How to write $r$ as a function of $\theta$?

Question

I have $$r(t) = 5^{1-t}13^t$$

and

$$\theta(t) = \arctan\left(\frac{4}{3}\right) (1-t) + \arctan\left(\frac{12}{5}\right)t$$

With $t$ going from $0$ to $1$.

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I want to rewrite the curve with $r$ as a function of $\theta$, so that $r(\theta)$ is the distance of the curve from the origin, at the angles in the range from $\theta=\arctan\left(\frac{4}{3}\right)\,\,$ to $\,\,\theta = \arctan\left(\frac{12}{5}\right)$.

I don't know how to do this, maybe something like $r(\theta(t))$ with chain rule? But I'm not taking any derivative yet so I don't think that applies...

I figured out that the start point of the curve is $(x,y) = (3,4)$ which is diagonally $5$ from the origin, and the end point of the curve is $(x,y) = (5,12)$ which is diagonally $13$ from the origin, they are pythagorean triples

Answer 1

You may solve for $t$ as a function of $\theta$

$$\theta(t) = \arctan\left(\frac{4}{3}\right) (1-t) + \arctan\left(\frac{12}{5}\right)t$$

$$ t=\frac {\theta- \arctan\left(\frac{4}{3}\right)}{ \arctan\left(\frac{12}{5}\right)-\arctan\left(\frac{4}{3}\right)}$$

Answer 2

Using $$\tan ^{-1}(a)\pm\tan ^{-1}(b)=\tan ^{-1}\left(\frac{a\pm b}{1\mp a b}\right)$$ you could simplify the result given in Mohammad Riazi-Kermani's answer to $$t=\frac{2 \left(\theta -\tan ^{-1}\left(\frac{4}{3}\right)\right)}{\tan ^{-1}\left(\frac{2016}{3713}\right)}$$ and rewrite $$r(t) = 5^{1-t}13^t=5\left(\frac{13}{5}\right)^t=5 \exp\left(t \log\left(\frac{13}{5}\right) \right)$$ that is to say $$r(\theta)=5 \exp\left(\frac{2 \left(\theta -\tan ^{-1}\left(\frac{4}{3}\right)\right)}{\tan ^{-1}\left(\frac{2016}{3713}\right)} \log\left(\frac{13}{5}\right) \right)$$