lecture 7: velocity, rate of change, and the...

Velocity Rate of Change The Derivative

Lecture 7: Velocity, Rate of Change, and the Derivative

VelocityRepresenting MotionAverage VelocityInstantaneous VelocityExample 25 – Computing Velocities from a GraphEstimating Slope by Averaging – Straddling

Rate of ChangeAverage Rate of ChangeInstantaneous Rate of ChangeExample 26 – Rate of Change of Area

The DerivativeThe Derivative at a PointMeanings of the Derivative at a PointExamples of DerivativesExample 27 – Calculating a Derivative at a Point

Clint Lee Math 112 Lecture 7: Velocity and Rate of Change 1/30


Representing Motion

Representing Motion

Motion Along a Line



Representing Motion

Representing Motion

Motion Along a Line

For an object moving along a straightline the position along the line relativeto some origin is given by s = f (t),



Representing Motion

Representing Motion

Motion Along a Line

For an object moving along a straightline the position along the line relativeto some origin is given by s = f (t),where s is a real number representingthe distance, positive to the right andnegative to the left, from the origin.



Representing Motion

Representing Motion

Motion Along a Line

For an object moving along a straightline the position along the line relativeto some origin is given by s = f (t),where s is a real number representingthe distance, positive to the right andnegative to the left, from the origin.The function f is the position functionof the object.



Representing Motion

Representing Motion

Motion Along a Line

For an object moving along a straightline the position along the line relativeto some origin is given by s = f (t),where s is a real number representingthe distance, positive to the right andnegative to the left, from the origin.The function f is the position functionof the object. The motion of the objectcan be represented in a diagram likethis.

0



Representing Motion

Representing Motion

Motion Along a Line

For an object moving along a straightline the position along the line relativeto some origin is given by s = f (t),where s is a real number representingthe distance, positive to the right andnegative to the left, from the origin.The function f is the position functionof the object. The motion of the objectcan be represented in a diagram likethis.

0

t=0

t=16t=2

t

s

s= f (t)

2 16

We get more information about the motion of the object by graphing fas a function of t, like this.



Average Velocity

Average Velocity

The velocity of a moving object is the displacement or change inposition per unit time.



Average Velocity

Average Velocity

The velocity of a moving object is the displacement or change inposition per unit time. Velocity can be positive, negative, or zero(even if the body is moving).



Average Velocity

Average Velocity

The velocity of a moving object is the displacement or change inposition per unit time. Velocity can be positive, negative, or zero(even if the body is moving).The speed of the object is the total distance traveled per unit time,which can be different than the velocity and is always positive.



Average Velocity

Average Velocity


Average Velocity



Average Velocity

Average Velocity


Average Velocity

The average velocity of the object whose position is given by s = f (t)over the time interval between t = t1 and t = t2 is



Average Velocity

Average Velocity


Average Velocity


vave =f (t2) − f (t1)

t2 − t1



Average Velocity

Average Velocity


Average Velocity


vave =f (t2) − f (t1)

t2 − t1=

f (t1) − f (t2)

t1 − t2



Average Velocity

Average Velocity


Average Velocity


vave =f (t2) − f (t1)

t2 − t1=

f (t1) − f (t2)

t1 − t2

The value (and sign) of the average velocity does not depend onwhich order we take for the times.



Average Velocity

Average Velocity as a Slope

The formula for the average velocityshould look familiar. It just represents theslope of the line joining the two points(t1, f (t1)) and (t2, f (t2)) on the graph ofthe position function.



Average Velocity


The formula for the average velocityshould look familiar. It just represents theslope of the line joining the two points(t1, f (t1)) and (t2, f (t2)) on the graph ofthe position function. With t2 after t1 itlooks like this.

t

s

t1 t2

s= f (t)



Average Velocity


The formula for the average velocityshould look familiar. It just represents theslope of the line joining the two points(t1, f (t1)) and (t2, f (t2)) on the graph ofthe position function. With t2 after t1 itlooks like this. With time t2 before t1, likethis.

t

s

t1t2

s= f (t)



Average Velocity


To make the formula for average velocitylook more familiar let t1 = a andt2 = a + h. If h > 0, then t2 is later than t1,and if h < 0, then t2 is earlier than t1.

