Water in cone problem solved with derivative and similar triangles

Water Filling Cone Solution
Find r using similar triangles. $\frac{r}{2}$ = $\frac{h}{6}$ r = $\frac{h}{6}$ $\frac{dV}{dt}$ is the rate water is poured in. $\frac{dV}{dt}$ = ⅔; h = 4; find $\frac{dh}{dt}$ Start with cone volume formula. Then substitue in value for r. V = ⅓ π r² h V = ⅓ π $(\frac{h}{3})$ ² h V = ⅓ π $\frac{1}{9}$ h³ V = (π h³) / 27 Take derivative. Solve for $\frac{dh}{dt}$ . $\frac{dV}{dt}$ = (π 3h² $\frac{dh}{dt}$ ) / 27 $\frac{dV}{dt}$ = (π h² $\frac{dh}{dt}$ ) / 9 Subsitute $\frac{dV}{dt}$ = ⅔ h = 4 $\frac{2}{3}$ = (π * 4² * $\frac{dh}{dt}$ ) / 9 6 = 16 π $\frac{dh}{dt}$ $\frac{dh}{dt}$ = $\frac{6}{16π}$ = $\frac{3}{8π}$ The water level rises at a rate of $\frac{3}{8π}$ inches per second when the water is 4 inches deep.
153

Water Filling Cone
Solution

Drawing of water filling a cone with r and h coordinates

Find r using similar triangles.

\frac{r}{2}

\frac{h}{6}

r =

\frac{h}{6}

\frac{dV}{dt}

is the rate water is poured in.

\frac{dV}{dt}

= ⅔; h = 4; find

\frac{dh}{dt}

Start with cone volume formula.
Then substitue in value for r.
V = ⅓ π r² h

V = ⅓ π $(\frac{h}{3})$ ² h

V = ⅓ π $\frac{1}{9}$ h³

V = (π h³) / 27

Take derivative. Solve for

\frac{dh}{dt}

.
$\frac{dV}{dt}$ = (π 3h² $\frac{dh}{dt}$ ) / 27

$\frac{dV}{dt}$ = (π h² $\frac{dh}{dt}$ ) / 9

Subsitute

\frac{dV}{dt}

= ⅔ h = 4

\frac{2}{3}

= (π * 4² *

\frac{dh}{dt}

) / 9
6 = 16 π

\frac{dh}{dt}

\frac{dh}{dt}

\frac{6}{16π}

\frac{3}{8π}

The water level rises at a rate
of

\frac{3}{8π}

inches per second when
the water is 4 inches deep.

153

Water Filling ConeSolution

Water Filling Cone
Solution