# DERIVATIVFES AS RATES OF CHANGE

A common use of rate of change is to describe the motion of an object moving in a straight line. In such problems, it is customary to use either a horizontal or a vertical line with a designated origin to represent the line of motion.

On such lines,

• movements in the forward direction considered to be in the positive direction
• movements in the backward direction is considered to be in the negative direction.

The function s(t) that gives the position (relative to the origin) of an object as a function of time t is called a position function.

It is denoted by s = f (t) .

The velocity and the acceleration at time t is denoted as

v(t) = ds/dt

a(t) = d2s/dt2

Notes :

Point 1 :

Speed is the absolute value of velocity regardless of direction and hence

speed = |v(t)| = |ds/dt|

Point 2 :

• When the particle is at rest then v(t) = 0 .
• When the particle is moving forward then v (t) > 0 .
• When the particle is moving backward then v(t) < 0 .
• When the particle changes direction, v(t) then changes its sign.

Point 3 :

If tc is the time point between the time points t1 and t2 (t1 < t< t1) where the particle changes direction then the total distance travelled from time t1 to time t2 is calculated as

|s(t1) - s(tc)| + |s(tc) - s(t2)|

Problem 1 :

The temperature T in celsius in a long rod of length 10 m, insulated at both ends is the function of length x given by

T = x(10 - x)

Prove that the rate of change of temperature at the midpoint of the rod is 0.

Solution :

T = x(10 - x) or T(x) = x (10 - x)

The length of the rod is 10 m, to find the temperature at the midpoint of the rod is at x = 5.

To find the derivative, we use the product rule.

T'(x) = x(0 - 1) + (10 - x )(1)

= - x + 10 - x

T'(x) = -2x + 10

Temperature at midpoint :

T'(5) = -2(5) + 10

= -10 + 10

T'(5) = 0

So, temperature at the midpoint of the rod is 0.

Problem 2 :

A person learnt 100 words for an English test. The number of words the person remembers in t days after learning is given by

W(t) = 100×(1− 0.1t)2 , 0  t 10

What is the rate at which the person forgets the words 2 days after learning?

Solution :

Number of word the person forgets after 2 days,

W'(2) = -20(1 − 0.1(2))

= -20(1 - 0.2)

= -20(0.8)

= -16

So, the number of words he forgets is 16.

Problem 3 :

A particle moves so that the distance moved is according to the law

s(t) = t3/3 - t2 + 3

At what time the velocity and acceleration is zero.

Solution :

s(t) = t3/3 - t2 + 3

To find velocity, we will find the first derivative.

s'(t) = 3t2/3 - 2t + 0

= t2 - 2t

When velocity = 0

t2 - 2t = 0

t(t - 2) = 0

t = 0 and t = 2

To find acceleration, we have to find the derivative of velocity function.

s'(t) = t2 - 2t

a(t) = s''(t) = 2t - 2

When acceleration = 0

2t - 2 = 0

t = 1

Problem 4 :

A particle is fired straight up from the ground to reach a height of s feet in t seconds where

s(t) = 128t - 16t2

i) Compute the maximum height of the particle reached ?

ii)  What is the velocity when the particle hits the ground ?

Solution :

s(t) = 128t - 16t2

i)  At the maximum height, the velocity v(t) of the particle is 0.

To find v(t), we have to find the derivative of s(t)

s'(t) = 128(1) - 16(2t)

v(t) = 128 - 32t

v(t) = 0

128 - 32t = 0

32t = 128

t = 128/32

t = 4

At 4 seconds, the particle reaches the maximum height.

Maximum height, when t = 4

s(t) = 128(4) - 16(4)2

= 512 - 16(16)

= 512 - 256

= 256 ft

ii) When the particle hits the ground, when s = 0

0 = 128t - 16t2

16t (8 - t) = 0

t = 0 and t = 8

It reaches the ground, when t = 0 and t = 8.

v(8) = 128 - 32(8)

= 128 - 256

= -128

The velocity when it reaches the ground is -128 ft/s

Problem 5 :

A particle moves along a horizontal line such that its position at any time t ≥ 0 is given by

s(t) = t3 − 6t2 + 9t + 1

where s is measured in meters and t in seconds?

(i) At what time the particle is at rest?

(ii) At what time the particle changes its direction?

(iii) Find the total distance travelled by the particle in the first 2 seconds.

Solution :

s(t) = t3 − 6t2 + 9t + 1

(i) When the particle is at rest, its velocity will become 0.

s(t) = t3 − 6t2 + 9t + 1

v(t) = 3t2 - 12t + 9 + 0

0 = 3t2 - 12t + 9

3(t2 - 4t + 3) = 0

3(t - 1)(t - 3) = 0

t = 1 and t = 3

(ii)

If 0 ≤ t < 1

v(t) = (t - 1)(t - 3)

say t = 0.5, then v(t) = + > 0

If 1 ≤ t < 3

say t = 2, then v(t) = - < 0

If t > 3

say t = 4, then v(t) = + > 0

Then the particle changes its direction in between 1 to 3 seconds.

(iii)  The total distance travelled by the particle from time t = 0 to t = 2 is given by,

= |s(0) − s(1)| + |s(1) − s(2)|

=|1− 5 | + | 5 − 3|

= 6 meters

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