Problem

The quantity of charge of $ Q $ in coulombs (C) t…

01:58

Problem 18 Medium Difficulty

If a tank holds 5000 gallons of water, which drains from the bottom of the tank in 40 minutes, then Torricelli's Law gives the volume $ V $ of water remaining in the tank after $ t $ minutes as

$ V = 5000 (1 - \frac {1}{40}t)^2 \space \space \space \space 0 \le t \le 40 $

Find the rate at which water is draining from the tank after (a) 5 min, (b) 10 min, (c) 20 min, and (d) 40 min. At what time is the water flowing out the fastest? The slowest? Summarize your findings.

Answer

a) 218.75$\frac{\text { gallons }}{\text {min}}$
b) 187.5$\frac{\text {gallons}}{\text {min}}$
c) 125$\frac{\text { gallons }}{\text { min }}$
d) 0$\frac{\text { gallons }}{\text {min}}$

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Video Transcript

Okay, So this problem involves tourists Eli's Law, which describes the volume of water remaining in the tank as a function of time. So we have this volume equation, and we're interested in finding the rate of change, the rate at which water is draining at various times. So we need the derivative of the function. So we'll use the chain rule first will leave the constant 5000 will bring down the two and will raise the inside to the first. And then we'll multiply by the derivative of the insight. Now we can multiply our Constance together 5000 times, two times negative 1/40 and we end up with the prime of tea equals negative 2 50 times, one minus 1/40 t. So we're going to use that function. And for part A, we're going to find the prime of five. So a substitute five in there when we get negative to 18.75 and the units would be gallons per minute. And for part B, we find the rate of change at 10 minutes. So the prime of 10 we substitute 10 into the equation and we get negative 1 87.5 Callens per minute. It's negative because of volume is decreasing for part C for 20 minutes, the prime of 20. And that gives us negative 1 25 gallons per minute Notice the rate is changing As the tank empties, the rate is slowing. And then part D. V, prime of 40 well, noticed that it only takes 40 minutes to empty the tank. So, do you have a guess as to what this is going to be when we substitute 40 into the equation? We end up with zero. The tank is empty. Okay, so what do we noticed? When is it the fastest? And when is it this lowest? It's the fastest at the beginning. It slows down as it goes along and as a tank empties slowest at the end.

Heather Z.

Oregon State University

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Problem

The quantity of charge of $ Q $ in coulombs (C) t…

Problem 18 Medium Difficulty

Answer

a) 218.75$\frac{\text { gallons }}{\text {min}}$
b) 187.5$\frac{\text {gallons}}{\text {min}}$
c) 125$\frac{\text { gallons }}{\text { min }}$
d) 0$\frac{\text { gallons }}{\text {min}}$

More Answers

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Calculus: Early Transcendentals

Catherine R.

Anna Marie V.

Heather Z.

Kristen K.

Precalculus Review - Intro

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a) 218.75$\frac{\text { gallons }}{\text {min}}$b) 187.5$\frac{\text {gallons}}{\text {min}}$c) 125$\frac{\text { gallons }}{\text { min }}$d) 0$\frac{\text { gallons }}{\text {min}}$

Calculus: Early Transcendentals

Catherine R.

Anna Marie V.

Heather Z.

Kristen K.

Precalculus Review - Intro

Functions on the Real Line - Overview

Catherine R.

Anna Marie V.

Heather Z.

Kristen K.

a) 218.75$\frac{\text { gallons }}{\text {min}}$
b) 187.5$\frac{\text {gallons}}{\text {min}}$
c) 125$\frac{\text { gallons }}{\text { min }}$
d) 0$\frac{\text { gallons }}{\text {min}}$