Respuesta :
Answer:
[tex]\frac{dV}{dt}=-475.0088\ cm.min^{-1}[/tex]
Explanation:
Given:
initial radius of the snowball, [tex]r_i=25\ cm[/tex]
final radius of the snowball, [tex]r_f=16\ cm[/tex]
time taken for the change, [tex]t=30\ min[/tex]
Now the rate of the change in radius of the snowball:
[tex]\frac{dr}{dt} =\frac{r_f-r_i}{t}[/tex]
[tex]\frac{dr}{dt}=\frac{16-30}{30}[/tex]
[tex]\frac{dr}{dt}=-\frac{7}{15}\ cm.min^{-1}[/tex] ................(1)
As we know that the volume of sphere:
[tex]V=\frac{4}{3}\times \pi.r^3[/tex]
Now differentiating Volume equation:
[tex]\frac{dV}{dt}=\frac{4}{3}\pi\times 3r^2\times (\frac{dr}{dt})[/tex] ...........................(2)
Now when the radius of the sphere is 9 cm:
[tex]\frac{dV}{dt}=\frac{4}{3}\pi\times 3\times9^2\times (-\frac{7}{15})[/tex]
[tex]\frac{dV}{dt}=-475.0088\ cm.min^{-1}[/tex] is the rate of change of volume when the radius is 9 cm.
The volume of the snowball is changing at a rate of -305.208 cm³/min the instant the radius is 9 cm.
According to the question, the rate of change of radius with time can be given by:
[tex]\frac{dr}{dt}=\frac{\Delta r}{t}[/tex]
where
Δr = final radius - initial radius = 16 - 25 = -9cm
and t = 30 minutes.
So,
[tex]\frac{dr}{dt}=-\frac{9}{30}= -\frac{3}{10} cm/min[/tex]
Now, the volume of the sphere is given by:
[tex]V=\frac{4}{3}\pi r^3[/tex]
where r = 9 cm is the radius of the sphere at the given moment
Taking time derivative of volume to calculate the rate of change of volume with time, we get:
[tex]\frac{dV}{dt}=\frac{4}{3}\pi\times3r^2\times\frac{dr}{dt}\\\\=4\times3.14\times 9^2\times(-\frac{3}{10})\\ \\\frac{dV}{dt}=-305.208cm^3/min[/tex] is the rate of change of volume when the radius is 9 cm.
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