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

Question

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:

Answer 1

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

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

which leads to

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

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

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Responding to comments:

$$ \theta (t) = \cot^{-1}\left(\frac{a-vt}{b}\right)$$

With $d(\cot^{-1}x)/dx = -1/(1+x^2)$,

$$\frac{d\theta}{dt} = \frac{\frac{v}{b}}{1+\left(\frac{a-vt}{b}\right)^2}= \frac{vb}{b^2+ (a-vt)^2}=\frac{vb}{L^2}=\frac{v}{L}\sin\theta$$

Answer 2

It's okay to not do inverse trig for this problem, we can stick with functions we do know how to take derivatives of. Apply chain rule to both sides of the equation. You'll get an L' term that doesn't appear in the final solution. Is there another relationship you can find in the picture that relates L' to other quantities?