Name: Rate of Change of Function: Example
Uploaded: 2019-07-19
Duration: 3 min 57 s
Description: In this video, we will use the rate of change to find the speed of the lower end of the ladder when it is sliding on the ground. Watch this video to know more.

Question

A particle moves along the curve

y = 2 x^{3} + 1

. Find the points on the curve at which the

y

-coordinate changes six times as fast as the

x

-coordinate.

Accepted Answer

For the given curve

$y = 2 x^{3} + 1$

On differentiating both sides with respect to time, we get

$\frac{d y}{d t} = \frac{d}{d t} (2 x^{3} + 1)$

$\Rightarrow \frac{d y}{d t} = 6 x^{2} \frac{d x}{d t} . . . . . . . (i)$

$∵$ The $y$ -coordinate is changes six times as fast as the $x$ -coordinate

$∴ \frac{d y}{d t} = 6 \frac{d x}{d t}$

On putting $\frac{d y}{d t} = 6 \frac{d x}{d t}$ in equation $(i)$ , we get

$6 \frac{d x}{d t} = 6 x^{2} \frac{d x}{d t}$

$\Rightarrow 1 = x^{2}$

$\Rightarrow x = \pm 1$

On putting the values of $x$ in the given curve, we get

For $x = - 1$

$\Rightarrow y = 2 {(- 1)}^{3} + 1 = - 1$

For $x = 1$

$\Rightarrow y = 2 {(1)}^{3} + 1 = 3$

Hence, there are two points on the curve which is $(- 1, - 1)$ and $(1, 3)$ .

A particle moves along the curve $y = 2 x^{3} + 1$ . Find the points on the curve at which the $y$ -coordinate changes six times as fast as the $x$ -coordinate.

Important Questions on Application of Derivatives

Learn and Prepare for any exam you want !

A particle moves along the curve y=2x3+1. Find the points on the curve at which the y-coordinate changes six times as fast as the x-coordinate.

Important Questions on Application of Derivatives

Learn and Prepare for any exam you want !

A particle moves along the curve $y = 2 x^{3} + 1$ . Find the points on the curve at which the $y$ -coordinate changes six times as fast as the $x$ -coordinate.