Taylor polynomial of degree 1 - solve for theta

1. Feb 24, 2012

oates151

1. The problem statement, all variables and given/known data

I was given the following problem, but I am having a hard time interpreting what some parts mean.

We're given the equation

sinθ+b(1+cos^2(θ)+cos(θ))=0

Assume that this equation defines θ as a function, θ(b), of b near (0,0). Computer the Taylor polynomial of degree one for θ(b) near b=0

2. Relevant equations

sinθ+b(1+cos^2(θ)+cos(θ))=0

3. The attempt at a solution

Not sure exactly what they mean by θ(b).

2. Feb 24, 2012

LCKurtz

Replace b by x for a moment so you are thinking of $\theta$ as a function of x, say f(x). Can you write the Taylor series for a function f(x) near x = 0? Do that first. Then replace $f$ by $\theta$ and $x$ by $b$. There will be a couple of things you will need to calculate from your equation.

3. Feb 24, 2012

oates151

So I took the derivative with respect to b and got

dθ/db= (1+cos^2(θ)+cos(θ))/(2bcos(θ)sin(θ)+sin(θ)-cos(θ))

knowing the general form of the taylor series, I used the value of dθ/db at b=0 with θ=0 to get a value of -3 and plugged it into the tayor series general form

θ(b)=θ(0)+(-3)(b) to get θ(b)=-3b, the same value I got when I took the taylor polynomial of this function and solved for θ in another step.

Is this right?

4. Feb 24, 2012

SammyS

Staff Emeritus
That looks right to me, but there is a small error in your θ'(b) .

θ'(b) = (1+cos^2(θ)+cos(θ))/(2bcos(θ)sin(θ)+bsin(θ)-cos(θ))