The parametric equations of a curve are
\(x = e^{-t} \cos t, \quad y = e^{-t} \sin t.\)
Show that \(\frac{dy}{dx} = \tan \left( t - \frac{1}{4} \pi \right).\)
Solution
First, find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) using the product rule:
\(\frac{dx}{dt} = e^{-t} \sin t - e^{-t} \cos t\)
\(\frac{dy}{dt} = e^{-t} \cos t - e^{-t} \sin t\)
Now, find \(\frac{dy}{dx}\) using \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\):
\(\frac{dy}{dx} = \frac{e^{-t} \cos t - e^{-t} \sin t}{e^{-t} \sin t - e^{-t} \cos t}\)
Simplify the expression:
\(\frac{dy}{dx} = \frac{\cos t - \sin t}{\sin t - \cos t}\)
\(\frac{dy}{dx} = \frac{\sin t - \cos t}{\cos t - \sin t}\)
Express \(\frac{dy}{dx}\) in terms of \(\tan t\):
\(\frac{dy}{dx} = \tan \left( t - \frac{1}{4} \pi \right)\)
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