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Calculus/Calculus #3

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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).

Scotto

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