Download presentation

Presentation is loading. Please wait.

1
**Differential Equations and Seperation of Variables**

2
**The population of the little town of Scorpion Gulch is now 1000 people**

The population of the little town of Scorpion Gulch is now 1000 people. The population is presently growing at about 5% per year. Write a differential equation that expresses this fact. Solve it to find an equation that expresses population as a function of time. Rate of change of the population Let P = population t years after the present. = .05P dP dt *label is people/yr

3
= .05P dP dt 4

4
**Must change up a little!!!! start over**

The population of the little town of Scorpion Gulch is now 1000 people. The population is presently growing at about 5% per year. Write a differential equation that expresses this fact. Solve it to find an equation that expresses population as a function of time. .To check accuracy of equation, find the population after one year given …..” The population of the little town of Scorpion Gulch is now 1000 people. The population is presently growing at about 5% per year” ; Use the equation we found to check if when t=1 the population comes out right! *when t = 1, P should equal 1050. (1000*.05= 50 plus original 1000) P = 1000 e0.05 t found equation *with this equation, however P = when t = 1. P should equal 1050. 4 Must change up a little!!!! start over

5
**The population of the little town of Scorpion Gulch is now 1000 people**

The population of the little town of Scorpion Gulch is now 1000 people. The population is presently growing at about 5% per year. Write a differential equation that expresses this fact. Solve it to find an equation that expresses population as a function of time. We must find a new constant (k) since 0.05 didn’t work!!! Constant of proportionality = kP dP dt We must find a new constant (k) since 0.05 didn’t work!!! *When t = 0, P = 1000 *when t = 1, P should equal 1050. (1000*.05= 50 plus original 1000)

6
**= kP dP dt *When t = 0, P = 1000 = kP dP dt**

*when t = 1, P should equal 1050. *Separate variables in = kP dP dt dP P = k dt solve ln | P | = k t + C

7
** P = j e k t Now have: |P| = Ce k t *When t = 0, P = 1000**

Solve for C using: *You can solve for C as soon as you find antiderivative if there is not an absolute value involved……otherwise solve with C intact AND substitute j = ± C allowing a drop of absolute value!!! *When t = 0, P = 1000 P = j e k t Solve for k (what we wanted) using above equation and *when t = 1, P = 1050.

8
**Law of exponential change!!**

therefore: P = 1000e ln t 21 20 In general: y = yo e k t where yo is original value at time, feet, or whatever = 0 Law of exponential change!!

Similar presentations

Presentation is loading. Please wait....

OK

6-5 Solving Absolute Value Equations

6-5 Solving Absolute Value Equations

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on 4g technology Ppt on art and craft movement furniture Ppt on conceptual art movement Free ppt on effective communication skills Ppt on bodybuilding Ppt on diode characteristics Ppt on employee motivation stories Free ppt on the road not taken Ppt on life insurance corporation of india Ppt online open channel