# Question applying Inverse trigonometry to show ${d\theta\over{dt}}={v\sin\theta\over L}$

There is a car moving along a straight road at the speed of $$v$$. There is a tree with the distance of $$L(t)$$ from the car. The angle the car is facing towards the tree is $$\theta(t)$$. Show that the following equation holds true: $${d\theta\over{dt}}={v\sin\theta\over L}$$

(This question is in another language from my book, so if the question isn't clear enough due to my bad translation, please ask for additional details in the comments)

I tried to use inverse trigonometric functions to solve this question since the book was talking about inverse trigonometric functions right before this question. I came up with this equation: $$\cos\theta(t) = {a-vt\over{L(t)}}$$ where $$a=L(t_{0})\cos\theta(t_{0})$$ and $$t_{0}$$ is the time when the car started moving towards the tree. Then, $$\theta(t)=\arccos{a-vt\over{L(t)}}$$ Therefore, $${d\theta\over{dt}}={d\over{dt}}\arccos{a-vt\over{L(t)}}$$ With intuition I thought that the above equation came pretty close to the question's given equation. But this is the point where I met a brick wall because I have no idea how to get the derivative of this particular $$\arccos$$. I learned about the inverse function theorem and have only dealt with much simpler functions like $$\arctan x$$.

Edit: The question didn't come with a picture or drawing, but this is what I imagine the picture would be like: • Could you draw the setup? – Botond Aug 28 '19 at 13:03
• @Botond Check out my edit - I tried my best to draw what I imagine the question would be talking about. Though I think my drawing is too bad to give any more insight to the question :( – linearAlg Aug 28 '19 at 13:16
• Nice picture. Like it! – Quanto Aug 28 '19 at 14:46

It is kind of cumbersome to work out the derivatives in your equation since there are three time-dependent variables $$\theta(t), t$$ and $$L(t)$$. Also, you want to avoid taking the derivative of the inverse function directly.

Instead, it'd be much easier to start with just two time-dependent variables $$\theta(t)$$ and $$t$$,

$$\cot \theta (t) = \frac{a-vt}{b}$$

where $$b$$ is the vertical distance to the tree, a constant.

Then, take the derivatives with $$d(\cot \theta)/d\theta = -1/\sin^2 \theta$$ to get

$$\frac{1}{\sin^2\theta}\frac{d\theta}{dt} = \frac{v}{b}$$

$$\frac{d\theta}{dt} = \frac{v}{b}\sin^2\theta = \frac{v}{L}\frac{L}{b}\sin^2\theta=\frac{v}{L}\sin\theta$$

where $$\sin\theta = b/L$$ is used.

$$\theta (t) = \cot^{-1}\left(\frac{a-vt}{b}\right)$$
With $$d(\cot^{-1}x)/dx = -1/(1+x^2)$$,
$$\frac{d\theta}{dt} = \frac{\frac{v}{b}}{1+\left(\frac{a-vt}{b}\right)^2}= \frac{vb}{b^2+ (a-vt)^2}=\frac{vb}{L^2}=\frac{v}{L}\sin\theta$$
• Wow that is such great insight to solving the question! Just one follow-up question though, could you get ${d\theta\over{dt}}$ using the inverse function theorem on arccot? Or is it not possible to get it by going down that route – linearAlg Aug 28 '19 at 14:32