If a hose filling up a cylindrical pool with a radius of 5 ft at 28 cubic feet per minute, how fast is the depth of the pool water increasing?

Question

Ezekiel Alexander · Accepted Answer

I got about #"0.357 ft/min"#.

To solve this, it's best to get organized and see what information you have.

You have these numbers at your disposal:

#r = "5 ft"#, the radius of the cylindrical pool
#(dV)/(dt) = ("28 ft"^3)/"min"#, the change in volume of water over time
#(dh)/(dt) = ???#, your goal: the change in height of the water over time

If you don't recall the equation for the volume of a cylinder, you can start from the area of a circle and derive it.

#A = A(r) = pir^2#

The volume of a cylinder is gotten by extruding a circle at constant radius so that it has a height.

Hence, what we have for the volume of a cylinder is

#\mathbf(V = pir^2h),#

where #r# is the radius of the pool and #h# is the height of the pool.

At this point, we have an equation that relates the volume to the height, and we have to solve for the change in height over time, #(dh)/(dt)#.

Next, we know that #h = h(t)#, but we should also acknowledge that #color(green)(V = V(h))#.

That means:

#\mathbf((dV)/(dt)) = (dV(h(t)))/(dt) = \mathbf((d)/(dh)[color(green)(V(h))]*stackrel("Chain Rule")stackrel("result")overbrace((dh)/(dt)))#

So by using the #(dy)/(dx)# notation, we can see the need for noting:

The implicit dependency of the change in volume on the height.

The implicit dependency of the change in height on the time.
To find #(dh)/(dt)# and to utilize the value for #(dV)/(dt)#, we thus have to take the derivative of the cylinder volume with respect to the height at constant radius:

#(dV(h))/(dt)#
#= d/(dh)stackrel(color(green)(V(h)))overbrace([pir^2h])*(dh)/(dt)#
#= stackrel("constant")overbrace(pir^2)d/(dh)[h] (dh)/(dt) = pir^2*1*(dh)/(dt)#
#= pir^2(dh)/(dt),#

which gives us the implicit change in height over time. Therefore:

#color(blue)((dh)/(dt))#
#= 1/(pir^2)(dV(h))/(dt)#
#= 1/(pi*("5" cancel("ft"))^cancel(2)) xx ("28 ft"^(cancel(3)^(1)))/"min"#
#= color(blue)((28)/(25pi) "ft/min" ~~ "0.357 ft/min")#

Christopher Allers · Answer

#0.357# ft per minute, rounded to three decimal places

Rate at which hose fills up the cylindrical pool#=28" cftm"^-1# Volume of the pool if water filled up to depth #d##=pir^2d#, Where #r# is the radius of the cylinder. Given volume of water filled per minute# =#Increase of volume of water in pool per minute. Equating the two. #28=pir^2d_"minute"# Where #d_"minute"# is increase in depth per minute. Inserting given values and solving for #d_"minute"# #d_"minute"=28/(pi5^2)# #=0.357# ft per minute, rounded to three decimal places

Samantha Adkison · Answer

undefined To find the rate at which the depth of the pool water is increasing, we can use the formula for the volume of a cylinder, $ V = \pi r^2 h $, where $ V $ is the volume, $ r $ is the radius, and $ h $ is the height (or depth) of the cylinder.

Taking the derivative of the volume with respect to time will give us the rate of change of volume with respect to time, which is the rate at which water is being added to the pool.

Given:
- Radius $ r = 5 $ ft
- Rate of change of volume $ \frac{dV}{dt} = 28 $ cubic feet per minute

Differentiating the volume formula with respect to time $ t $ gives:
$$ \frac{dV}{dt} = \pi r^2 \frac{dh}{dt} $$

We can solve this equation for $ \frac{dh}{dt} $, the rate at which the depth of the water is increasing:
$$ \frac{dh}{dt} = \frac{1}{\pi r^2} \frac{dV}{dt} $$

Substituting the given values:
$$ \frac{dh}{dt} = \frac{1}{\pi (5)^2} \times 28 $$
$$ \frac{dh}{dt} = \frac{28}{25\pi} $$

So, the rate at which the depth of the pool water is increasing is $ \frac{28}{25\pi} $ feet per minute.