t

s

t1 t2

s= f (t)



Average Velocity


To make the formula for average velocitylook more familiar let t1 = a andt2 = a + h. If h > 0, then t2 is later than t1,and if h < 0, then t2 is earlier than t1. Likethis.

t

s

a a+h

h

s= f (t)



Average Velocity



Alternate Formula for Average Velocity

The formula for the average velocity canbe written

vave =

t

s

a a+h

h

s= f (t)



Average Velocity



Alternate Formula for Average Velocity

The formula for the average velocity canbe written

vave =f (a + h) − f (a)

h

t

s

a a+h

h

s= f (t)



Instantaneous Velocity


The instantaneous velocity of a movingobject can be found by taking the averagevelocity over shorter and shorter timeintervals. This means we take t2 closer tot1, from either side of t1.

t

s

a a+h

h

s= f (t)





The instantaneous velocity of a movingobject can be found by taking the averagevelocity over shorter and shorter timeintervals. This means we take t2 closer tot1, from either side of t1. Equivalently, thismeans taking h closer and closer to zero,from above or below.

t

s

a a+h

h

s= f (t)






The instantaneous velocity of an objectmoving along a straight line withposition function s = f (t) at time t = a is

t

s

a a+h

h

s= f (t)







v(a)

t

s

a a+h

h

s= f (t)







v(a) = limh→0

f (a + h) − f (a)

h

t

s

a a+h

h

s= f (t)







v(a) = limh→0

f (a + h) − f (a)

h

The instantaneous velocity is the slope ofthe tangent line to the graph of theposition function at t = a, as shown in thegraph.

t

s

a

s= f (t)



Example 25 – Computing Velocities from a Graph


The graph shows the position functions = f (t) of an object moving along astraight line with the distance, s, from theorigin given in metres and the time, t, inseconds.

(a) Compute the average velocity ofthe object over time intervals:

t = 10 to t = 14t = 6 to t = 10

2

4

6

8

10

4 8 12 t (s)

s (m)

s= f (t)

(c) Estimate the instantaneous velocity of the object at t = 10 byaveraging the two average velocities in part (a).

(d) Estimate the instantaneous velocity by computing the slope ofthe tangent to the graph of the position function at t = 10.




Solution: Example 25(a) & (b)

Reading off the graph we have






f (6) = − 1






f (6) = − 1, f (10) = 2






f (6) = − 1, f (10) = 2, f (14) = 10






f (6) = − 1, f (10) = 2, f (14) = 10

For the interval from t = 10 to t = 14 take a = 10 and h = 4, so that






f (6) = − 1, f (10) = 2, f (14) = 10


vave =f (14) − f (10)

4= 2






f (6) = − 1, f (10) = 2, f (14) = 10


vave =f (14) − f (10)

4= 2

For the interval from t = 6 to t = 10 take a = 10 and h = −4, so that






f (6) = − 1, f (10) = 2, f (14) = 10


vave =f (14) − f (10)

4= 2


vave =f (6) − f (10)

−4=

3

4






f (6) = − 1, f (10) = 2, f (14) = 10


vave =f (14) − f (10)

4= 2


vave =f (6) − f (10)

−4=

3

4

Then an estimate of the instantaneous velocity at t = 10 is theaverage of these two values






f (6) = − 1, f (10) = 2, f (14) = 10


vave =f (14) − f (10)

4= 2


vave =f (6) − f (10)

−4=

3

4


v(10)






f (6) = − 1, f (10) = 2, f (14) = 10


vave =f (14) − f (10)

4= 2


vave =f (6) − f (10)

−4=

3

4


v(10) ≈1

2

(

2 +3

4

)

=11

8= 1.375 m/s




Solution: Example 25(c)

The tangent to the graph of the positionfunction at t = 10 is shown.

2

4

6

8

10

4 8 12 t (s)

s (m)

s= f (t)






Reading from the graph the tangent linegoes through the points (6.6,−2.0) and(14.0, 6.9).

