Question

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

Answer 1

`\begin{aligned} (dy)/(dx) &= (\cos(t) + \sin(t))/(\cos(t) - \sin(t)) \end{aligned}` given that
`a \\e 0` and that
`\cos(t) - \sin(t) \\e 0`.

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} (dy)/(dx) = \quad \text{$(dy/dt)/(dx/dt)$ given that $(dx)/(dt) \\e 0$} \end{aligned}`.

Find
`\begin{aligned} (dx)/(dt) \end{aligned}` and
`\begin{aligned} (dy)/(dt) \end{aligned}` as follows:

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

`\begin{aligned} (dx)/(dt) \\e 0 \end{aligned}` as long as
`a \\e 0` and
`\cos(t) - \sin(t) \\e 0`.

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

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

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