[MATH 100] 2-Derivative as a Rate of Change

7/31/2019 [MATH 100] 2-Derivative as a Rate of Change

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The Derivative as Rate of Change

Mathematics 100

Institute of Mathematics

Math 100 (Inst. of Mathematics) The Derivative as Rate of Change 1 / 19
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Outline

1 The Derivative as slope of tangent line

2 The Derivative as instantaneous rate of change

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The Derivative as slope of tangent line

1. Find the equation of the tangent line to the curve y = x3 4 at the

point (2,4).

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point (2,4).Solution:

Solving for the derivative of the function, we have

dy

dx = 3x2.

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point (2,4).Solution:

Solving for the derivative of the function, we have

dy

dx = 3x2.

When x = 2,dy

dx= 3(2)2 = 12. So 12 is the slope of the tangent line at

the point (2,4). And using the point-slope form, the equation of the

tangent line is given by

y 4 = 12(x 2).

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2. Find an equation of the normal line to the curve y = 4x2 8x at

the point (2, 0).

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the point (2, 0).

Solution:

dy

dx= 8x 8.

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the point (2, 0).

Solution:

dy

dx= 8x 8.

When x = 2,dy

dx= 8(2) 8 = 8. The normal line is a perpendicular

line to the tangent line. So the slope of the normal line is 1

8. Using

the point-slope form, the equation is given by

y = 1

8(x 2).

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3. Find an equation of the line tangent to the curvey = 3x2 4x and perpendicular to the line 2x y+ 3 = 0.

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Solution:

dy

dx= 6x 4.

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Solution:

dy

dx= 6x 4.

The line 2x y+ 3 = 0 has slope m= 2. So take m1 = 1

2

to be the slope of the

line perpendicular to it. Hence

dy

dx= 6x 4 =

1

2 x =

7

12

y = 3

7

12

2 4

7

12

=

21

16.

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Solution:

dy

dx= 6x 4.

The line 2x y+ 3 = 0 has slope m= 2. So take m1 = 1

2

to be the slope of the

line perpendicular to it. Hence

dy

dx= 6x 4 =

1

2 x =

7

12

y = 3

7

12

2 4

7

12

=

21

16.

Using the point-slope form, the equation is given by

y+21

16=

1

2

x

7

12

.

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The Derivative as instantaneous rate of change

Definition:

Given y = f(x), the average rate of change of y with respect to x on

the interval [x, x1] is given by

y

x=

f(x1) f(x)

x1 x.

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If a free-falling object is dropped from a height of 100 feet, its height h

at time t is given by h(t) = 16t2 + 100, where h is measured in feetand t in seconds. Find the average rate of change of the height over

the following intervals:

1. [1,2]


Th D i i i f h
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1. [1,2] We first determine the height at t = 1 and t= 2.

h(1) = 16(1)2 + 100 = 84 h(2) = 16(2)2 + 100 = 36


Th D i ti i t t t f h
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1. [1,2] We first determine the height at t = 1 and t= 2.

h(1) = 16(1)2 + 100 = 84 h(2) = 16(2)2 + 100 = 36

So for the interval [1,2], the object falls from a height of 84 to 36ft. Thus the average rate of change is

ht

= 36 842 1

= 48 ft/sec.


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2. [1,1.5]


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2. [1,1.5]h(1.5) = 16(1.5)2 + 100 = 64. Hence

h

t=

64 84

1.5 1= 40 ft/sec.


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2. [1,1.5]h(1.5) = 16(1.5)2 + 100 = 64. Hence

h

t=

64 84

1.5 1= 40 ft/sec.

3. [1,1.1]


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2. [1,1.5]h(1.5) = 16(1.5)2 + 100 = 64. Hence

h

t=

64 84

1.5 1= 40 ft/sec.

3. [1,1.1]

h(1.1) = 16(1.1)2 + 100 = 80.64. Henceh

t=

80.64 84

1.

1 1= 33.6 ft/sec.




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2. [1,1.5]h(1.5) = 16(1.5)2 + 100 = 64. Hence

h

t=

64 84

1.5 1= 40 ft/sec.

3. [1,1.1]

h(1.1) = 16(1.1)2 + 100 = 80.64. Henceh

t=

80.64 84

1.

1 1= 33.6 ft/sec.

t 1 0.5 0.1 0.01 0.001 0.0001 0

dhdt

48 40 33.6 32.16 32.016 32.0016 32


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As we take smaller and smaller values for t, it seems reasonable toconclude that the instantaneous rate of change of the height when

t = 1 is 12 32 ft/sec.


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As we take smaller and smaller values for t, it seems reasonable toconclude that the instantaneous rate of change of the height when

t = 1 is 12 32 ft/sec.

Definition:

The instantaneous rate of change of y at x is the limit of the averagerate of change on the interval [x, x1] :

limx1x

y

x= lim

x1x

f(x1) f(x)

x1 x.

which is just the same as the limit definition of the derivative of f(x).


