Name: Derivative of Functions in Parametric Form
Uploaded: 2019-07-19
Duration: 8 min 37 s
Description: Learn everything you'll need to know about derivatives of functions in parametric forms with this quick and easy lesson that'll cover the concept comprehensi...

Question

The parametric equations of a curve are $x = \ln (\tan t), y = 2 \sin 2 t$ for $0 < t < \frac{π}{2} .$

Hence, show that at the point where $x = 0$ the tangent is parallel to the $x$ -axis.

Accepted Answer

Given, the parametric equations of a curve are $x = \ln (\tan t),$ and $y = 2 \sin 2 t$

We need to show that the tangent is parallel to the $x$ -axis at the point where $x = 0$ .

We know that when a curve is given in parametric form, in terms of the parameter $t$ we can use the chain rule to find $\frac{d y}{d x}$ in terms of $t$ .

Since $x = \ln (\tan t)$

$\Rightarrow \frac{d x}{d t} = \frac{d}{d t} (\ln (\tan t))$

$= \frac{1}{\tan t} \times \frac{d}{d t} (\tan t)$

$= \frac{\sec^{2} t}{\tan t} . . . . . (i)$ $[∵ \frac{d}{d t} (\tan t) = \sec^{2} t]$

And, $y = 2 \sin 2 t$

$\Rightarrow \frac{d y}{d t} = \frac{d}{d t} (2 \sin 2 t)$

$= 2 \frac{d}{d t} (\sin 2 t)$

$= 2 \cos 2 t \times \frac{d}{d t} (2 t)$

$= 4 \cos 2 t . . . . . (i i)$ $[∵ \frac{d}{d x} (x^{n}) = n x^{n - 1}]$

Now, applying the chain rule, we have

$\frac{d y}{d x} = \frac{d y}{d t} \times \frac{d t}{d x}$

$= (4 \cos 2 t) \times (\frac{1}{\frac{\sec^{2} t}{\tan t}})$ [Using $(i)$ , and $(i i)$ ]

$= 4 \cos 2 t \times (\frac{\tan t}{\sec^{2} t})$

$= 4 \cos 2 t \times (\frac{\sin t}{\cos t \times \frac{1}{\cos^{2} t}})$ $[∵ \tan t = \frac{\sin t}{\cos t}, \sec t = \frac{1}{\cos t}]$

$= 4 \cos 2 t \times \sin t \cos t$

$= 2 \cos 2 t \times (2 \sin t \cos t)$

$= 2 \sin 2 t \cos 2 t$ $[∵ \sin 2 A = 2 \sin A \cos A]$

$= \sin 4 t$

Hence, $\frac{d y}{d x} = \sin 4 t . . . . . (i i i)$ .

Now, when $x = 0$

$x = \ln (\tan t)$

$\Rightarrow \ln (\tan t) = 0$

$\Rightarrow \tan t = 1$

$\Rightarrow t = \frac{π}{4}$ .

Thus, we need to find the gradient of the tangent at $t = \frac{π}{4}$

$\frac{d y}{d x}]_{t = \frac{π}{4}} = \sin 4 (\frac{π}{4}) = sinπ = 0$

Since the gradient of any line parallel to the $x$ -axis is zero.

Hence, the tangent to the given curve at the point where $x = 0$ is parallel to the $x$ -axis.

The parametric equations of a curve are $x = \ln (\tan t), y = 2 \sin 2 t$ for $0 < t < \frac{π}{2} .$

Hence, show that at the point where $x = 0$ the tangent is parallel to the $x$ -axis.

Important Questions on Differentiation

Learn and Prepare for any exam you want !

The parametric equations of a curve are x=lntant, y=2sin2t for 0<t<π2. Hence, show that at the point where x=0 the tangent is parallel to the x -axis.

Important Questions on Differentiation

Learn and Prepare for any exam you want !

The parametric equations of a curve are $x = \ln (\tan t), y = 2 \sin 2 t$ for $0 < t < \frac{π}{2} .$

Hence, show that at the point where $x = 0$ the tangent is parallel to the $x$ -axis.