A conical tank has a circular base with radius 5 ft and height 12 ft. If water is flowing out of the tank at a rate of 3 ft^3/min, how fast is the height of the water changing when the height is 7 ft?

Question

Paisley Johnson · Accepted Answer

#0.112253 #ft /sec. [to #6# decimal places]

We are given that #dV#/dt#=3#...........#[1]# [ the rate of change of volume with respect to time, t. Assuming the cone is uniform with straight sides, then although #r and h# are variables their ratio will remain constant,i.e. #r/5=h/12#, so #r=[5h]/12#.............#[2]# Volume of such a cone is #V=1/3pir^2h#.........#[3]#, Substituting #r #from ......#[2],# in......... #[3]# we obtain, Volume of cone #=pi/3[[5h]/12]^2h,# = #[25pih^3]/[3[144]]#.......#[4]# differentiating #[4]# with respect to time #[t]# implicitly, #dV/dt=[25pih^3]/[3[144]##[dh/dt][h^3]#=#[[75pih^2]/[3[144]]]dh/dt#, since #d/dt[h^3]=3h^2#. We know #dV/dt=3#, So #3=[75pih^2]/[3[144]##dh/dt#, and when #h=7# #3=[[75]49pi/[3[144]]dh/dt# and so #dh/dt=9[144]/[[75][49pi]# #dh/dt=0.112253# ft /sec#.

Samantha Adkison · Answer

undefined To find the rate at which the height of the water is changing when the height is 7 ft, we can use related rates and the formula for the volume of a cone.

$$ V = \frac{1}{3} \pi r^2 h $$

$$ \frac{dV}{dt} = \frac{1}{3} \pi (2rh \frac{dr}{dt} + r^2 \frac{dh}{dt}) $$

Given:
$$ r = 5 \text{ ft} $$
$$ h = 7 \text{ ft} $$
$$ \frac{dV}{dt} = -3 \text{ ft}^3/\text{min} $$
$$ \frac{dr}{dt} = 0 \text{ ft/min} $$

Solve for $ \frac{dh}{dt} $:

$$ -3 = \frac{1}{3} \pi (2(5)(7)(0) + 5^2 \frac{dh}{dt}) $$

$$ -3 = \frac{1}{3} \pi (25 \frac{dh}{dt}) $$

$$ -9 = 25 \frac{dh}{dt} $$

$$ \frac{dh}{dt} = -\frac{9}{25} $$

So, the height of the water is changing at a rate of $ -\frac{9}{25} $ ft/min when the height is 7 ft.