2

$$f(\theta) = \theta \sin(\theta)$$

Derived to:

$$f'(\theta) = \sin(\theta) + \theta \cos(\theta)$$

How do I solve it from here?

janmarqz
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user10984587
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  • Well, what's the generic equation for a tangent line to $f(\theta)=\theta \cdot \sin(\theta)$ at the point $(\theta, f(\theta))$? – Sinister Cutlass Nov 12 '15 at 19:25
  • Not entirely sure. Can't say my class has reviewed that yet. – user10984587 Nov 12 '15 at 19:30
  • The slope of the tangent to the graph of $f$ is given by the derivative of $f$. Hence you want to find the $\theta$ such that $\sin\theta + \theta\cos\theta = 1$. For instance $\theta = \pi/2$. Check for other solutions. – Simon S Nov 12 '15 at 19:31

1 Answers1

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The generic tangent line to f at the point $(\theta_{0},f(\theta_{0}))$ is $$T(\theta)=f(\theta_{0})+f'(\theta_{0})(\theta-\theta_{0})=\theta_{0}\cdot \sin(\theta_{0})+(\sin(\theta_{0})+\theta_{0}\cos(\theta_{0}))(\theta-\theta_{0})$$

You need $\theta_{0}$ such that $f(\theta_{0})-f'(\theta_{0})\theta_{0}=0$ and $f'(\theta_{0})=1$; that is, $f'(\theta_{0})=1$ and $f(\theta_{0})=\theta_{0}$. So, you need to solve: $$\sin(\theta_{0})+\theta_{0}\cos(\theta_{0})=1$$ and $$\theta_{0}\sin(\theta_{0})=\theta_{0}$$ for $\theta$. To satisfy the second equation, you clearly need $\theta=\frac{\pi}{2}+2n\pi$ for any integer $n$, since this is the set of angles which have a sine value of $1$. If you plug this set of angles into the first equation, you see that since each has a cosine value of $0$, the first equation automatically holds. Thus, your answer is the set $\{\frac{\pi}{2}+2n\pi|n\in \mathbf{Z}\}$... or just $\frac{\pi}{2}$ if you are restricting your answer to angles between $0$ and $2\pi$.

Sinister Cutlass
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