If $x=t^2\sin3t$ and $y=t^2\cos3t$, find $\frac{dy}{dx}$ in terms of $t$

Question

If $x=t^2\sin3t$ and $y=t^2\cos3t$, find $\frac{dy}{dx}$ in terms of $t$. This is how I tried solving it:

$$ \frac{dx}{dt} = 2t\sin3t + 3t^2\cos3t \\ \frac{dy}{dt} = 2t\cos3t - 3t^2\sin3t \\ \frac{dy}{dx} = \frac{2t\cos3t - 3t^2\sin3t}{2t\sin3t + 3t^2\cos3t} $$

But the answer listed is:

$$ \frac{2-3t\tan3t}{2\tan3t+3t} $$

Is my answer incorrect, or can I simplify it even more?

Answer 1

You are on the right trace. Just divide $t\cos 3t$ in the numerator and denominator of $\frac{dy}{dx}$.

Answer 2

Your answer and the given answer are the same: just divide the numerator and denominator of your answer by $t\cos 3t$ to find the other answer.

Answer 3

Your answer is correct, $\frac{2t\cos3t - 3t^2\sin3t}{2t\sin3t + 3t^2\cos3t}=\frac{t\cos{3t}}{t\cos{3t}}\frac{2-3t\frac{\sin{3t}}{\cos{3t}}}{2\frac{\sin{3t}}{\cos{3t}}+3t}=\frac{2-3t\tan3t}{2\tan3t+3t}$

Answer 4

What you did is entirely correct. To show that it is equivalent to the second solution, divide numerator and denominator by $t \cos 3t$: $$\frac{dy}{dx} = \frac{2t\cos3t - 3t^2\sin3t}{2t\sin3t + 3t^2\cos3t}\cdot \frac{\frac 1{t\cos 3t}}{\frac{1}{t\cos 3t}} = \frac{2-3t\tan 3t}{t\tan 3t+ 3t}$$