Refer a friend and earn $50 when they subscribe to an annual planRefer Now

Get the answer to your homework problem.

Try Numerade Free for 30 Days

Like

Report

For the following problems, set up and solve the differential equations.A 1000 -liter tank contains pure water and a solution of 0.2 kg saltL is pumped into the tank at a rate of 1 $\mathrm{L} /$ min and is drained at the same rate. Solve for total amount of salt in the tank at time $t$ .

$u(t)=200\left(1-e^{-\frac{t}{1000}}\right)$

Calculus 2 / BC

Chapter 4

Introduction to Differential Equations

Section 5

First-order Linear Equations

Differential Equations

Oregon State University

Baylor University

University of Nottingham

Idaho State University

Lectures

01:11

In mathematics, integratio…

06:55

In grammar, determiners ar…

04:43

A tank containing 10 kilog…

06:43

A $1000$ L tank contains 5…

05:08

03:50

03:48

A tank contains 1000 $\mat…

04:03

A 2000 -liter cistern is e…

06:02

Two large tanks, each hold…

06:10

Use $E q .(4)$ and Torrice…

05:57

Water flows into a tank at…

04:28

A tank contains 1000 L of …

So we're going to start this problem by setting up a formula for the rate of change of the amount of salt in the tank with respect to time. So we know that this race is going to be given to us by the rate of salt going in minus the rate of salt going out. So that's what you tell us that the S d T will be equal to. We know that one of the solutions in Compton is going to a pump insult at a rate of 0.2 killer grounds for leader times 20 liters per minute. Well, give us kilograms per minute. Plus, our other solution is pumped in iterated 0.5 kilograms for leader times five liters per minute will again give us units of leaders per minute or accuse me kilograms for a minute in minus. Uh, the salt out will be. We know we're pumping it out at a rate of 25 liters per minute. But the concentration of the assault that's being pumped out is going to be equal to the amount of salt in the tank. I'm divided by the volume of tanks and so when we simplify this, Um, we can actually go ahead and substitute in, um, our salt the amount of salt in the tank is equal to the concentration times the volume because that's going to give us units of kilograms. So the rate of change of salt in the tank with respect to time, um, is going to be equal to 0.0 4 to 5, um, minus, uh, 25 times concentration times the volume over the volume so we can see that this is going to cancel. And we know that the rate of change of the concentration with time is going to be equal to the amount of salt divided by the volume. So essentially, we're just going to divide every term here by the volume. So when we do that, we get the CD T is equal to 0.0 for 4 to 5. Should have been not for five to minus 25 times a concentration over the volume. And, um, if we are to calculate the rate of change of volume with respect to time, we're going to find that the volume is actually a constant 1000 leaders. You can calculate this on your own if you want to check me. But I've already calculated it. Um And so essentially, we're just going to plug in 1000 for V. So when we do that, um, we get DC DT is equal to I'm in this step, actually already divided by 1000. I'm sorry if your number was a little bit off there, That would explain why that was my bad. I actually already divided that I'm so that's equal to 0.425 minus, um, 0.25 c so we can go ahead and simple for this to DC DT is going to be equal to a 0.17 minus C over 40. And so I'll go ahead and start a new page here, but we can see that we're going to be able to use our separation of variables here to integrate. Um, And so when we separate our variables, we're going to end up with one over 10.17 minus C. D. C is equal to won over 40 duty. And so when we integrate both sides of our equation here we get the negative natural log of 0.17 minus seat is equal to one over 40 teeth. And so when we solve for C, you will find that C is equal to 0.17 minus a e to the negative one over 40 teeth. And so to solve for our constant A we're going to plug in or initial condition. We know that at times zero see is going to be equal Thio 10 kilograms divided by 1000 liters. So that's going to be, um, 0.1 And so when we plug this in, we get the point. No one is equal to 0.17 minus a E to zero power, and we solved for A and get that A's equal 2.16 So our final equation is our concentration is equal to 0.17 minus 0.16 a e to the negative or he's not a a 0.16 e to the negative one over 40 teeth, Um, and that's given to us in kilograms per liter.

View More Answers From This Book

Find Another Textbook

In mathematics, integration is one of the two main operations in calculus, w…

In grammar, determiners are a class of words that are used in front of nouns…

A tank containing 10 kilograms of salt dissolved in 1000 liters of water has…

A $1000$ L tank contains 500 L of water with a salt concentration of 10 $\ma…

A tank contains 1000 $\mathrm{L}$ of pure water. Brine that contains 0.05 $\…

A 2000 -liter cistern is empty when water begins flowing into it (at $t=0$ )…

Two large tanks, each holding 100 L of liquid, are interconnected by pipes, …

Use $E q .(4)$ and Torricelli's Law $[E q .(5)]$.

At $t=0,$ a c…

Water flows into a tank at the variable rate of $R_{\text { in }}=$ 20$/(1+t…

A tank contains 1000 L of brine with 15 $\mathrm{kg}$ of dissolved salt. Pur…

01:13

In the following exercises, use partial fractions to find the power series o…

00:58

State whether each of the following series converges absolutely, conditional…

00:24

For the following sequences, plot the first 25 terms of the sequence and sta…

03:36

For the following problems, find the general solution to the differential eq…

03:34

For the following problems, find the solution to the initial value problem.<…

00:25

For the following exercises, each set of parametric equations represents a l…

01:23

For which $p > 0$ does the series $\sum_{n=1}^{\infty} \frac{n^{p}}{2^{n}…

03:22

07:17

For the following exercises, approximate the integrals using the midpoint ru…

The following exercises explore applications of annuities.Calculate the …

92% of Numerade students report better grades.

Try Numerade Free for 30 Days. You can cancel at any time.

Annual

0.00/mo 0.00/mo

Billed annually at 0.00/yr after free trial

Monthly

0.00/mo

Billed monthly at 0.00/mo after free trial

Earn better grades with our study tools:

Textbooks

Video lessons matched directly to the problems in your textbooks.

Ask a Question

Can't find a question? Ask our 30,000+ educators for help.

Courses

Watch full-length courses, covering key principles and concepts.

AI Tutor

Receive weekly guidance from the world’s first A.I. Tutor, Ace.

30 day free trial, then pay 0.00/month

30 day free trial, then pay 0.00/year

You can cancel anytime

OR PAY WITH

Your subscription has started!

The number 2 is also the smallest & first prime number (since every other even number is divisible by two).

If you write pi (to the first two decimal places of 3.14) backwards, in big, block letters it actually reads "PIE".

Receive weekly guidance from the world's first A.I. Tutor, Ace.

Mount Everest weighs an estimated 357 trillion pounds

Snapshot a problem with the Numerade app, and we'll give you the video solution.

A cheetah can run up to 76 miles per hour, and can go from 0 to 60 miles per hour in less than three seconds.

Back in a jiffy? You'd better be fast! A "jiffy" is an actual length of time, equal to about 1/100th of a second.