2

4

6

8

10

4 8 12 t (s)

s (m)

s= f (t)






Reading from the graph the tangent linegoes through the points (6.6,−2.0) and(14.0, 6.9). So the slope of the tangent lineis

mT =6.9 − (−2.0)

14.9 − 6.6

2

4

6

8

10

4 8 12 t (s)

s (m)

s= f (t)







mT =6.9 − (−2.0)

14.9 − 6.6= 1.20 m/s

2

4

6

8

10

4 8 12 t (s)

s (m)

s= f (t)







mT =6.9 − (−2.0)

14.9 − 6.6= 1.20 m/s

Since the slope of the tangent at t = 10equals the instantaneous velocity att = 10, the instantaneous velocity att = 10 is estimated to be 1.20 m/s.

2

4

6

8

10

4 8 12 t (s)

s (m)

s= f (t)



Estimating Slope by Averaging – Straddling


In Example 25(b) we estimated the slopeof the tangent line, and hence theinstantaneous velocity, by averaging theslopes of two secant lines. The calculationis

v(10) ≈1

2

(

f (14) − f (10)

4+

f (6) − f (10)

−4

) 2

4

6

8

10

4 8 12 t (s)

s (m)

10

s= f (t)






v(10) ≈1

2

(

f (14) − f (10)

4+

f (6) − f (10)

−4

)

=f (14) − f (6)

8

2

4

6

8

10

4 8 12 t (s)

s (m)

10

s= f (t)






v(10) ≈1

2

(

f (14) − f (10)

4+

f (6) − f (10)

−4

)

=f (14) − f (6)

8

2

4

6

8

10

4 8 12 t (s)

s (m)

10

s= f (t)

This means we can estimate the slope of the tangent line at a point bycomputing the slope of a secant line joining two points on either sideof the point, at an equal distance on either side. See the diagram.






v(10) ≈1

2

(

f (14) − f (10)

4+

f (6) − f (10)

−4

)

=f (14) − f (6)

8

2

4

6

8

10

4 8 12 t (s)

s (m)

10

s= f (t)

This means we can estimate the slope of the tangent line at a point bycomputing the slope of a secant line joining two points on either sideof the point, at an equal distance on either side. See the diagram. This isstraddling. The straddling secant line has a slope at least as close to the slopeof the tangent as either of the one-sided secant lines.



Average Rate of Change

Increments

If quantity y is a function of quantity x, y = f (x),




Increments

If quantity y is a function of quantity x, y = f (x), and, if x changesfrom x1 to x2, the increment in x is

∆x = x2 − x1




Increments

If quantity y is a function of quantity x, y = f (x), and, if x changesfrom x1 to x2, the increment in x is

∆x = x2 − x1

and y changes from f (x1) to f (x2) for an increment of

∆y = f (x2) − f (x1)






The average rate of change of y with respect to x when x changesfrom x1 to x2 is

average rate of change







average rate of change =f (x2) − f (x1)

x2 − x1








x2 − x1=

∆y

∆x








x2 − x1=

∆y

∆x

This rate of change can also be described as the change in y per unitchange in x.



Instantaneous Rate of Change


As x2 gets closer and closer to x1, the average rate of changeapproaches the instantaneous rate of change of y with respect to x atx = x1.





As x2 gets closer and closer to x1, the average rate of changeapproaches the instantaneous rate of change of y with respect to x atx = x1. As x2 → x1, the increment ∆x → 0.






Instantaneous Rate of Change at x1

instantaneous rate of change = limx2→x1

f (x2) − f (x1)

x2 − x1






Instantaneous Rate of Change at x1

instantaneous rate of change = limx2→x1

f (x2) − f (x1)

x2 − x1= lim

∆x→0

∆y

∆x





Letting x1 = a and ∆x = h, we have





Letting x1 = a and ∆x = h, we have x2 = x1 + ∆x





Letting x1 = a and ∆x = h, we have x2 = x1 + ∆x = a + h.






Instantaneous Rate of Change at a







The definition of instantaneous rate of change can be written

instantaneous rate of change







The definition of instantaneous rate of change can be written

instantaneous rate of change = limh→0

f (a + h) − f (a)

h



Example 26 – Rate of Change of Area


The area A of a square with side length s is A = f (s) = s2.

(a) Determine the average rate of change of the area of a squarewhen the side length changes from

s = 5 m to s = 6 ms = 5 m to s = 5.5 ms = 5 m to s = 5.1 m

Give the units of the rate of change.

