If $x$ and $y$ are connected parametrically by the given equation, then without eliminating the parameter, find $\frac{d y}{d x}$ .
$x = a (cos θ + θ sin θ), y = a (sin θ - θ cos θ)$

Open in App

Solution

The given equations are $x = a (cos θ + θ sin θ), y = a (sin θ - θ cos θ)$
Then, $\frac{d x}{d θ} = a [\frac{d}{d θ} cos θ + \frac{d}{d θ} (θ sin θ)] = a [- sin θ + θ \frac{d}{d θ} (sin θ) + sin θ \frac{d}{d θ} (θ)]$
$= a [- sin θ + θ cos θ + sin θ] = a θ cos θ$
& $\frac{d y}{d θ} = a [\frac{d}{d θ} (sin θ) - \frac{d}{d θ} (θ cos θ)] = a [cos θ - (θ \frac{d}{d θ} (cos θ) + cos θ . \frac{d}{d θ} (θ))]$
$= a [cos θ + θ sin θ - cos θ]$
$= a θ sin θ$
$∴ \frac{d y}{d x} = \frac{(\frac{d y}{d θ})}{(\frac{d x}{d θ})} = \frac{a θ sin θ}{a θ cos θ} = tan θ$

Suggest Corrections

Similar questions

Find $\frac{d y}{d x}$ , if x and y are connected parametrically by the equations given in questions without eliminating the parameter.

$x = a (c o s θ + θ s i n θ), y = a (s i n θ - θ c o s θ)$

x

y

\frac{d y}{d x}

x = a cos θ, y = b cos θ

If x and y are connected parametrically by the equation, without eliminating the parameter, find.

x = a cos θ, y = b cos θ

Find $\frac{d y}{d x}$ , if x and y are connected parametrically by the equations given in questions without eliminating the parameter.

$x = a c o s θ, y = b c o s θ$

x

y

\frac{d y}{d x}

x = a sec θ, y = b tan θ

Join BYJU'S Learning Program