Solving the Equation 29θ - 17.7cosθ = 0

[jaredmt]

Homework Statement

29θ - 17.7cosθ = 0
find θ

The Attempt at a Solution

is there a specific way to do this without a calculator?

 

[cepheid]

one thing I can think of...differentiate both sides.

edit: well, even then, if you want an actual numerical answer for theta, I guess you'd need a calculator, unless you can do inverse trig functions in your head or use a table.

 

[gabbagabbahey]

one thing I can think of...differentiate both sides.

edit: well, even then, if you want an actual numerical answer for theta, I guess you'd need a calculator, unless you can do inverse trig functions in your head or use a table.

Hmmm... I don't think that taking the derivative is especially helpful here... Taking the derivative of 29 \theta -17.7Cos(\theta)=0 gives you the same thing as taking the derivative of 29 \theta -17.7Cos(\theta)=65967575865.79999678+68768i. Just because a value of \theta satisfies the equation f'(\theta)=0 does not necessarily mean that the same value of \theta satisfies the equation f(\theta)=0. After all, the anti derivative is only unique up to a constant. (i.e. f'(\theta)=0 \Rightarrow f(\theta)=constant)

One useful method for estimating the root of the equation does involve derivatives however. If you plug pi/6 into the equation f(\theta)=29 \theta -17.7Cos(\theta) you will see that f(pi/6) is pretty close to zero. You could then taylor series expand the function about the point pi/6 and keep say the first 2 or 3 terms of the expansion, then find the root closest to pi/6 of the resulting polynomial.

Other than that, I would just plug it into your calculator.

 

[jaredmt]

ok sorry, i can use a calculator for finding sines and cosines but i just meant that i can't plug the equation directly and get the answer like that. only because my calculator isn't sophisticated enough for that haha

 

[gabbagabbahey]

Yes, taylor expanding the function to second order should give you a quadratic which you can solve with the help of a basic calculator, and give you a good estimate for theta (in fact, in this case exanding to second order gives a value of theta that is accurate to 6 decimal places!)

 

[jaredmt]

wow, did you know this from experience or something? i would never have thought of using taylor series for anything other than programming applications. Thanks for the help, I'm going to try this and see how it works out

 

[gabbagabbahey]

We use taylor series approximations a lot in physics. They are very useful for getting approximate solutions for complicated functions.

 

[cepheid]

Taylor series occurred to me as well, but I wasn't sure if it was a good suggestion or not. As for the differentiation thing...yeah, I was being an idiot. Thanks for pointing that out gabbagabbahey.