(b) Determine the instantaneous rate of change of the area of thesquare with respect to its side length when the side length iss = 5 m.




Solution: Example 26(a)

The average rates of change are






f (6) − f (5)

6 − 5=

36 − 25

1= 11






f (6) − f (5)

6 − 5=

36 − 25

1= 11

f (5.5) − f (5)

5.5 − 5=

30.25 − 25

0.5= 10.5






f (6) − f (5)

6 − 5=

36 − 25

1= 11

f (5.5) − f (5)

5.5 − 5=

30.25 − 25

0.5= 10.5

f (5.1) − f (5)

5.1 − 5=

26.01 − 25

0.1= 10.1






f (6) − f (5)

6 − 5=

36 − 25

1= 11

f (5.5) − f (5)

5.5 − 5=

30.25 − 25

0.5= 10.5

f (5.1) − f (5)

5.1 − 5=

26.01 − 25

0.1= 10.1

In each case the units of the rate of change are m2/m or just metres.




Solution: Example 26(b)

The instantaneous rate of change is given by






lim∆s→0

f (5 + ∆s) − f (5)

∆s






lim∆s→0

f (5 + ∆s) − f (5)

∆s= lim

∆s→0

(5 + ∆s)2− 25

∆s






lim∆s→0

f (5 + ∆s) − f (5)

∆s= lim

∆s→0

(5 + ∆s)2− 25

∆s

= lim∆s→0

25 + 10∆s + ∆s2− 25

∆s






lim∆s→0

f (5 + ∆s) − f (5)

∆s= lim

∆s→0

(5 + ∆s)2− 25

∆s

= lim∆s→0

25 + 10∆s + ∆s2− 25

∆s

= lim∆s→0

(10 + ∆s)






lim∆s→0

f (5 + ∆s) − f (5)

∆s= lim

∆s→0

(5 + ∆s)2− 25

∆s

= lim∆s→0

25 + 10∆s + ∆s2− 25

∆s

= lim∆s→0

(10 + ∆s) = 10






lim∆s→0

f (5 + ∆s) − f (5)

∆s= lim

∆s→0

(5 + ∆s)2− 25

∆s

= lim∆s→0

25 + 10∆s + ∆s2− 25

∆s

= lim∆s→0

(10 + ∆s) = 10

This agrees with the trend in the values found in part (a).




Solution: Example 26 Comment

Consider the change in the area of thesquare when the side length increases froms = 5 m to s = 6 m. This is shown in thediagram, where the the shaded portionsrepresent the change in area.





Consider the change in the area of thesquare when the side length increases froms = 5 m to s = 6 m. This is shown in thediagram, where the the shaded portionsrepresent the change in area.

s = 5

∆s = 1





Consider the change in the area of thesquare when the side length increases froms = 5 m to s = 6 m. This is shown in thediagram, where the the shaded portionsrepresent the change in area. This area ismade up of three parts, two rectangles anda small square. The combined area is2 × 5 × 1 + 1 × 1 = 11.

s = 5

∆s = 1





Consider the change in the area of thesquare when the side length increases froms = 5 m to s = 6 m. This is shown in thediagram, where the the shaded portionsrepresent the change in area. This area ismade up of three parts, two rectangles anda small square. The combined area is2 × 5 × 1 + 1 × 1 = 11. And thecorresponding average rate of change of thearea is 2 × 5 + 1 = 11.

s = 5

∆s = 1






s = 5

∆s = 1

When the side length changes from s = 5 m to s = 5.5 m the diagramlooks like this.






s = 5

∆s = 0.5

When the side length changes from s = 5 m to s = 5.5 m the diagramlooks like this. The change in area is 2 × 5 × 0.5 + 0.5 × 0.5 = 5.25,






s = 5

∆s = 0.5

When the side length changes from s = 5 m to s = 5.5 m the diagramlooks like this. The change in area is 2 × 5 × 0.5 + 0.5 × 0.5 = 5.25,and the average rate of change is 2 × 5 + 0.5 = 10.5.






s = 5

∆s = 0.5


When the side length changes from s = 5 m to s = 5.1 m the diagramlooks like this.







