calculator technique for solving volume flow rate problems in calculus

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 Solutionrh=410r=25h 

Volume of water inside the tankV=13πr2hV=13π(25h)2hV=475πh3

Differentiate both sides with respect to timedVdt=425πh2dhdt2.15=425πh2dhdt 

When h = 3.5 m2.15=425π(3.52)dhdtdhdt=0.3492m/min           answer

Solution by Calculator

ShowClick here to show or hide the concept behind this technique

MODE → 3:STAT → 3:_+cX2

X Y0 010 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 dhdt=dV/dtA for

this problem.

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

Volume of water inside the bowlV=13πh2(3r−h)V=13πh2[3(6)−h]V=13π(18h2−h3) 

Differentiate both sides with respect to timedVdt=13π(36h−3h2)dhdt 

When h = 2 inches, dh/dt = 0.1273 inch/secdVdt=13π[36(2)−3(22)](0.1273)x=7.9985in3/sec           answer

Calculator Technique

MODE → 3:STAT → 3:_+cX2

X Y0 06 π62

12 0

AC → 0.1273 × 2y-caret = 7.9985           answer

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