Solution:
Now We can say that:
Volume of sphere is given by:
[tex]V=\frac43 \pi r^3[/tex]
Now taking derivative on both side we get;
[tex]\frac{d}{dt}(V)=\frac{d}{dt}\frac43 \pi r^3[/tex]
[tex]\frac{d}{dt}(V)=\frac43 \pi\times 3\times r^2\frac{d}{dt} r\\\\\frac{d}{dt}(V)=4\pi\times r^2\frac{d}{dt} r[/tex]
But  [tex]r= 14\ cm[/tex] and [tex]\frac{d}{dt}r=4 \ cm/min[/tex] so we get;
[tex]\frac{d}{dt}(V)=4\pi\times 14^2 \times 4[/tex]
[tex]\frac{d}{dt}(V)=3136 \ \pi \ cm^3/min[/tex]
Rate of change of volume = [tex]3136 \ \pi \ cm^3/min[/tex]
Hence the rate of change in volume is [tex]3136 \ \pi \ cm^3/min[/tex].
Applying implicit differentiation, it is found that the volume is changing at a rate of 9852 cubic centimetres per minute.
The volume of a sphere of radius r is given by:
[tex]V = \frac{4\pi r^3}{3}[/tex]
Applying implicit differentiation, the rate of change is given by:
[tex]\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}[/tex]
We are given that:
[tex]\frac{dr}{dt} = 4, r = 14[/tex]
Then
[tex]\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}[/tex]
[tex]\frac{dV}{dt} = 4\pi 14^2(4)[/tex]
[tex]\frac{dV}{dt} = 9852[/tex]
The volume is changing at a rate of 9852 cubic centimetres per minute.
A similar problem is given at https://brainly.com/question/24158553