When the side length changes from s = 5 m to s = 5.1 m the diagramlooks like this. The change in area is 2 × 5 × 0.1 + 0.1 × 0.1 = 1.01,




Solution: Example 26 Comment continued

In general, the average rate of change of the area of the square whenthe side length changes from s = 5 to s = 5 + ∆s is






average rate of change of area = 2 × 5 + ∆s







Of course, as ∆s → 0, this average rate of change approaches 2 × 5 m,the instantaneous rate of change of the area with respect to the sidelength. As the change in the side length gets smaller, the contributionof the small square to the rate of change in the area gets smaller,eventually becoming negligible.








This value is twice the side length of the square, or, alternatively, onehalf of the perimeter of the square. So we can say








This value is twice the side length of the square, or, alternatively, onehalf of the perimeter of the square. So we can say

instantaneous rate of change ofarea of square with respect to sidelength

=1

2perimeter of the square





This is true for a square of side length a, since






lim∆s→0

f (a + ∆s) − f (a)

∆s






lim∆s→0

f (a + ∆s) − f (a)

∆s= lim

∆s→0

(a + ∆s)2− a2

∆s






lim∆s→0

f (a + ∆s) − f (a)

∆s= lim

∆s→0

(a + ∆s)2− a2

∆s

= lim∆s→0

a2 + 2a∆s + ∆s2− a2

∆s






lim∆s→0

f (a + ∆s) − f (a)

∆s= lim

∆s→0

(a + ∆s)2− a2

∆s

= lim∆s→0

a2 + 2a∆s + ∆s2− a2

∆s

= lim∆s→0

(2a + ∆s)






lim∆s→0

f (a + ∆s) − f (a)

∆s= lim

∆s→0

(a + ∆s)2− a2

∆s

= lim∆s→0

a2 + 2a∆s + ∆s2− a2

∆s

= lim∆s→0

(2a + ∆s)

= 2a






lim∆s→0

f (a + ∆s) − f (a)

∆s= lim

∆s→0

(a + ∆s)2− a2

∆s

= lim∆s→0

a2 + 2a∆s + ∆s2− a2

∆s

= lim∆s→0

(2a + ∆s)

= 2a =1

2perimeter of square






lim∆s→0

f (a + ∆s) − f (a)

∆s= lim

∆s→0

(a + ∆s)2− a2

∆s

= lim∆s→0

a2 + 2a∆s + ∆s2− a2

∆s

= lim∆s→0

(2a + ∆s)

= 2a =1

2perimeter of square

To pursue this idea further, express the area of a square as a functionof the distance x of each side of the square from its centre. Show thatthe rate of change of the area of the square with respect to x equalsthe perimeter of the square, and draw a diagram to understand howthis result comes about.




Other Rates of Change

Slope and Velocity as Rates of Change






The slope of a line is the rate of change of the y-coordinate ofpoints on the line with respect to the x-coordinate.






The slope of a line is the rate of change of the y-coordinate ofpoints on the line with respect to the x-coordinate.

The velocity of a moving object is the rate of change of theposition of the object with respect to time.



The Derivative at a Point


Derivative at a Point






For a function f defined at x = a, the derivative of f at a is

f ′(a) 1

2







f ′(a) = limx→a

f (x) − f (a)

x − a1

2







f ′(a) = limx→a

f (x) − f (a)

x − a1

= limh→0

f (a + h) − f (a)

h2



Meanings of the Derivative at a Point








There are three interpretations, or applications, of the derivative at apoint

The slope of the tangent line to the graph of f at the point(a, f (a)).This is also called the slope of the linearization of f at a, or justthe slope of the curve y = f (x) at (a, f (a)).








The (instantaneous) velocity at time t = a of a moving objectwhose position at time t is given by s = f (t).








The (instantaneous) velocity at time t = a of a moving objectwhose position at time t is given by s = f (t).

The (instantaneous) rate of change of the quantity y with respectto the quantity x at x = a, when y = f (x).



Examples of Derivatives


We have already computed or estimated derivatives.






In Example 22 for the function f (x) = x3 we found that f ′(1) = 3.In this example the derivative gave us the slope, and so theequation of the line tangent to y = x3 at the point (1, 1).







In Example 25 we estimated f ′(10) to give the instantaneousvelocity of a moving object whose position is given by s = f (t).The estimate was based on the graph of the function f .







