Asked by Z32
A spherical snowball is melting in such a way that its diameter is decreasing at rate of 0.2 cm/min. At what rate is the volume of the snowball decreasing when the diameter is 16 cm.
Answers
Answered by
Christiaan
The diameter of the sphere is decreasing at 0.2 cm/min. This means that its radius is decreasing by 0.1 cm/min.
The formul
The formul
Answered by
Christiaan
The diameter of the sphere is decreasing at 0.2 cm/min. This means that its radius is decreasing by 0.1 cm/min. We also know its current radius is 8 cm.
The formula for the volume of a sphere is:
V= (4/3)*Pi*r^3
So, we can calculate the volume difference in 1 minute, by subtracting its volume after one minute from its current volume:
Vc-Vf = [(4/3)*Pi*(8 cm)^3] -
[(4/3)*Pi*(8-0.1 cm)^3]
= 2144.66 cm^3- 2065.24 cm^3= 79.42 cm^3
This means its volume is decreasing by 79.42 cm^3/min
The formula for the volume of a sphere is:
V= (4/3)*Pi*r^3
So, we can calculate the volume difference in 1 minute, by subtracting its volume after one minute from its current volume:
Vc-Vf = [(4/3)*Pi*(8 cm)^3] -
[(4/3)*Pi*(8-0.1 cm)^3]
= 2144.66 cm^3- 2065.24 cm^3= 79.42 cm^3
This means its volume is decreasing by 79.42 cm^3/min
Answered by
Z32
Hmm. I tried that and it wasn't correct.
Answered by
Andora
The answer is 80.4248 cm^3/min.
(1/3)(4π)(1/8)(3(16)²)(-0.2)
(1/3)(4π)(1/8)(3(16)²)(-0.2)
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