If \(x = f(t)\) and \(y = g(t)\) are differentiable functions of \(t\), so that \(y\) is a function of \(x\) and \(\frac{dx}{dt} \ne 0\), then prove that \(\frac{dy}{dx} = \frac{\left(\frac{dy}{dt}\right)}{\left(\frac{dx}{dt}\right)}\)Hence find \(\f…

Answer: Proof as below; for \(y = at^2,\ x = 2at\), \(\frac{dy}{dx} = t\).

Let \(\delta x\) and \(\delta y\) be the increments in \(x\) and \(y\) corresponding to a small increment \(\delta t\) in \(t\). Since \(x = f(t)\) is differentiable, it is continuous, so \(\delta x \to 0\) as \(\delta t \to 0\). Also \(\delta x \ne 0\) for small \(\delta t\) since \(\frac{dx}{dt} \ne 0\).

Now \(\frac{\delta y}{\delta x} = \frac{\delta y/\delta t}{\delta x/\delta t}\).

Taking limits as \(\delta t \to 0\) (so that \(\delta x \to 0\)):

\(\lim_{\delta x \to 0}\frac{\delta y}{\delta x} = \frac{\lim_{\delta t \to 0}\frac{\delta y}{\delta t}}{\lim_{\delta t \to 0}\frac{\delta x}{\delta t}}\)

Since \(x\) and \(y\) are differentiable functions of \(t\) and \(\frac{dx}{dt} \ne 0\), the limits on the right exist, so \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\).

Hence: \(y = at^2 \Rightarrow \frac{dy}{dt} = 2at\); \(x = 2at \Rightarrow \frac{dx}{dt} = 2a \ne 0\).

\(\therefore \frac{dy}{dx} = \frac{2at}{2a} = t\).