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Differentiation
2.1
2.2
2.3
2.4
2.5
2.6
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2. 2.1 The Derivative and the Tangent Line Problem Objectives
Find the slope of the tangent line to a curve at a point.
Use the limit definition to find the derivative of a function.
Understand the relationship between differentiability and continuity.
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3. The Tangent Line Problem
Calculus grew out of four major problems that European mathematicians were working on during the seventeenth century.
1. The tangent line problem
2. The velocity and acceleration problem
3. The minimum and maximum problem
4. The area problem
Each problem involves the notion of a limit, and calculus can be introduced with any of the four problems.

4. Definition of the Derivative of a Function
(x+Δx, f(x+Δx))
(x, f(x))

5. The Derivative of a Function
The most common notations that are used to denote the derivative of y = f(x) are
The notation dy/dx is read as “the derivative of y with respect to x” or simply “dy, dx.” Using limit notation, you can write

6. Example 1 – The Slope of the Graph of a Linear Function
Find the slope of the graph of f(x) = 2x – 3 at the point (2, 1).
To find the slope of the graph of f at any given value for x use the definition of the derivative

7. Example 3 – Finding the Derivative by the Limit Process
Find the derivative of f(x) = x3 + 2x.
= 3x2 + 2

8. Example 4 – Finding the Derivative by the Limit Process
Given
Find
Then find the slopes of the graph f at the points (1,1) and (4,2). Discuss the end behavior of f at (0,0).
Slope at (1,1) is
Slope at (4,2) is
The slope at (0,0) doesn’t exist since the slope from the right and left at (0,0) for the original function isn’t the same value.
Discuss the end behavior of f at (0,0).

9. Example 5 – Finding the Derivative by the Limit Process
Find the derivative of y with respect to t for the function y = 2/t.

10. Graphs are NOT Differentiable at:
Points of discontinuity
Vertical Tangent Lines
Sharp Points

11. Copyright © Cengage Learning. All rights reserved.
2.1 Summary
Objectives
Find the slope of the tangent line to a curve at a point.
Use the limit definition to find the derivative of a function.
Understand the relationship between differentiability and continuity.
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12. 2.2 Basic Differentiation Rules and Rates of Change Objectives
Find the derivative of a function using the Constant Rule.
Find the derivative of a function using the Power Rule.
Find the derivative of a function using the Constant Multiple Rule.
Find the derivative of a function using the Sum and Difference Rules.
Find the derivatives of the sine function and of the cosine function.
Use derivatives to find rates of change.
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13. The Constant Rule Given function f(x) = c find the derivative
Figure 2.14

14. The Power Rule f(x) = x3 f'(x) = 3x2 y = x10 dy/dx = 10x9 f(x) = x1/2
Yesterday we found derivatives of functions like the following using the definition of the derivative. Please look for a pattern, rule, shortcut,…. for the following derivative
f(x) = x3
f'(x) = 3x2
y = x10
dy/dx = 10x9
f(x) = x1/2
f'(x) = ½ x–1/2
y = x–2
y' = –2x–3

15. Example 1 – Finding the Derivative
f(x) = x7
f'(x) = 7x6
y = x–3
dy/dx = –3x–4
f(x) = x5/2
f'(x) = 5/2x3/2
y = x
y' = 1x0 = 1
f(x) = 6
f'(x) = 0

16. Example 2 – Using the Power Rule
Re-write
Differentiate
Simplify

17. The Constant Multiple Rule
The Constant Multiple Rule and the Power Rule can be combined into one rule. The combination rule is

18. Example 3 – Using the Constant Multiple Rule

19. Ex 4 – Find the Slope of a Graph
Find the slope of the graph of f(x) = x4 when
a) x = b) x = c) x = 1
The slope is found by evaluating the derivative at the given values
What is the derivative of f(x) =
The slope at x = -1
f’(-1) = 4(-1)3 = -4
f′(x) = 4x3
b) The slope at x = 0
f’(0) = 4(0)3 = 0
c) The slope at x = 1
f’(1) = 4(1)3 = 4

20. Ex 5 – Find the Equation of a Tangent Line
Find the equation of the tangent line to the graph of f(x) = x2 when x = -2
We will use the point-slope formula after finding the point and the slope for the tangent line
The Slope: m = f′(-2)
f′ (x) = 2x
f′ (-2) = -4
m = -4
The Point: (-2, f(-2))
(-2, 4)
Find the equation of the line passing through (-2,4) with a slope of -4
y – 4 = –4(x + 2)
y = –4x – 4

