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 = e^{2 t}, y = 1 + 2 t e^{t} .

Find the equation of the normal to the curve at the point where

t = 0 .

Accepted Answer

Given, the parametric equations of a curve are $x = e^{2 t},$ and $y = 1 + 2 t e^{t}$ .

We need to find the equation of the normal to the curve at the point where $t = 0$ .

At $t = 0, x = e^{2 (0)} = 1,$ and $y = 1 + 2 (0) e^{0} = 1$

We know that the gradient of the normal to the curve $y = f (x)$ at the point $(x_{1}, y_{1})$ is given by $- \frac{1}{{\frac{d y}{d x}]}_{(x_{1}, y_{1})}}$

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 $x = e^{2 t}$

$\Rightarrow \frac{d x}{d t} = \frac{d}{d t} (e^{2 t})$

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

$= 2 e^{2 t} . . . . . (i)$ $[∵ \frac{d}{d t} (t^{n}) = n t^{n - 1}, n \in R]$

And, $y = 1 + 2 t e^{t}$

$\Rightarrow \frac{d y}{d t} = \frac{d}{d t} (1 + 2 t e^{t})$

$= \frac{d}{d t} (1) + 2 \frac{d}{d t} (t e^{t})$

$= 0 + 2 [t \frac{d}{d t} (e^{t}) + e^{t} \frac{d}{d t} (t)]$ $[∵ \frac{d}{d x} (u v) = u \frac{d v}{d x} + v \frac{d u}{d x}]$

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

Now, applying the chain rule, we have

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

$= (2 t e^{t} + 2 e^{t}) \times (\frac{1}{2 e^{2 t}})$ [Using $(i)$ , and $(i i)$ ]

$= \frac{(t + 1)}{e^{t}}$ $. . . . . (i i i)$

Now, we need to find the value of $\frac{d y}{d x}$ at $t = 0$

Thus,

${\frac{d y}{d x}]}_{(1, 1)} = \frac{(0 + 1)}{e^{0}} = 1$

Hence, the gradient of the normal to the given curve at the point where $t = 0$ is $- 1$ .

Now, at point $t = 0$ , we have $(x, y) = (1, 1)$

Hence, the equation of the normal to the given curve at the point $(1, 1)$ with gradient $- 1$ is given by

$(y - 1) = - 1 (x - 1)$

$\Rightarrow y - 1 = - x + 1$

$\Rightarrow x + y = 2$ which is the required equation of the normal.

The parametric equations of a curve are $x = e^{2 t}, y = 1 + 2 t e^{t} .$ Find the equation of the normal to the curve at the point where $t = 0 .$

Important Questions on Differentiation

Learn and Prepare for any exam you want !

The parametric equations of a curve are x=e2t, y=1+2tet. Find the equation of the normal to the curve at the point where t=0.

Important Questions on Differentiation

Learn and Prepare for any exam you want !

The parametric equations of a curve are $x = e^{2 t}, y = 1 + 2 t e^{t} .$ Find the equation of the normal to the curve at the point where $t = 0 .$