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 $x = t + 4 \ln t, y = t + \frac{9}{t}$ for $t > 0 .$

The curve has one stationary point. Find the $y$ -coordinate of this point and determine whether it is a maximum or a minimum point.

Accepted Answer

Given, the parametric equations of a curve are $x = t + 4 \ln t,$ , and $y = t + \frac{9}{t} for t > 0$ .

We know that the stationary point is a point on a curve where the gradient is zero.

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 $x = t + 4 \ln t$

$\Rightarrow \frac{d x}{d t} = \frac{d}{d t} (t + 4 \ln t)$

$= \frac{d}{d t} (t) + 4 \frac{d}{d t} (\ln t)$

$= 1 + \frac{4}{t} . . . . (i)$ $[∵ \frac{d}{d t} (\ln t) = \frac{1}{t}]$

And, $y = t + \frac{9}{t}$

$\Rightarrow \frac{d y}{d t} = \frac{d}{d t} (t + 9 t^{- 1})$

$= 1 + 9 \times (- t^{- 2})$ $[∵ \frac{d}{d t} (t^{n}) = n t^{n - 1}, n \in R]$

$= 1 - \frac{9}{t^{2}} . . . . . (i i)$

Now, applying the chain rule, we have

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

$= (1 - \frac{9}{t^{2}}) \times (\frac{t}{t + 4})$ [Using $(i)$ , and $(i i)$ ]

$= \frac{t^{2} - 9}{t (t + 4)} . . . . . (i i i)$

Now, $\frac{d y}{d x} = 0$

$\Rightarrow \frac{t^{2} - 9}{t (t + 4)} = 0$

$\Rightarrow t^{2} - 9 = 0$

$\Rightarrow t = \pm 3$ .

At $t = 3$ , we have

$x = 3 + 4 \ln 3$ , and $y = 3 + \frac{9}{3} = 3 + 3 = 6$

Now, let us find $\frac{d^{2} y}{d x^{2}} = \frac{d^{2} y}{d t^{2}} \times \frac{d^{2} t}{d x^{2}}$

Since $\frac{d y}{d t} = 1 - \frac{9}{t^{2}}$

$\Rightarrow \frac{d^{2} y}{d t^{2}} = \frac{d}{d t} (1 - 9 t^{- 2})$

$= \frac{d}{d t} (1) - 9 \frac{d}{d t} (t^{- 2})$

$= 0 - 9 \times (- 2 t^{- 3})$

$= 18 t^{- 3} . . . . . (i v)$

Also, $\frac{d^{2} x}{d t^{2}} = \frac{d}{d t} (\frac{d x}{d t})$

$= \frac{d}{d t} (1 + 4 t^{- 1})$

$= \frac{d}{d t} (1) + 4 \frac{d}{d t} (t^{- 1})$

$= 0 + 4 \times (- t^{- 2})$

$= - \frac{4}{t^{2}} . . . . (v)$

Hence, $\frac{d^{2} y}{d x^{2}} = \frac{d^{2} y}{d t^{2}} \times \frac{d^{2} t}{d x^{2}}$

$= - 18 t^{- 3} \times (\frac{1}{- \frac{4}{t^{2}}})$

$= \frac{9}{2} t^{- 1}$

And, the value of $\frac{d^{2} y}{d x^{2}}$ at $t = 3$ is $\frac{3}{2}$ , which is greater than $0$ .

Hence, $t = 3$ or $y = 6$ is a point of minima.

The parametric equations of a curve are $x = t + 4 \ln t, y = t + \frac{9}{t}$ for $t > 0 .$

The curve has one stationary point. Find the $y$ -coordinate of this point and determine whether it is a maximum or a minimum point.

Important Questions on Differentiation

Learn and Prepare for any exam you want !

The parametric equations of a curve are x=t+4ln t, y=t+9t for t>0. The curve has one stationary point. Find the y -coordinate of this point and determine whether it is a maximum or a minimum point.

Important Questions on Differentiation

Learn and Prepare for any exam you want !

The parametric equations of a curve are $x = t + 4 \ln t, y = t + \frac{9}{t}$ for $t > 0 .$

The curve has one stationary point. Find the $y$ -coordinate of this point and determine whether it is a maximum or a minimum point.