21. The Sum and Difference Rules
Example 6:

22. Proof of the Derivative of Sinx
Find the Derivative of sinx using the definition of the Derivative
Reminder from Ch 1
Reminder from section 2.1
Extra Credit for proving derivatives of all six trig functions
= cosx

23. Derivatives of the Sine and Cosine Functions
Example 7:

24. Rates of Change
You have seen how the derivative is used to determine slope.
The derivative can also be used to determine the rate of change of one variable with respect to another.
Applications involving rates of change occur in a wide variety of fields.
A few examples are population growth rates, production rates, water flow rates, velocity, and acceleration.

25. Rates of Change
A common use for 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, movement to the right (or upward) is considered to be in the positive direction, and movement to the left (or downward) is considered to be in the negative direction.

26. Rates of Change
The function s that gives the position (relative to the origin)
of an object as a function of time t is called a position
function.
If, over a period of time t, the object changes its position by the amount s = s(t + t) – s(t), then, by the familiar formula
the average velocity is

27. Example 9A – Finding Average Velocity of a Falling Object
If a billiard ball is dropped from a height of 100 feet, its height s at time t is given by the position function
s = –16t Position function
where s is measured in feet and t is measured in seconds. Find the average velocity over the time interval [1,2].
For the interval [1, 2], the object falls from a height of
s(1) = –16(1) = 84 feet to a height of
s(2) = –16(2) = 36 feet.
The average velocity is

28. Example 9(b) –
cont’d
If a billiard ball is dropped from a height of 100 feet, its height s at time t is given by the position function
s = –16t Position function
where s is measured in feet and t is measured in seconds. Find the average velocity over the time interval [1,1.5].
For the interval [1, 1.5], the object falls from a height of feet to a height of 64 feet.
The average velocity is
Note that the average velocities are negative, indicating that the object is moving downward.

29. Rates of Change
In general, if s = s(t) is the position function for an object moving along a straight line, the velocity of the object at time t is
In other words, the velocity function is the derivative of the position function.
Velocity can be negative, zero, or positive.
The speed of an object is the absolute value of its velocity.
Speed cannot be negative.

30. Rates of Change
The position of a free-falling object (neglecting air resistance) under the influence of gravity can be represented by the equation
where s0 is the initial height of the object, v0 is the initial velocity of the object, and g is the acceleration due to gravity.
On Earth, the value of g is approximately –32 feet per second per second or –9.8 meters per second per second.

31. Example 10 – Using the Derivative to Find Velocity
At time t = 0, a diver jumps from a platform
diving board that is 32 feet above the water
(see Figure 2.21). The position of the diver
is given by
s(t) = –16t2 + 16t Position function
where s is measured in feet and t is
measured in seconds.
When does the diver hit the water?
b. What is the diver’s velocity at impact?
Figure 2.21

32. Example 10(a) – Solution When does the diver hit the water?
s(t) = –16t2 + 16t Position function
To find the time t when the diver hits the water, let s = 0 and solve for t.
–16t2 + 16t + 32 = Set position function equal to 0.
–16(t + 1)(t – 2) = Factor.
t = –1 or Solve for t.
Because t ≥ 0, choose the positive value to conclude that
the diver hits the water at t = 2 seconds.

33. Example 10(b) – Solution b. What is the diver’s velocity at impact?
cont’d
b. What is the diver’s velocity at impact?
s(t) = –16t2 + 16t Position function
The velocity at time t is given by the derivative
s(t) = –32t + 16.
So, the velocity at time is t = 2 is s(2) = –32(2) + 16
= –48 feet per second.

34. Add on to Notes #A
To estimate the height of a building, a stone is dropped from the top of the building into a pool of water at ground level. How high is the building if the splash is seen 7.0 seconds after the stone is dropped? Use the position function for free-falling objects given below. (Round your answer to one decimal place.) s(t) = –4.9t2 + v0t + s0
s(t) = –4.9t2 + v0t + s0
Dropped v0 = 0
s(t) = –4.9t2 + v0t + s0
Dropped v0 =
Position Equation:
s(t) = –4.9t2 + s0
The height is zero when t = 7.0 sec:
S(7.0) = 0
–4.9(7.0)2 + s0 = 0
s0 = meters
The volume of a cube with sides of length s is given by V = s3. Find the rate of change of the volume with respect to s when s = 3 centimeters.
Rate of Change = Derivative
V = s3
dv/ds = 3s2
dv/ds = 3(3)2 = 27

