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Sagot :
Given :
A cylindrical tank with radius 3m is being filled with water at a rate 4m³/min.
To Find :
How fast is the height of the water increasing.
Solution :
We know, volume of cylinder is given by :
[tex]V = \pi r^2h[/tex] ....1)
Also, we have given that, [tex]\dfrac{dV}{dt} = 4 \ m^3/min[/tex] .
Differentiating equation 1) w.r.t t we get :
[tex]\dfrac{dV}{dt} =3^2 \pi \dfrac{ dh }{dt}\\\\\dfrac{dh}{dt} = \dfrac{4}{3^2 \times \pi}\\\\\dfrac{dh}{dt} = 0.141 \ m/min[/tex]
Therefore, the height of water is increasing at a rate of 0.141 m/min.
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