Question

a conical container has a radius of 10 fts and height of 15 ft. soda is poured into the container at the rate of 20TT (pi) cubic ft by per min. at what rate is the height if the soda increasing when the height of the soda is 4 ft

Answer 1

So we know that the rate of soda pouring is a change in volume per time,
\[\frac{\mathrm{d}V}{\mathrm{d}t} = 20 \pi ft^3/\text{min}\]
And we know that the volume the soda takes up is equal to the Total volume of the cone minus the volume of the cone above the soda line
\[V_{soda}=V_{cone} - \frac{1}{3}B_r(H-h)\]
where
\[B = \pi R^2=100 \pi ft^2
\\ H = 15ft
\\ V_{cone} = \frac{1}{3} \pi R^2 H=500 \pi ft^3\]
and
\[B_r = \pi r^2\]
We also know that the triangle formed by (H-h) and r is similar to the triangle formed by H and R
so
\[\frac{H}{R} = \frac{H-h}{r}\]
from which we can solve for r in terms of h
\[r=\frac{R(H-h)}{H}\]
Then the volume of the soda becomes
\[V_{soda}=\frac{1}{3}\pi R^2H - \frac{1}{3}\pi r^2(H-h)
\\ \
\\= \frac{1}{3}\pi R^2H - \frac{1}{3}\pi \left( \frac{R(H-h)}{H} \right)^2(H-h)
\\ \
\\ =\frac{1}{3}\pi R^2H \left(1- \left( \frac{H-h}{H}\right)^3 \right)
\\ \
\\ = V_{cone} \left(1- \left( 1-\frac{h}{H}\right)^3 \right)
\\ \
\\ = V_{cone} - V_{cone} \left( 1-\frac{h}{H}\right)^3 \]

Phew! Almost there! Only a derivative left.