Question

A cylindrical tank with radius 3 m is being filled with water at a rate of 4 m3/min. How fast is the height of the water increasing (in m/min)

Answer 1

Answer:

Let R be the radius of the tank, H(t) the height of the water at time t, and V(t) the volume of the water. The quantities V(t), R and H(t) are related by the equation

V(t) = πR²H(t) ___(1)

The rate of increase of the volume is the derivative with respect to time,

dV/dt

and the rate of increase of the height is

dH/dt

We can therefore restate the given and the unknown as follows

Given:

dV/dt = 4m³/min

Unknown:

dH/dt

Now we take derivative of each side of (1) with respect to t:

dV/dt = πR²dH/dt

So

dH/dt = 1/πR²dV/dt

Substituting R = 3 m and dV/dt = 4m³/min we have

dH/dt = 1/π(3)² ⋅ 4 = 4/9π

Answer: the height of the water increasing at a rate of

dH/dt = 4/9π²

≈ 0.14 m/min

Answer 2

Height of the water increasing is at rate of  $#(dh)/(dt)=3/(25 pi)m/(min)#$

How to solve?

With related rates, we need a function to relate the 2 variables, in this case it is clearly volume and height. The formula is:

$#V=pi r^2 h#$

There is radius in the formula, but in this problem, radius is constant so it is not a variable. We can substitute the value in:

$#V=pi (5m)^2 h#$

Since the rate in this problem is time related, we need to implicitly differentiate wrt (with respect to) time:

$#(dV)/(dt)=(25 m^2) pi (dh)/(dt)#$

In the problem, we are given $#3(m^3)/min# which is #(dV)/(dt)#.$

So we need to substitute this in:

$#(dh)/(dt)=(3m^3)/(min (25m^2) pi)=3/(25 pi)m/(min)#$

Hence,  Height of the water increasing is at rate of  $#(dh)/(dt)=3/(25 pi)m/(min)#$

Formula used:

$#V=pi r^2 h#$

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