If $x=ae^{t}(\sin t+\cos t)$ and $y=ae^{t}(\sin t-\cos t)$, then prove that $\dfrac{dy}{dx}=\dfrac{x+y}{x-y}$.

CBSE Class 12 · 2019 · DifferentiationLong Answer

x = a e^{t} (sin t + cos t)

and

y = a e^{t} (sin t - cos t)

, then prove that

\frac{d y}{d x} = \frac{x + y}{x - y}

Answer: Proved

Differentiate each w.r.t.

t

(product rule on

e^{t} \cdot (\dots)

\frac{d x}{d t} = a e^{t} [(sin t + cos t) + (cos t - sin t)] = 2 a e^{t} cos t

\frac{d y}{d t} = a e^{t} [(sin t - cos t) + (cos t + sin t)] = 2 a e^{t} sin t

Hence

\frac{d y}{d x} = \frac{sin t}{cos t} = tan t

. Now observe:

x + y = 2 a e^{t} sin t, x - y = 2 a e^{t} cos t \Rightarrow \frac{x + y}{x - y} = tan t

\frac{d y}{d x} = \frac{x + y}{x - y}

CBSE Class 12 Board · 2019 · Subjective