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The related rates problem is a problem in which we know one of the rates of change at a given instant or given time, say dx/dt which is x with respect to time 't' and dy/dt which is 'y' with respect to time 't'.Another important use of the Chain Rule is to find the rates of change of two or more related variables that are changing with respect to time.

For example, when water is drained out from a cylindrical drum, the volume 'V', the radius 'r' and height 'h' are all function of time 't'.

$V = \pi r^{2}h$

Differentiate implicitly with respect to time to find the related rates,

$\frac{\text{d} }{\text{d}t}[V] =\frac{\text{d}}{\text{d}t}[\pi r^{2} h]$

$\frac{\text{d} }{\text{d}t}[V] =\pi\frac{\text{d}}{\text{d}t}[\pi r^{2} h]$

$\frac{\text{d} }{\text{d}t}[V] =\pi[2rh\frac{\text{dr}}{\text{d}t}+ r^{2}\frac{\text{d}h}{\text{d}t}]$

This equation is of the rate of change of volume 'V' is related to the rate of change of radius and height.

$\frac{\text{d}x}{\text{d}t}$ = 20 cm/sec ( given )

$ V = x^{3}$

Now differentiate the above equation with respect to 't'

$\frac{\text{d}}{\text{d}t}(v) = 3x^{2}.\frac{\text{d}x}{\text{d}t} $

$\frac{\text{d}v}{\text{d}t} = 3x^{2}.(20)$

$\frac{\text{d}v}{\text{d}t} = 60x^{2}$

As length of a cube = x = 6 cm

$\frac{\text{d}v}{\text{d}t} = 60(6)^{2}$

$\frac{\text{d}v}{\text{d}t}= 2,160 cm^{3}/sec $

Thus the volume of the cube is increasing at the rate of $2,160 cm^{3}/sec $ when the length of the cube is 6 cm.

Rate of change of y co-ordinate = 8 (rate of change of x co-ordinate)

$\frac{\text{d}y}{\text{d}t} = 8 $\frac{\text{d}x}{\text{d}t}

We have, 6y = $x^{3} $ + 2

Now differentiate the above equation with respect to time 't'

6$\frac{\text{d}y}{\text{d}t} = 3x^{2}\frac{\text{d}x}{\text{d}t} $

6 (8 $\frac{\text{d}x}{\text{d}t}) = 3x^{2}\frac{\text{d}x}{\text{d}t} $

48 $\frac{\text{d}x}{\text{d}t}) = 3x^{2}\frac{\text{d}x}{\text{d}t} $

48 = 3$x^{2}$

$x^{2}$ = 48/3

$x^{2}$ = 16

$x = \pm $4

When x = 4 $\Rightarrow 6y = 4^{3}$ + 2 = 66

6y = 66

y = 11

and when x = -4 $\Rightarrow 6y =(-4)^{3} + 2 = -64 + 2 = -62 $ 6y = -62

y = -31/3

So, the required points are (4, 11) and (-4, 31/3).

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