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The radius of a sphere is increasing at a rate of 5 mm/s. how fast is the volume increasing when the diameter is 60 mm? evaluate your answer numerically. (round the answer to the nearest whole number.)

Respuesta :

The diameter of the sphere is 60 mm which means its radius is 30mm. The radius of a sphere is increasing at a rate of 5 mm/s. 
The formula for the radius when the time is x second would be:
r(t)= 30+ 5x
dr/dt= 5

The speed of volume increase of the sphere volume would be derived from sphere volume: 
V(r)=4/3* pi * r^3
dV/dt= 4/3 * pi * 3r^2 * (dr/dt) =4/3* pi * 3(30)^2 * 5 =  18000 pi= 56571 mm^3/s
ernikhil

The rate of change of volume of the sphere is [tex]56520\;\rm{mm/sec}[/tex].

According to the question, the radius of a sphere is increasing at a rate of [tex]5 mm/s[/tex] and the diameter is [tex]60 mm[/tex].

The volume of the sphere is, [tex]V=\dfrac{4}{3}\pi r^3[/tex] where, [tex]r[/tex] is the radius of the sphere.

Differentiate the volume of the sphere both sides with respect to [tex]t[/tex] as-

[tex]\dfrac{dV}{dt}=\dfrac{4\pi}{3}\dfrac{dr^3}{dt}\\\dfrac{dV}{dt}=\dfrac{4\pi}{3}\times 3r^2\dfrac{dr}{dt}\\\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}[/tex]

The radius of the sphere is-

[tex]r=\dfrac{60}{2}\\r=30mm[/tex]

Substitute the value of the parameter as-

[tex]\dfrac{dV}{dt}=4\pi\times (30)^2\times 5\\\dfrac{dV}{dt}=62.8\times 900\\\dfrac{dV}{dt}=56520\;\rm{mm/sec}[/tex]

Hence, the rate of change of volume of the sphere is [tex]56520\;\rm{mm/sec}[/tex].

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