How to find $\frac{dP}{dt}$ when $P=r\cos t$ and $r\sin t =2te^r$ (Boas, 3ed, Chapter 4, Problem 7.2)

Question

We want to find $\frac{dP}{dt}$ (not $\frac{\partial P}{\partial t}$), with $$P=r\cos t$$ $$r\sin t =2te^r$$

The book usually takes the differentials like this:

$$ dP = \cos t\,dr-r\sin t\,dt$$ $$ \sin t\,dr+r\cos t\,dt = 2e^r\,dt+2te^r\,dr $$

Then eliminating $dr$ (with help from Mathematica) gives:

$$ \frac{dP}{dt}= \frac{2e^r\cos t - r + 2r\,t\,e^r\sin t} {\sin t-2te^r} ~~~\textrm{(wrong)}$$

Boas gives a very different answer: $$ \frac{dP}{dt}= \frac{2e^r\cos t-r+r^2\sin^2t} {(1-r)\sin t} ~~~\textrm{(correct)} $$

EDIT after getting 2 answers (thanks!): I plugged in different $(r,t)$ values in the two formulas and got different answers... but then $r$ and $t$ are not independent.

Answer 1

Using $2te^r=r\sin t$ in, you will have the same answer.

Answer 2

Both formulas (correct) and (wrong) are the same. Just use $2te^r = r\sin(t)$ so $2rte^r \sin(t) = r^2 \sin^2(t)$