First of all, I am not asking for help with homework - I aced all three semesters of calculus when I took it 120 years ago. It’s just that I haven’t used it since then.

Here’s the problem I’ve been pondering: describing the velocity and acceleration of a piston in a car. Using fairly standard notation, I’ve got a triangle with sides a, b and c; with angles A, B and C opposite the respective sides. “a” represents the rod length, “b” half of the crankshaft stroke, and “c” the position of the piston wrist pin. Angle “A” is the angle of the crankshaft.

The law of cosines says:

a^2 = b^2 + c^2 - 2*b*c*cosA

Rearranging terms:

0 = c^2 + c*(-2*b*cosA) + (b^2 - a^2)

Solving the quadratic:

c = ((-(-2*b*cosA)±((-2*b*cosA)^2 - 4*1*(b^2 - a^2))^.5)/(2*1)

(provided I got all of the parentheses right), or:

c = b*cosA ± (a^2 - b^2 + b^2*cosA*cosA)^.5

Taking the derivative with respect to time (indicated by “’”):

c’ = -b*sinA*A’ ± (.5*(a^2 - b^2 + b^2*cosA*cosA)^-.5)*b^2*2*cosA*-sinA*A’

or:

c’ = -b*sinA*A’ ± b^2*(a^2 - b^2 + b^2*cosA*cosA)^-.5)*cosA*-sinA*A’

(Here’s where I get doubtful):

c" = -b*(sinA*A" + cosA*A’*A’) ± b^2*((-.5*(a^2 - b^2 + b^2*cosA*cosA)^-1.5)*b^2*2*cosA*-sinA*A’) cosA-sinA*A’ + (a^2 - b^2 + b^2

*cosA*cosA)^-.5)

*-sinA*-sinA

*A’*cosA)^-.5)

*A’ + (a^2 - b^2 + b^2*cosA*cosA*-cosA

*A’*cosA)^-.5)

*A’ + (a^2 - b^2 + b^2*cosA*cosA*-sinA*A"

I’m only interested in constant RPM, so A" = 0:

c" = -b*cosA*A’*A’ ± b^2*((-.5*(a^2 - b^2 + b^2*cosA*cosA)^-1.5)*b^2*2*cosA*cosA*sinA*sinA*A’ A’ + ((a^2 - b^2 + b^2cosA*cosA)^-.5)

*sinA*sinA

*A’*cosA)^-.5)

*A’ + ((a^2 - b^2 + b^2*cosA*cosA*-cosA*A’*A’)

Well, did I get it right?