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About Scotto
Expertise
Any kind of mathematics (calculus, analysis, game theory, linear approximation, finite differences, linear regression, linear programming, numerical analysis, probability, statistics, etc.). I also have answered some questions in Physics (mass, momentum, falling bodies), Chemistry (charge, reactions, symbols, molecules), and Biology.

Experience
Experience in the area: I have tutored students in all areas of mathematics for over 20 years. Education/Credentials: BSand MS in Mathematics from Oregon State University, where I completed sophomore course in Physics and Chemistry. I received both degrees with high honors. Awards and Honors: I have passed Actuarial tests 100, 110, and 135.

Publications
Maybe not a publication, but I have respond to well oveer 3000 questions on the PC. That's around 2,000 in basic math and 1,000 in advanced math.

Education/Credentials
I aquired well over 40 hours of upper division courses. This was well over the number that were required. I graduated with honors in both my BS and MS degree from Oregon State University. I was allowed to jump into a few junior level courses my sophomore year.

Awards and Honors
I have been nominated as the expert of the month several times. All of my scores right now are at least a 9.8 average (out of 10).

Past/Present Clients
My past clients have been students at OSU, students at the college in South Seattle, referals from a company, friends and aquantenances, people from my church, and people like you.

You are here: Experts > Teens > Homework/Study Tips > Calculus > calculus

Calculus - calculus

Expert: Scotto - 11/5/2009

Question
A particle moves along the x-axis so that its x-coordinate at time t is given by x=ae^-kt+be^kt. Show that its acceleration is proportional to x.

Answer
It is known that the derivative of e^kt = ke^kt.
The acceleration is two derivatives of the distance.
This means the second derivative is ak²e^-kt + bk²e^kt = k²(ae^-kt + be^kt) = k²x.

Note that if the exponet is negative, the 1st derivative is negtative,
but the 2nd derivative is positive again. In this way, it doesn't matter what the sign of k is.

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