MATH
newexpgrowth

# newexpgrowth - Exponential growth and decay section 7.5...

• Notes
• 31

This preview shows pages 1–11. Sign up to view the full content.

Exponential growth and decay section 7.5

This preview has intentionally blurred sections. Sign up to view the full version.

What do the bacteria in a culture the internet traffic the human population, and the capital in a bank have in common?

This preview has intentionally blurred sections. Sign up to view the full version.

They can all be represented by a mathematical function that grows exponentially. What do the bacteria in a culture the internet traffic the human population, and the capital in a bank have in common?
A function y=f(t) has “exponential growth” if it grows at a rate proportional to its size: dy dt = ky for some k ! ( dy / dt ) y = k ! The relative growth rate of y is a constant. If the constant k is negative, we talk about “exponential decay”

This preview has intentionally blurred sections. Sign up to view the full version.

dy dt = ky We need to study the differential equation where k is a fixed constant.
Let y=e 3t . y satisfies the differential equation dy/dt=3y. True False

This preview has intentionally blurred sections. Sign up to view the full version.

The most general solution of the differential equation dy/dt=3y is where C is any constant. A) C e 3t B) e 3t + C Question
Solving the differential equation dy/dt= 6y. dy dt = 6 y ! dy = 6 ydt ! dy y = 6 dt ! ln | y | = 6 t + c ! | y | = e 6 t + c ! | y | = e c e 6 t ! y = ± e c e 6 t ! y = C e 6 t This is a new constant, call it C There are infinitely many solutions, one for each value of the constant C.

This preview has intentionally blurred sections. Sign up to view the full version.

Find y such that dy/dt= 6y and y(1)=3e 6 y ( t ) = C e 6 t ! y (1) = Ce 6 .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern