Note on Information in the Equation of the Tangent Line

Knowing the equation of the tangent line to the graph of f(x) at x = a is exactly the same as knowing f(a) and f ‘ (a)

y ( )f a ( )f ' a ( )x aIf we know a, f(a) and f ‘ (a) then the equation of the tangent line is:

Conversely if y = m x + b is known to be the tangent line at x =a thenm = f ‘(a) and f(a) = ma+b

Example: If y = 3 x-7 is the equation of the tangent line to the graph of f(x) atx = 2 what are f(2) and f ‘ (2)?

Ans: f ‘ (2) = slope of tan line = 3; f(2) = 3(2)-7 = -1

Responses to Some QuestionsLimits Involving infinity

limx

( )f x 2

lim +x 3

( )f x

lim -x 3

( )f x

Calculationslim

x

1

x0

limx 0

1

xDoes Not Exist

limx

x limx ( )

x

limx ( )

1

x0

limx ( )

x2 lim

x x

2

Rational Functions

limx

7 x3

4 x2

x 1

100 2 x3

Divide Numerator and Denominator by x^n where n is the largest degree of numerator and denominator

7 x3

4 x2

x 1

x3

100 2 x3

x3

limx

limx

7 x3

x3

4 x2

x3

x

x3

1

x3

74

x

1

x2

1

x3

limx

100

x3

2

100

x3

2 x3

x3

= 7

-2

May Have To Compute Degree

limx

( )3 x2

1 ( )1 x

x ( )1 3 x

Degree of numerator = 3, Degree of denominator = 2

3 x2

1

x2

1 x

x

1 3 x

x

x

x

1

x

limx

=

(3) (-1)

(1) (-3) 1

x

limx

= X

limx

limx

Other possibility

Degree of numerator = 2, Degree of denominator = 3

3 x2

1

x2

1 x

x

1 3 x

x

x

x

1

xlimx

=

(1)(3) (0)

(3) (-1)

limx

= 0

limx

x ( )1 3 x

3 x2

( )1 x

General Principles

• If degree of numerator is larger than degree of denominator then limit as n->infinity does not exist

• If degree of numerator is less than degree of denominator then limit as n->infinity is 0

• If degree of numerator = degree of denominator then the limit can easily be calculated by dividing both numerator and denominator by by x^n where n the highest degree.

Lecture 7

Derivatives as Rates

Average Rate of Change

• If f is a function of t on the interval [a,b] then the average rate of change of f on the interval [a,b] with respect to t is

fav

( )f b ( )f ab a

When the units of f(t) are distance and those of t are time the average rate of change is called the average velocity

Need only the values a,b, f(a), f(b)

• A particle moves along a line so that its distance from the origin at time 4 seconds is 9 feet and its distance from the origin at time t = 10 seconds is 2 feet. What is its average velocity over the interval [4,10]?

Here a = 4 sec., f(4) = 9 ft., b = 10 sec, f(10) = 2 ft.

Average velocity = (2 – 9)/(10-4) ft/sec = -7/6 ft/sec

Average Velocity is Slope of Secant Line

fav

( )f b ( )f ab a

Motion on a Line

x(-1) = 0

• Suppose a particle is moving along the x-axis and its x-coordinate at time t seconds is given by feet.

x(0) = -1 x(1) = 0 x(2) = 3 x(3) = 8

( )x t t2

1

Previous Example

( )x t t2

1t in seconds, x in feet over the interval [-1,3]

xav

( )x 3 ( )x -13 ( )-1

8

4ft/sec= = 2 ft/sec

The average velocity over the interval [-1,3] is 2 ft per second

What is the velocity at t = 2?

• Can’t calculate average velocity over interval [2,2] (dividing by 0).

• Can calculate average velocity over [2,2+h]

( )x 2 h ( )x 2

h

The limiting average velocity as h -> o must be the “instantaneous” velocity at time t = 2.

Instantaneous Velocity

• If f(t) is the position on a line of a particle at time t then the instantaneous velocity of the particle at time t =a is

limh 0

f(a+h)-f(a)

h

Which is f ‘ (a)

Example

• A particle moves along the x-axis so that its position at time hours is

( )x t1 t

2

1 t2

miles

What is its instantaneous velocity at time t = 3 hours?At what times is it moving to the left of the origin and at what times is it moving to the right?

solution

x ' (t) 4t

( )1 t2

2

x ' (3) 3

25miles/hr

x ‘ is positive for t < 0 so for t < 0 the particle is moving in the positive direction which is to the right. It is negative for time t > 0 which means that for time t >0 itIs moving to the left

( )x t1 t

2

1 t2

Projectile Motion

• After t seconds the height, in feet of a projectile fired straight upward from the ground is

( )h t 16 t2

100 ta. What is the (upward) velocity 3 seconds after it is fired?

b. What is the (upward) velocity when it hits the ground?

c. At what time does the projectile stop ascending and begin to return to Earth?

Solution:Let v(t) = velocity at time t.

v(t) = h’ (t) = 32 t 100 ft/sec

(a)At time t = 3 the velocity is v(3) = -32*3 + 100 = 4 ft/sec

(b) The projectile stops rising when its velocity is reduced to 0. That is when -32 t + 100 = 0 or when t = 100/32 sec.

(c) The projectile returns to Earth when its height is 0 i.e. when 0 =

( )h t 16 t2

100 t

0 =( -16 t + 100) t has solutions t = 0 and t = 100/16

The time t = 0 is when it was launched, t = 100/16 is when it returned. The velocity then is v(100/16) = -100 ft/sec

Problem on Average Velocity

• A driver travels the first 40 miles of a trip to Louisville at 80 miles per hour and the remaining 30 miles at 60 miles per hour. What was her average velocity for the trip?

fav

( )f b ( )f ab a