Question

If x = f(t) & y = g(t) are diff f^n of t so that y is diff f^n of x & $\frac{dx}{dt} \neq 0$ then PT $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ Hence find $\frac{dy}{dx}$ if x = sint & y = cost

Answer 1

$\frac{dy}{dx} = -\tan t$

Answer 2

Here's the solution to the problem:

Part 1: Proof

Given that $x = f(t)$ and $y = g(t)$ are differentiable functions of $t$, and $y$ is a differentiable function of $x$. We are also given that $\frac{dx}{dt} \neq 0$.

Since $y$ is a differentiable function of $x$, and $x$ is a differentiable function of $t$, by the chain rule for functions of a single variable, we have: $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$ Given that $\frac{dx}{dt} \neq 0$, we can divide both sides of the equation by $\frac{dx}{dt}$: $\frac{dy/dt}{dx/dt} = \frac{dy}{dx}$ Rearranging the terms, we get: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ This proves the required formula for the derivative of parametric functions.

Part 2: Application

Given the parametric equations: $x = \sin t$ $y = \cos t$

We need to find $\frac{dy}{dx}$. First, we find the derivatives of $x$ and $y$ with respect to $t$: $\frac{dx}{dt} = \frac{d}{dt}(\sin t) = \cos t$ $\frac{dy}{dt} = \frac{d}{dt}(\cos t) = -\sin t$ Now, using the formula $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ (assuming $\frac{dx}{dt} = \cos t \neq 0$): $\frac{dy}{dx} = \frac{-\sin t}{\cos t}$ $\frac{dy}{dx} = -\tan t$

Thus, if $x = \sin t$ and $y = \cos t$, then $\frac{dy}{dx} = -\tan t$.