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2.2 Summary
Objectives
Find the derivative of a function using the Constant Rule.
Find the derivative of a function using the Power Rule.
Find the derivatives of the sine function and of the cosine function.
Use derivatives to find rates of change.
The average velocity is
d/dx (sin x) = cosx
d/dx (cos x) = -sinx
Position Function: s(t) = ½ gt2 + v0t + s0
Velocity Function: v(t) = s′(t)
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36. 2.3 Product and Quotient Rules and Higher-Order Derivatives Objectives
Find the derivative of a function using the Product Rule.
Find the derivative of a function using the Quotient Rule.
Find the derivative of a trigonometric function.
Find a higher-order derivative of a function.
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37. The Product Rule

38. Example 1 – Using the Product Rule
Find the derivative of
3x – 2x2
5 + 4x
3 – 4x 4
h′(x) = (3x – 2x2)(4) + (5 + 4x)(3 – 4x)
= 12x – 8x – 8x – 16x2
= –24x2 + 4x + 15

39. Example 2 – Using the Product Rule
Find the derivative of y = 3x2sinx
3x2
sinx
y′ = (3x2)(cosx) + (sinx)(6x) 6x
cosx
= 3x2 cosx + 6x sinx
Find the derivative of y = 2x cosx – 2 sinx 2x
cosx
y′ = (2x)(-sinx) + (cosx)(2) – 2 cosx 2
-sinx
= –2x sinx + 2 cosx – 2 cosx
= –2x sinx

40. Example 3 – Using the Quotient Rule
Find the derivative of
Lo: x2 + 1
Hi: 5x – 2
dLo: 2x
dHi: 5 y′

41. Example 4 – Using the Quotient Rule
Find the equation of the tangent line to the graph of f(x) at x = -1
We need the point and slope of the tangent line at x = -1
The point: find f(-1)
The point is (-1,1)
The slope: find f′(-1)
Lo: x + 5
Hi: 3 – 1/x
= 3 – x-1
dLo: 1
dHi: 1x-2
= 1/x2
The equation of the tangent line at (-1,1) is y = 1

42. Ex 5 – Not every Quotient needs Quotient Rule
Find the derivative of each

43. Proof of the Derivative of Tanx
Find the Derivative of tan x using the derivatives of sin x and cos x
Reminder from section 2.2
d(sin x) = cos x
d(cos x) = -sin x
Extra Credit for proving derivatives of all six trig functions

44. Proof of the Derivative of Cscx
Find the Derivative of csc x using the derivatives of sin x and cos x
Reminder from section 2.2
d(sin x) = cos x
d(cos x) = -sin x
Extra Credit for proving derivatives of all six trig functions

45. Derivatives of Trigonometric Functions
10 bonus points in the test category for proving the derivatives of all six trig functions.
Example 6

46. Higher-Order Derivatives
Higher-order derivatives are denoted as follows.

47. Higher-Order Derivatives
Just as you can obtain a velocity function by differentiating a position function, you can obtain an acceleration function by differentiating a velocity function.
Another way of looking at this is that you can obtain an acceleration function by differentiating a position function twice.

48. Example 10 – Finding the Acceleration Due to Gravity
Because the moon has no atmosphere, a falling object on the moon encounters no air resistance. In 1971, astronaut David Scott demonstrated that a feather and a hammer fall at the same rate on the moon. The position function for each of these falling objects is given by
s(t) = –0.81t2 + 2
where s(t) is the height in meters and t is the time in seconds. What is the ratio of Earth’s gravitational force to the moon’s?
s(t) = –0.81t Position function
s'(t) = –1.62t Velocity function
s"(t) = – Acceleration function
So, the acceleration due to gravity on the moon is –1.62 meters per sec per sec

49. Ex 11 – use the nderiv button on calculator
A population of 490 bacteria is introduced into a culture and grows in number according to the equation below, where t is measured in hours. Find the rate at which the population is growing when t = 3. (Round your answer to two decimal places.)
( )
( )
Answer: bacteria/hour

