# A spherical star is collapsing in size, while remaining spherical.  When its radius is one million kilometres, the radius is decreasing at the rate of 500 km per sec.. Find a) the rate of decrease of its volume b) the rate of decrease of its surface area

We can use differentiation to tackle this kind of problem.

For a sphere,

V = Volume and A= Area,

`V = (4pir^3)/3 and A = 4pir^2`

now we can differentiate these two functions with respect to t, to find the rates.

Differentiating V,

`(dV)/(dt) = (4pi)/3*3r^2*(dr)/(dt)`

`(dV)/(dt) = 4pir^2(dr)/(dt)`

Differentiating...

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We can use differentiation to tackle this kind of problem.

For a sphere,

V = Volume and A= Area,

`V = (4pir^3)/3 and A = 4pir^2`

now we can differentiate these two functions with respect to t, to find the rates.

Differentiating V,

`(dV)/(dt) = (4pi)/3*3r^2*(dr)/(dt)`

`(dV)/(dt) = 4pir^2(dr)/(dt)`

Differentiating A,

`(dA)/(dt) = 4pi*2r*(dr)/(dt)`

`(dA)/(dt) = 8pir(dr)/(dt)`

When,

` r = 10^6 km, (dr)/(dt) = -500 kms^-1`

Therefore the rate of volume shrinkage,

`(dV)/(dt) = 4pi(10^6)^2*(-500) = -6.283*10^15`

So the rate of volume shrinkage is 6.283 x 10^15 km^3s^-1

The rate of area shrinkage,

`(dA)/(dt) = 8pi*10^6*(-500) = -1.257*10^10`

Therefore the rate of area shrinkage is 1.257 x 10^10 km^2s^-1

Approved by eNotes Editorial Team