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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 #3

Calculus - Calculus #3

Expert: Scotto - 11/5/2009

Question

Problem #3

See the attachment - I have started this and I'm stuck on where to go from what I got.

Thanks!

Answer
The length of the rod is 1200 cm.
The radius of the wheel is 40 cm.

Draw a ling (or picture a line) from A that is perpendicular with the x-axis.
The length of this vertical line is 40•sin(Θ).
The length on the x axis is 40•cos(Θ).

Now 40•sin(Θ)/1200 is the sin(α), so α = arcsin(40•sin(Θ)/1200).

Call the place straight below A on the x-axis B.
The length from B to P is 1200•cos(α).

Thus, the entire length of from 0 to P is 40•sin(Θ) + 1200•cos(arcsin(40•sin(Θ)/1200)).

To answer a, note that it is moving at 360 revolutions per minute.
This is the same as 6 revolutions per second.
In radians per second, multiply (6 rev/sec) by 2π rad/rev and get 12•π•rad/sec.

This is the answer to b.

The answer to c is dP/dΘ, which has a chain rule.
In this case we have f(g(h(Θ))) where f = cos(), g = arcsin(), and h=sin().

Note to do this differential, the derivative of cos(x) is -sin(x)
where x = arcsin(40•sin(Θ)/1200).
The derivative of arcsin(x) = sqrt(1/(1-x²)) where x = 40•sin(Θ)/1200.
The derivative of sin(Θ) is cos(Θ).

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