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2.3 Summary
Objectives
Find the derivative of a function using the Product Rule.
Find the derivative of a function using the Quotient Rule.
Find the derivative of a trigonometric function.
Find a higher-order derivative of a function.
d(sinx) = cosx
d(tanx) = sec2x
d(secx) = secxtanx
d(cosx) = –sinx
d(cotx) = –csc2x
d(cssx) = –cscxcotx
Position Function: s(t) = ½ gt2 + v0t + s0
Velocity Function: v(t) = s′(t)
Acceleration Function: a(t) = v′(t) = s′′(t)
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51. Copyright © Cengage Learning. All rights reserved.
2.4
The Chain Rule
Objectives
Find the derivative of a composite function using the Chain Rule.
Find the derivative of a function using the General Power Rule.
Simplify the derivative of a function using algebra.
Find the derivative of a trigonometric function using the Chain Rule.
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52. The Chain Rule
This text has yet to discuss one of the most powerful differentiation rules—the Chain Rule.
This rule deals with composite functions and adds a surprising versatility to the rules discussed in the two previous sections.
For example, compare the functions shown below.

53. Example 2 – Using the Chain Rule
Find dy/dx for y = (x2 + 1)6
You could take the binomial and raise it the sixth power then take the derivative but that would be time consuming.
Using the Chain Rule is much quicker and efficient method for finding the derivative.
Think of the Chain Rule as the reverse of the order of operations.
By order of operations we would first find the value of
x (junk)6
dy/dx = 6(junk)5 d(x2 + 1)
= 6(x2 + 1)5(2x)
= 12x(x2 + 1)5

54. Example 3 & 4 – Using the Chain Rule
Find the derivative of f(x) = (3x – 2x2)3.
Order of operations:
3x – 2x (junk)3
f′(x) = 3(junk)2 d(3x – 2x2)
= 3(3x – 2x2)2(3 – 4x)
Find the derivative of
Order of operations:
x2 – (junk)2/3
y′ = 2/3(junk)–1/3 d(x2 – 1)
= 2/3(x2 – 1)–1/3 (2x)

55. Example 5 – Using the Chain Rule
Find the derivative of
First re-write the function
Order of operations:
2t – (junk)–2
g′(t) = 14(junk)–3 d(2t – 3)
= 14(2t – 3)–3 (2)

56. Simplifying Derivatives
The next three examples illustrate some techniques for simplifying the “raw derivatives” of functions involving products, quotients, and composites.

57. Example 6 – Combining Several Rules
What is the GCF?

58. Example 7 – Combining Several Rules

59. Example 8 – Combining Several Rules

60. Trigonometric Functions and the Chain Rule
The “Chain Rule versions” of the derivatives of the six trigonometric functions are as follows.
d(sin u) = cos u du
d(tan u) = sec2 u du
d(sec u) = sec u tan u du
d(cos u) = –sin u du
d(cot u) = –csc2 u du
d(csc u) = –csc u cot u du

61. Example 9 – Applying the Chain Rule to Trigonometric Functions
Find the derivative of each:
y = sin 2x
y = cos(x – 1)
y = tan 3x
y′ = cos(2x) d(2x)
Order of operations:
2x sin(junk)
y′ = 2cos 2x
y′ = –sin(x – 1) d(x - 1)
y′ = – sin(x – 1)
Order of operations:
x – 1 cos(junk)
y′ = sec2(3x) d(3x)
Order of operations:
3x tan(junk)
y′ = 3sec2(3x)

62. Ex 10 – More Complicated Find the derivative of each: y = cos 3x2
y′ = –sin (3x2) d(3x2)
y′ = –6x sin (3x2)
cos(3) is a #
(just a coef of x2)
y′ = 2cos(3)x
y = cos (9x2)
y′ = –sin (9x2) d(9x2)
y′ = –18x sin (9x2)
y = (cos x)2
y′ = 2(cos x)1 d(cos x)
y′ = –2 cos x sin x
y = (cos x)1/2
y′ = ½ (cos x)–1/2 d(cos x)

63. Ex 11 – Even More Complicated
Find the derivative of f(t) = sin3 4t
= (sin (4t))3
Order of operations
4t sin(junk) (stuff)3
Derivative in reverse order
3(stuff) cos(junk) d(4t)
f′(t) = 3(sin (4t))2 d(sin (4t))
= 3 sin2(4t) cos(4t) d(4t)
= 3 sin2(4t) cos(4t) (4)
f′(t) = 12 sin2(4t) cos(4t)

