## Rate Of Change

Definition

If a quantity y depends on and varies with quantity x the rate of change of y with respect to x is dy/dx.

In other words,we can define rate-of-change as the ratio of the rate-of-change in the input values and rate-of-change in output values.

Rate of change can be positive or negative.

Now consider the above diagram to understand the term rate-of-change much better.Here the water level is raised as time increases. Let us consider "h" be the water level and "t" be time. So we can write the rate- of-change for this as dh/dt. Now let us see some example problems to understand this concept.

Example 1:

The length of a iron rod in meters at θ° C is given as L = 5 θ + 4 θ². Determine the rate-of-change of length when the temperature is 100° C and 200° C.

Solution:

Here "L" represents length of the iron rod and θ represents temperature. Since the length of the rod is changing when temperature changes. So the rate of change will be dL/dθ.

L = 5 θ + 4 θ²

Differentiating with respect to θ.

dL/dθ = 5 (1) + 4 (2 θ)

dL/dθ  = 5 + 8 θ

When θ = 100° C

dL/dθ = 5 + 8 (100)

= 5 + 800

= 805 m/° C

When  θ = 200° C

dL/dθ = 5 + 8 (200)

= 5 + 1600

= 1605 m/° C

Example 2:

Air is being pumped into a spherical balloon so that its volume increases at a rate of 200 cm³/s. How fast is the radius of the balloon increasing when the radius is 50 cm.

Solution:

Here volume is increasing 200 cm³/s as time increasing . So dv/ds = 200 cm³/s. Now we need to find increasing rate of radius as time increasing.

That is we need to find dr/ds.

Volume of sphere V = (4/3) Π r³

dv/ds = (4/3) Π (3r²) dr/ds

200 = (4/3) Π 3 (50)² dr/ds

200  = 4 Π (2500) dr/ds

(200/2500) = 4 Π (dr/ds)

(2/25 x 4Π) = dr/ds

1/50Π = dr/ds

dr/ds = (1/50 Π) cm³/s

 Questions Solution

 (1) The radius of a circular plate is increasing in length at 0.01 cm per second. What is the rate at which the area is increasing when the radius is 13 cm? Solution (2) A square plate is expanding uniformly each side is increasing at the constant rate of 1.5 cm/min. Find the rate at which  the area is increasing when the side is 9 cm. Solution (3) A stone thrown into still water causes a series of concentric ripples. If the radius of outer ripple is increasing at the rate of 5 cm/sec,how fast  is the area of the distributed water increasing when the outer most ripple has the radius of 12 cm/sec. Solution (4) The radius of a spherical balloon is increasing at the rate of 4 cm/sec. Find  the rate of increases of the volume and surface area when the radius is 10 cm. Solution (5)  A balloon which remains spherical is being inflated be pumping in 90 cm³/sec. Find the rate at which the surface area of the balloon is increasing when the radius is 20 cm. Solution

Related Topics

Rate of Change to First principal