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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