related rates | ap calculus bc,12th grade

Related Rates

Covid-19 has led the world to go through a phenomenal transition .

E-learning is the future today.

Stay Home , Stay Safe and keep learning!!!

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.

Examples on Related Rates

Example 1 : An edge of a cube is increasing at the rate of 20 cm/sec. How fast the volume of the cube is increasing when the edge is 6 cm long?
Solution : Let the 'x' be the length of the cube and 'V' be its volume at any time 't'. Then,
$\frac{\text{d}x}{\text{d}t}$ = 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.

Example 2 : A particle moves along the curve, 6y = $x^{3} $ + 2. find the points on the curve at which the y-coordinate is changing 8 times as fast as the x- coordinate.
Solution : Let the required point be P(x,y)
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).

12th grade math


Covid-19 has affected physical interactions between people.

Don't let it affect your learning.