Examples
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Examples

At time t= 0, a diver jumps from a diving board that is 32 feet high.The position of the diver is given by h(t) = 16t2 + 16t+ 32, where

h(t) is measured in feet and t is measured in seconds.

a) When does the diver hit the water?


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Examples

At time t= 0, a diver jumps from a diving board that is 32 feet high.The position of the diver is given by h(t) = 16t2 + 16t+ 32, where

h(t) is measured in feet and t is measured in seconds.

a) When does the diver hit the water?

Solution:

To find the time at which the diver hits the water, we let h(t) = 0 andsolve for t:

0 = 16t2 + 16t+ 32

= 16(t2 t 2)

= 16(t+ 1)(t 2)

t= 1 or 2

The diver will hit the water after t= 2 seconds.


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Examples

b) What is the divers velocity at impact?


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Examples

b) What is the divers velocity at impact?

Solution:

The velocity at time t is given by the derivative

v(t) = h

(t) = 32t+ 16.

Therefore the velocity at time t= 2 is

v(2) = 32(2) + 16 = 48 ft/sec


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Examples

c) What is the divers acceleration at any time t?


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Examples

c) What is the divers acceleration at any time t?

Solution:

Acceleration is the derivative of velocity with respect to time. Hence

a(t) = v(t) = 32 ft/sec2

is the divers acceleration at any time t.


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p

A ball is thrown vertically upward from the ground with an initial velocity

of 64 ft/sec. If the positive direction of the distance from its starting

point is upwards, the equation of motion is s(t) = 16t2 + 64t. Let t

seconds be the time that has elapsed since the ball was thrown and sfeet be the distance of the ball from the starting point at t seconds.

a) Find the instantaneous velocity and speed of the ball at the end of

3 seconds.


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p

Solution:

Solving for the velocity, we get

v(t) = s(t) = 32t+ 64

So when t = 3,

v(3) = 32(3) + 64 = 32 ft/sec.

Since speed at time t is |v(t)|, the speed of the ball at the end of 3seconds is 32 ft/sec.


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p

b) How many seconds does it take the ball to reach its highest point?

How high will the ball go?


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p

b) How many seconds does it take the ball to reach its highest point?

How high will the ball go?

Solution:

At maximum height, the velocity of the object is 0. So we equate

v(t) = 0 and solve for t.

0 = 32t+ 64= 32(t 2)

t = 2

So the ball will reach its highest point after t = 2 seconds. And the

maximum height is given by

s(2) = 16(2)2 + 64(2) = 64 ft.


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p

c) How many seconds does it take the ball to reach the ground?


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c) How many seconds does it take the ball to reach the ground?

Solution:

Equate s(t) = 0 and solve for t.

0 = 16t2 + 64t= 16t(t 4)

t= 0,4

At t= 0, the ball is in its initial position. So the ball will reach theground after t= 4 seconds.


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d) Find the instantaneous velocity of the ball when it reaches the

ground.


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d) Find the instantaneous velocity of the ball when it reaches the

ground.

Solution:

Velocity upon impact is given by

v(4) = 32(4) + 64 = 64 ft/sec.


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A particle is moving along a horizontal line according to the equation

s(t) = 2t3 4t2 + 2t 1, where s(t) represents distance from the

starting point. Determine the intervals of time when the particle ismoving to the right and when it is moving to the left. Also determine

the instant when the particle reverses its direction.


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Solution:


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Solution:

v(t) = s(t) = 6t2 8t+ 2 = 2(3t2 4t+ 1) = 2(3t 1)(t 1) = 0.


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Solution:

v(t) = s(t) = 6t2 8t+ 2 = 2(3t2 4t+ 1) = 2(3t 1)(t 1) = 0.

A particle reverses direction at time t where the velocity changes from

a positive to a negative value, or a negative to a positive value. Hence

the particle reverses its direction when t=1

3and t = 1.


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Solution:

v(t) = s(t) = 6t2 8t+ 2 = 2(3t2 4t+ 1) = 2(3t 1)(t 1) = 0.

A particle reverses direction at time t where the velocity changes from

a positive to a negative value, or a negative to a positive value. Hence

the particle reverses its direction when t=13

and t = 1.

When v is positive, we say that the particle is moving to the right.

Otherwise, it is moving to the left.


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Solution (Cont.):Looking at the table below,

Interval v(t)

x< 13 +

x = 13 013 < x< 1

x = 1 0

x> 1 +


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Solution (Cont.):Looking at the table below,

Interval v(t)

x< 13 +

x = 13 013 < x< 1

x = 1 0

x> 1 +

It shows that the particle is moving to the right whent (0,1/3) (1,+) and moving to the left when t (1/3,1).

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[MATH 100] 2-Derivative as a Rate of Change

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