In Example 25 we estimated f ′(10) to give the instantaneousvelocity of a moving object whose position is given by s = f (t).The estimate was based on the graph of the function f .

In Example 26 for the function A = f (s) = s2 we found thatf ′(5) = 10, and, more generally, that f ′(a) = 2a. In this examplethe derivative gave us the rate of change of the area A of a squarewith respect to its side length s.



Example 27 – Calculating a Derivative at a Point


Letf (x) =

x

x − 1

(a) Calculate f ′(3) using Formula 1 .

(b) Calculate f ′(3) using Formula 2 .

(c) Calculate f ′(a) for any number a in the domain of f using

Formula 2 .





First note that f (3) = 3/2. Then applying Formula 1 with

f (x) =x

x − 1and a = 3 gives

f ′(3) =






f (x) =x


f ′(3) = limx→3

x

x − 1−

3

2x − 3






f (x) =x


f ′(3) = limx→3

x

x − 1−

3

2x − 3

= limx→3

(

2x − 3(x − 1)

2(x − 1)(x − 3)

)






f (x) =x


f ′(3) = limx→3

x

x − 1−

3

2x − 3

= limx→3

(

2x − 3(x − 1)

2(x − 1)(x − 3)

)

= limx→3

(

2x − 3x + 3

2(x − 1)(x − 3)

)






f (x) =x


f ′(3) = limx→3

x

x − 1−

3

2x − 3

= limx→3

(

2x − 3(x − 1)

2(x − 1)(x − 3)

)

= limx→3

(

2x − 3x + 3

2(x − 1)(x − 3)

)

= limx→3

(

−(x − 3)

2(x − 1)(x − 3)

)






f (x) =x


f ′(3) = limx→3

x

x − 1−

3

2x − 3

= limx→3

(

2x − 3(x − 1)

2(x − 1)(x − 3)

)

= limx→3

(

2x − 3x + 3

2(x − 1)(x − 3)

)

= limx→3

(

−(x − 3)

2(x − 1)(x − 3)

)

= limx→3

(

−1

2(x − 1)

)






f (x) =x


f ′(3) = limx→3

x

x − 1−

3

2x − 3

= limx→3

(

2x − 3(x − 1)

2(x − 1)(x − 3)

)

= limx→3

(

2x − 3x + 3

2(x − 1)(x − 3)

)

= limx→3

(

−(x − 3)

2(x − 1)(x − 3)

)

= limx→3

(

−1

2(x − 1)

)

= −1

4





Now applying Formula 2 with f (x) =x


f ′(3) =







f ′(3) = limh→0

3 + h

(3 + h) − 1−

3

2

h







f ′(3) = limh→0

3 + h

(3 + h) − 1−

3

2

h= lim

h→0

1

h

(

3 + h

2 + h−

3

2

)







f ′(3) = limh→0

3 + h

(3 + h) − 1−

3

2

h= lim

h→0

1

h

(

3 + h

2 + h−

3

2

)

= limh→0

(

6 + 2h − (6 + 3h)

2h(2 + h)

)







f ′(3) = limh→0

3 + h

(3 + h) − 1−

3

2

h= lim

h→0

1

h

(

3 + h

2 + h−

3

2

)

= limh→0

(

6 + 2h − (6 + 3h)

2h(2 + h)

)

= limh→0

(

−h

2h(2 + h)

)







f ′(3) = limh→0

3 + h

(3 + h) − 1−

3

2

h= lim

h→0

1

h

(

3 + h

2 + h−

3

2

)

= limh→0

(

6 + 2h − (6 + 3h)

2h(2 + h)

)

= limh→0

(

−h

2h(2 + h)

)

= limh→0

(

−1

2(2 + h)

)







f ′(3) = limh→0

3 + h

(3 + h) − 1−

3

2

h= lim

h→0

1

h

(

3 + h

2 + h−

3

2

)

= limh→0

(

6 + 2h − (6 + 3h)

2h(2 + h)

)

= limh→0

(

−h

2h(2 + h)

)

= limh→0

(

−1

2(2 + h)

)

= −1

4







f ′(3) = limh→0

3 + h

(3 + h) − 1−

3

2

h= lim

h→0

1

h

(

3 + h

2 + h−

3

2

)

