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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 > Calc Related Rates

Calculus - Calc Related Rates

Expert: Scotto - 11/5/2009

Question
The volume of a cube is expanding at a constant rate of 2cm^3/sec. How fast is the length of an edge changing when t=1? When t=5? Why the difference?

Answer
The volume V(t) is equal to the side s(t) to the third.
That is, V(t) = s³(t).

Given this, the change in volume is dV/dt = 3s²(t)(ds/dt).

We are given that dV/dt = 2, so 2 = 3s²(t)(ds/dt).

That means ds/dt = 2/(3s²(t)).

As t increases, our volume is increasing, so s(t) is increasing as well.

As V(t) increases, so does s(t). This means that since s(t) is in the denominator,
ds/dt would be increasing even slower.

When t=1, ds/dt=2/3. When t=5, there are 10 cm³.
This means the side length is the cube root(10).
This says ds/dt = 2/(3*cuberoot²(10)), or (2/3)(10^(-2/3)).

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