Question

If x = a cos theta and y = b sin theta , find dy by dx

Answer 1

I'm denoting theta by z
so , x = acosz
now, dx/dz = -asinz .....eq (1)
and y = bsinz
so, dy/dz = bcosz ......eq(2)

Now, dividing eq(2) by eq (1)
we get, dy/dx = -b/a cotz. ( z for theta)

Answer 2

The value of dy by dx $\bold{=\frac{-b \cot \theta}{a}}$.

Given data suggest that there are two equation with x and y, which is to be differentiated as dy with respect to dx.  

The given equation as x = a cos θ and y = b sin θ. The differentiation as $\frac{d y}{d x}$ is given by differentiating individual by θ. Thereby, $\frac{d x}{d \theta}=\frac{d}{d \theta(a \cos \theta)}$.  

$\begin{array}{l}{\Rightarrow \frac{\mathrm{d} \mathrm{x}}{\mathrm{d} \theta}=\frac{\mathrm{a} \mathrm{d}}{\mathrm{d} \theta \cos \theta}} \\ \\ {\Rightarrow \frac{\mathrm{d} \mathrm{x}}{\mathrm{d} \theta}=\mathrm{a}(-\sin \theta)} \\ \\{\Rightarrow \frac{\mathrm{d} \mathrm{x}}{\mathrm{d} \theta}=-\mathrm{a} \sin \theta}\end{array}$.

Similarly, $\frac{d y}{d \theta}=\frac{d}{d \theta(b \sin \theta)}$  

$\Rightarrow \frac{\mathrm{d} \mathrm{y}}{\mathrm{d} \theta}=\mathrm{b} \cos \theta$

$\Rightarrow \text { Now, we have } \frac{\mathrm{d} \mathrm{x}}{\mathrm{d} \theta} \text { and } \frac{\mathrm{dy}}{\mathrm{d} \theta}$

So, $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\left(\frac{\mathrm{d} \mathrm{y}}{\mathrm{d} \theta}\right)}{\left(\frac{\mathrm{d} \mathrm{x}}{\mathrm{d} \theta}\right)}$

$\begin{array}{l}{\Rightarrow \frac{d y}{d x}=\frac{(b \cos \theta)}{(-a \sin \theta)}} \\ \\{\Rightarrow \frac{d y}{d x}=\frac{-b \cot \theta}{a}}\end{array}$