64. Copyright © Cengage Learning. All rights reserved.
2.4 Summary
Objectives
Find the derivative of a composite function using the Chain Rule.
Find the derivative of a function using the General Power Rule.
Simplify the derivative of a function using algebra.
Find the derivative of a trigonometric function using the Chain Rule.
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65. Copyright © Cengage Learning. All rights reserved.
2.5
Implicit Differentiation
Objectives
Distinguish between functions written in implicit form and explicit form.
Use implicit differentiation to find the derivative of a function.
Introduction:
Notes on whiteboard:
Find d/dx [3x2] d/dx [4x3 + 4y5 – 3x2] d/da [4y2 – 2x5 + 3a4]
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66. Implicit and Explicit Functions
Most functions have been expressed in explicit form For example, in the equation
the variable y is explicitly written as a function of x.
Explicit form

67. Implicit and Explicit Functions
Suppose you were asked to find dy/dx for this equation. You could begin by writing y explicitly as a function of x and then differentiating.
This strategy works whenever you can solve for the function explicitly.
You cannot, however, use this procedure when you are unable to solve for y as a function of x.

68. Implicit and Explicit Functions
For instance, how would you find dy/dx for the equation
where it is very difficult to express y as a function of x explicitly? To find dy/dx, you can use implicit differentiation.

69. Implicit and Explicit Functions
To understand how to find dy/dx implicitly, you must realize that the differentiation is taking place with respect to x.
This means that when you differentiate terms involving x alone, you can differentiate as usual.
However, when you differentiate terms involving y, you must apply the Chain Rule, because you are assuming that y is defined implicitly as a differentiable function of x.

70. Example 1 – Differentiating with Respect to x

71. Example 1 – Differentiating with Respect to x
cont’d

72. Implicit Differentiation

73. Example 2 – Implicit Differentiation
Find dy/dx given that y3 + y2 – 5y – x2 = –4.
1. Differentiate both sides of the equation with respect to x.
2. Collect the dy/dx terms on the left side of the equation and move all other terms to the right side of the equation.
3. Factor dy/dx out of the left side of the equation.
4. Solve for dy/dx by dividing by (3y2 + 2y – 5).

74. Implicit Differentiation
To see how you can use an
implicit derivative, consider
the graph shown in Figure 2.27.
From the graph, you can see that
y is not a function of x. Even so,
the derivative found in Example 2
gives a formula for the slope of the
tangent line at a point on this graph.
The slopes at several points on the
graph are shown below the graph.
y3 + y2 – 5y – x2 = –4
Figure 2.27

75. Ex 4 – Finding the Slope of a Graph
Find the slope of the tangent line to the graph of x2 + 4y2 = 4 at the point
2x + 8y dy/dx = 0
8y dy/dx = –2x
dy/dx = -2x/8y
dy/dx = -x/4y

76. Example 5 – Finding the Slope of a Graph Implicitly
Find the slope of the graph of 3(x2 + y2)2 = 100xy at the point (3, 1).
100x y
100
dy/dx

77. Example 7 Find the Second Derivate Implicitly
Given x2 + y2 = 25, find
Now find the derivative of
dy/dx = -x/y using quotient rule
2x + 2y dy/dx = 0
2y dy/dx = -2x
dy/dx = -2x/2y
dy/dx = -x/y
Hi: -x
Lo: y
dHi: -1
dLo: dy/dx = -x/y

78. Ex 8 – Find a Tangent Line to a Graph
Find the tangent line to the graph given by x2(x2 + y2) = y2 at the point

79. Copyright © Cengage Learning. All rights reserved.
2.5 Summary
Objectives
Distinguish between functions written in implicit form and explicit form.
Use implicit differentiation to find the derivative of a function.
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80. Copyright © Cengage Learning. All rights reserved.
2.6
Related Rates
Objectives
Find a related rate.
Use related rates to solve real-life problems.
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81. Finding Related Rates
The Chain Rule can be used to find dy/dx implicitly. Another important use of the Chain Rule is to find the rates of change of two or more related variables that are changing with respect to time.

82. Finding Related Rates
For example, when water is drained out of a conical tank (see Figure 2.33), the volume V, the radius r,
and the height h of the water level are
all functions of time t.
Knowing that these variables are related
by the equation
Figure 2.33

83. Finding Related Rates
you can differentiate implicitly with respect to t to obtain the related-rate equation
From this equation you can see that the rate of change of
V is related to the rates of change of both h and r.

