Find dy/dx given x = e^tsin t, y = e^tcos t
Find $\dfrac{dy}{dx}$ given $x = e^{t}\sin t$, $y = e^{t}\cos t$.
Explanation of Method
The logarithmic spiral, whose tangent makes a constant angle.
Step 1: Differentiate with respect to $t$
Product rule on each, with the exponential reproducing itself: $x'_{t} = e^{t}\sin t + e^{t}\cos t = e^{t}\bigl(\sin t+\cos t\bigr)$.
\[ y'_{t} = e^{t}\cos t - e^{t}\sin t = e^{t}\bigl(\cos t-\sin t\bigr). \]
Step 2: Form the ratio
The factor $e^{t}$ cancels — the spiral's growth does not affect its slope: $\frac{dy}{dx} = \frac{\cos t-\sin t}{\sin t+\cos t}$.
Step 3: Recognise a tangent
Dividing numerator and denominator by $\cos t$ and using the subtraction formula for the tangent: \[\frac{dy}{dx} = \frac{1-\tan t}{1+\tan t} = \frac{\tan\frac{\pi}{4}-\tan t}{1+\tan\frac{\pi}{4}\tan t} = \tan\!\left(\frac{\pi}{4}-t\right).\]
Step 4: Verification
\[t = 0:\quad (x,y) = (0,1),\quad \frac{dy}{dx} = \frac{1-0}{0+1} = 1 = \tan\frac{\pi}{4} \;\;\checkmark\]
\[t = \frac{\pi}{4}:\quad \frac{dy}{dx} = \tan 0 = 0 \text{ — horizontal tangent} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\frac{dy}{dx} = \dfrac{\cos t-\sin t}{\sin t+\cos t} = \tan\!\left(\dfrac{\pi}{4}-t\right)\,}\]
Remark
Remark. In polar form the curve is $\rho = e^{t}$ with polar angle $\tfrac{\pi}{2}-t$, the logarithmic spiral. Its defining property is that the tangent meets the radius vector at a constant angle — here $45^{\circ}$, which is why the answer is a tangent of $\tfrac{\pi}{4}-t$ rather than something more complicated. The spiral is equiangular, and the cancelling $e^{t}$ is the computational shadow of that self-similarity.