Calculator Technique for Solving Volume Flow Rate Problems in Calculus

Jhun Vert's picture

Flow Problems in Calculus Solved by a Formula from Physics | Differential Calculus

The following models of CASIO calculator may work with this method: fx-570ES, fx-570ES Plus, fx-115ES, fx-115ES Plus, fx-991ES, and fx-991ES Plus.

The following calculator keys will be used for the solution

Name Key Operation
Shift shift.jpg SHIFT
Mode mode-setup.jpg MODE
Name Key Operation
Stat 1-stat.jpg SHIFT → 1[STAT]
AC ac-off.jpg AC


This is one of the series of post in calculator techniques in solving problems. You may also be interested in my previous posts: Calculator technique for progression problems and Calculator technique for clock problems; both in Algebra.

Flow Rate Problem
Water is poured into a conical tank at the rate of 2.15 cubic meters per minute. The tank is 8 meters in diameter across the top and 10 meters high. How fast the water level rising when the water stands 3.5 meters deep.

Traditional Solution
flow-rate-cone.gif$\dfrac{r}{h} = \dfrac{4}{10}$

$r = \frac{2}{5}h$

Volume of water inside the tank
$V = \frac{1}{3}\pi r^2 h$

$V = \frac{1}{3}\pi (\frac{2}{5}h)^2 h$

$V = \frac{4}{75}\pi h^3$

Differentiate both sides with respect to time
$\dfrac{dV}{dt} = \frac{4}{25}\pi h^2 \dfrac{dh}{dt}$

$2.15 = \frac{4}{25}\pi h^2 \dfrac{dh}{dt}$

When h = 3.5 m
$2.15 = \frac{4}{25}\pi (3.5^2) \dfrac{dh}{dt}$

$\dfrac{dh}{dt} = 0.3492 \, \text{m/min}$           answer

Solution by Calculator


flow-rate-cone-caltech.gifMODE → 3:STAT → 3:_+cX2

0 0
10 π42
5 π22

AC → 2.15 ÷ 3.5y-caret = 0.3492           answer

To input the 3.5y-caret above, do
3.5 → SHIFT → 1[STAT] → 7:Reg → 6:y-caret

What we just did was actually v = Q / A which is the equivalent of $\dfrac{dh}{dt} = \dfrac{dV/dt}{A}$ for this problem.

Water is being poured into a hemispherical bowl of radius 6 inches at the rate of x cubic inches per second. Find x if the water level is rising at 0.1273 inch per second when it is 2 inches deep?

Traditional Solution
flow-rate-hemisphere.gifVolume of water inside the bowl
$V = \frac{1}{3}\pi h^2(3r − h)$

$V = \frac{1}{3}\pi h^2 [ \, 3(6) - h \, ]$

$V = \frac{1}{3}\pi (18h^2 - h^3)$

Differentiate both sides with respect to time
$\dfrac{dV}{dt} = \frac{1}{3}\pi (36h - 3h^2) \dfrac{dh}{dt}$

When h = 2 inches, dh/dt = 0.1273 inch/sec
$\dfrac{dV}{dt} = \frac{1}{3}\pi [ \, 36(2) - 3(2^2) \, ] (0.1273)$

$x = 7.9985 \, \text{in}^3\text{/sec}$           answer

Calculator Technique
flow-rate-hemisphere-caltech.gifMODE → 3:STAT → 3:_+cX2

0 0
6 π62
12 0

AC → 0.1273 × 2y-caret = 7.9985           answer

I hope you enjoy this post. Next time you solve problems involving flow rate, try to use this calculator technique to save time.

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