Answer :
Answer:
The radius is decreasing at 4 mm/s
Explanation:
The volume of a sphere is:
[tex]V = 4/3*\pi *r^3[/tex] So, when the volume is 972π mm^3 the radius r is:
r = 9mm
Now, the change rate is given by the derivative:
[tex]dV/dt = 4/3*\pi *3*r^2*dr/dt[/tex]
Where: dV/dt = -324π mm^2/s
r = 9mm
Solving for dr/dt:
dr/dt = -4mm/s
The rate at which the radius of the snowball is decreasing when the volume of the snowball is 972π mm³ is 1mm/s
The snowball in question is spherical in nature. The formula for finding the volume of the snowball is expressed as:
[tex]V = \frac{4}{3} \pi r^3\\[/tex]
The rate at which the volume is decreasing is expressed as:
[tex]\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt}[/tex]
[tex]\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}[/tex]
Given the following parameters
V = 972π mm³
[tex]\frac{dV}{dt} = 324 \pi mm^3/s[/tex]
Get the radius of the ball;
[tex]972\pi = \frac{4}{3} \pi r^3\\972 = \frac{4}{3} r^3\\4r^3 = 972 \times 3\\ 4r^3 =2916\\r^3 = \frac{2916}{4}\\r^3= 729\\r=\sqrt[3]{729}\\r= 9mm[/tex]
Get the rate at which the radius is decreasing:
[tex]324 \pi= 4\pi (9)^2 \frac{dr}{dt}\\324=324 \frac{dr}{dt}\\ \frac{dr}{dt} = \frac{324}{324}\\ \frac{dr}{dt} = 1mm/s[/tex]
Hence the rate at which the radius of the snowball is decreasing when the volume of the snowball is 972π mm³ is 1mm/s
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