Question

If \( x = f(t) \) and \( y = g(t) \) are differentiable functions of t so that y is a differentiable function of x and \( \frac{dx}{dt} \neq 0 \), then prove that \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \). Hence find \( \frac{dy}{dx} \) if \( x = \sin t \) and \( y = \cos t \).

Hint:

Chain rule for parametric: \( \frac{dy}{dx} = \frac{y'(t)}{x'(t)} \).

Answer 1

Proof: Since \( y = g(t) \), \( x = f(t) \), \( \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} \).
\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}, \frac{dx}{dt} \neq 0. \] Hence: \( x = \sin t \), \( y = \cos t \).
\[ \frac{dx}{dt} = \cos t, \frac{dy}{dt} = -\sin t. \] \[ \frac{dy}{dx} = \frac{-\sin t}{\cos t} = -\tan t. \] Answer: Proved; \( \frac{dy}{dx} = -\tan t \).