= limh→0

(

6 + 2h − (6 + 3h)

2h(2 + h)

)

= limh→0

(

−h

2h(2 + h)

)

= limh→0

(

−1

2(2 + h)

)

= −1

4

Of course, we get the same result using the two formulas. In some

cases the algebra is easier using Formula 1 , but we will use Formula

2 in most cases.






x − 1for a general a gives

f ′(a) =







f ′(a) = limh→0

a + h

(a + h) − 1−

a

a − 1

h







f ′(a) = limh→0

a + h

(a + h) − 1−

a

a − 1

h= lim

h→0

1

h

(

a + h

a − 1 + h−

a

a − 1

)







f ′(a) = limh→0

a + h

(a + h) − 1−

a

a − 1

h= lim

h→0

1

h

(

a + h

a − 1 + h−

a

a − 1

)

= limh→0

(

(a + h)(a − 1) − a(a − 1 + h)

h(a − 1)(a − 1 + h)

)







f ′(a) = limh→0

a + h

(a + h) − 1−

a

a − 1

h= lim

h→0

1

h

(

a + h

a − 1 + h−

a

a − 1

)

= limh→0

(

(a + h)(a − 1) − a(a − 1 + h)

h(a − 1)(a − 1 + h)

)

= limh→0

(

a2− a + ah − h −

(

a2− a + ah

)

h(a − 1)(a − 1 + h)

)







f ′(a) = limh→0

a + h

(a + h) − 1−

a

a − 1

h= lim

h→0

1

h

(

a + h

a − 1 + h−

a

a − 1

)

= limh→0

(

(a + h)(a − 1) − a(a − 1 + h)

h(a − 1)(a − 1 + h)

)

= limh→0

(

a2− a + ah − h −

(

a2− a + ah

)

h(a − 1)(a − 1 + h)

)

= limh→0

(

a2− a + ah − h − a2 + a − ah

h(a − 1)(a − 1 + h)

)







f ′(a) = limh→0

a + h

(a + h) − 1−

a

a − 1

h= lim

h→0

1

h

(

a + h

a − 1 + h−

a

a − 1

)

= limh→0

(

(a + h)(a − 1) − a(a − 1 + h)

h(a − 1)(a − 1 + h)

)

= limh→0

(

a2− a + ah − h −

(

a2− a + ah

)

h(a − 1)(a − 1 + h)

)

= limh→0

(

a2− a + ah − h − a2 + a − ah

h(a − 1)(a − 1 + h)

)

= limh→0

(

−h

h(a − 1)(a − 1 + h)

)







f ′(a) = limh→0

a + h

(a + h) − 1−

a

a − 1

h= lim

h→0

1

h

(

a + h

a − 1 + h−

a

a − 1

)

= limh→0

(

(a + h)(a − 1) − a(a − 1 + h)

h(a − 1)(a − 1 + h)

)

= limh→0

(

a2− a + ah − h −

(

a2− a + ah

)

h(a − 1)(a − 1 + h)

)

= limh→0

(

a2− a + ah − h − a2 + a − ah

h(a − 1)(a − 1 + h)

)

= limh→0

(

−h

h(a − 1)(a − 1 + h)

)

= limh→0

(

−1

(a − 1)(a − 1 + h)

)







f ′(a) = limh→0

a + h

(a + h) − 1−

a

a − 1

h= lim

h→0

1

h

(

a + h

a − 1 + h−

a

a − 1

)

= limh→0

(

(a + h)(a − 1) − a(a − 1 + h)

h(a − 1)(a − 1 + h)

)

= limh→0

(

a2− a + ah − h −

(

a2− a + ah

)

h(a − 1)(a − 1 + h)

)

= limh→0

(

a2− a + ah − h − a2 + a − ah

h(a − 1)(a − 1 + h)

)

= limh→0

(

−h

h(a − 1)(a − 1 + h)

)

= limh→0

(

−1

(a − 1)(a − 1 + h)

)

= −1

(a − 1)2




Solution: Example 27(c) – Continued

Substituting a = 3 into the formula for the derivative just derivedgives






f ′(3) = −1

22= −

1

4






f ′(3) = −1

22= −

1

4

which agrees with our calculations in parts (a) and (b).


lecture 7: velocity, rate of change, and the...

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