84. Example 1 – Two Rates That Are Related
Suppose x and y are both differentiable functions of t and are related by the equation y = x2 + 3.
Find dy/dt when x = 1, given that dx/dt = 2 when x = 1.
Differentiate both sides of the equation with respect to t.
When x = 1 and dx/dt = 2, you have

85. Problem Solving with Related Rates

86. Example 2 – Ripples in a Pond
A pebble is dropped into a calm pond, causing ripples in the form of concentric circles. The radius r of the outer ripple is increasing at a constant rate of 1 foot per second. When the radius is 4 feet, at what rate is the total area of the disturbed water changing?
Given:
Find:
When:
dr/dt = 1 ft/sec
dA/dt
r = 4 ft
Equation:
dA/dt = 2πr dr/dt
dA/dt = 2π(4)(1)
A = πr2
dA/dt = 8π

87. Example 3 – An Inflating Balloon
Air is being pumped into a spherical balloon (see Figure 2.35) at a rate of 4.5 cubic feet per minute. Find the rate of change of the radius when the radius is 2 feet.
Given:
Find:
When:
r = 2
Equation:
9/2 = 4π(2)2 dr/dt
dr/dt = 9/32π ≈ 0.09 ft/min
Figure 2.35

88. Ex 4 – Sand Falling into a Conical Pile
At a sand & gravel plant, sand is falling off a conveyor and onto a conical pile at a rate of 6 cubic feet per minute. The diameter of the base of the cone is approximately three times the altitude. At what rate is the height of the pile changing when the pile is 4 feet high?
Given:
Find:
When:
dV/dt = 6 ft/min
dh/dt
h = 4 ft
Equation:
V = 1/3 πr2h
dV/dt = 9/4 πh2 dh/dt
V = 1/3 π(9/4h2)h
6 = 9/4 π(4)2 dh/dt
d = 3h
V = 3/4 πh3
6 = 36π dh/dt
2r = 3h
dh/dt = 1/6π ft/min
r = 3/2h
r2 = 9/4h2

89. Ex 5 – Fishing dx/dt = -2.9 ft/sec dθ/dt x = 25 ft Sin θ = 10/x
A fish is reeled in at a rate of 2.9 feet per second from a point 10 feet above the water (see figure). At what rate is the angle between the line and the water changing when there is a total of 25 feet of line out? (Round your answer to three decimal places.)
Given:
Find:
When:
dx/dt = -2.9 ft/sec
dθ/dt
x = 25 ft
Equation:
Sin θ = 10/x
Cos θ dθ/dt = –10x–2 dx/dt
Sin θ = 10x–1
When:
x = 25 ft
dθ/dt ≈ rad/sec

90. Ex 6a – Ladder Problem dx/dt = 2 ft/sec dy/dt x = 7 ft x2 + y2 = 252
A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2 feet per second.
(a) What is the velocity of the top of the ladder when the base is 7 feet from the wall?
Given:
Find:
When:
dx/dt = 2 ft/sec
dy/dt
x = 7 ft
Equation:
x2 + y2 = 252
2x dx/dt + 2y dy/dt = 0
When:
2(7)(2) + 2(24)dy/dt = 0
y = 24 ft
dy/dt = -7/12 ft/sec
x = 7 ft

91. Ex 6 – Ladder Problem dx/dt = 2 ft/sec dA/dt x = 7 ft
A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2 feet per second.
(b) Consider the triangle formed by the side of the house, ladder, and the ground. Find the rate at which the area of the triangle is changing when the base of the ladder is 7 feet from the wall.
Given:
Find:
When:
dx/dt = 2 ft/sec
dA/dt
x = 7 ft
dy/dt = -7/12 ft/sec
←from part a
Equation:
When:
A = ½ xy
y = 24 ft
dA/dt = 527/24 ft2/sec
x = 7 ft

92. Ex 6 – Ladder Problem dx/dt = 2 ft/sec dθ/dt x = 7 ft Sin θ = x/25
A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2 feet per second.
(c) Find the rate at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 7 feet from the wall.
Given:
Find:
When:
dx/dt = 2 ft/sec
dθ/dt
x = 7 ft
Equation:
When:
Sin θ = x/25
Cos θ dθ/dt = 1/25 dx/dt
Sin θ = 1/25 x
(24/25) dθ/dt = (1/25) (2)
y = 24 ft
dθ/dt = 1/12 rad/sec
x = 7 ft

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2.6 Summary
Objectives
Find a related rate.
Use related rates to solve real-life problems.
Given:
Find:
When:
Equation:
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