Question

Find dy/dx x=a(cost +sint) , y=a(sint-cost)​

Answer 1

Answer:

$\begin{aligned} \frac{dy}{dx} &= \frac{\cos(t) + \sin(t)}{\cos(t) - \sin(t)} \end{aligned}$ given that $a \ne 0$ and that $\cos(t) - \sin(t) \ne 0$.

Step-by-step explanation:

The relation between the $y$ and the $x$ in this question is given by parametric equations (with $t$ as the parameter.)

Make use of the fact that:

$\begin{aligned} \frac{dy}{dx} = \quad \text{ \frac{dy/dt}{dx/dt} given that \frac{dx}{dt} \ne 0 } \end{aligned}$.

Find $\begin{aligned} \frac{dx}{dt} \end{aligned}$ and $\begin{aligned} \frac{dy}{dt} \end{aligned}$ as follows:

$\begin{aligned} \frac{dx}{dt} &= \frac{d}{dt} [a\, (\cos(t) + \sin(t))] \\ &= a\, (-\sin(t) + \cos(t)) \\ &= a\, (\cos(t) - \sin(t))\end{aligned}$.

$\begin{aligned} \frac{dx}{dt} \ne 0 \end{aligned}$ as long as $a \ne 0$ and $\cos(t) - \sin(t) \ne 0$.

$\begin{aligned} \frac{dy}{dt} &= \frac{d}{dt} [a\, (\sin(t) - \cos(t))] \\ &= a\, (\cos(t) - (-\sin(t))) \\ &= a\, (\cos(t) + \sin(t))\end{aligned}$.

Calculate $\begin{aligned} \frac{dy}{dx} \end{aligned}$ using the fact that $\begin{aligned} \frac{dy}{dx} = \text{ \frac{dy/dt}{dx/dt} given that \frac{dx}{dt} \ne 0 } \end{aligned}$. Assume that $a \ne 0$ and $\cos(t) - \sin(t) \ne 0$:

$\begin{aligned} \frac{dy}{dx} &= \frac{dy/dt}{dx/dt} \\ &= \frac{a\, (\cos(t) + \sin(t))}{a\, (\cos(t) - \sin(t))} \\ &= \frac{\cos(t) + \sin(t)}{\cos(t) - \sin(t)}\end{aligned}$.