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Derivative as a Rate of Change
Francis Joseph H. CampenaMathematics Department
De La Salle Univerisity
October 21, 2014
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Rectilinear Motion
Here we call the motion of a particle on a line to be a rectilinearmotion.We consider one direction on the line to be positive and theopposite direction to be negative. For simplicity, we assume thatthe particle is moving on a horizontal line, with distance to theright as positive and distance to left as negative. We select somepoint on the line and denote it by the letter O. Let f be thefunction determining the directed distance of the particle from Oat any particular time. Here, let s meters (m) be the directeddistance of the particle from O at t seconds (s). Then f is afunction defined by s = f (t) which gives the directed distancefrom the point O to the particle at a particular instant.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Rectilinear Motion
Definition
If f is a function given by the equation
s = f (t)
and a particle is moving along a line such that s is the number ofunits in the directed distance of the particle from a fixed point onthe line at t units of time, then the instantaneous velocity of theparticle at t units of time is v units of velocity, where
v = f ′(t)⇔ v =ds
dt
if it exists.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Rectilinear Motion
Example
An object moves along a coordinate line so that its position ssatisfies s = 2t2 − 12t + 8. Determine the velocity of the objectwhen t = 1 and when t = 6. When is the velocity 0? When is itpositive?
There is a technical distinction between velocity and speed.Velocity has a sign associated with it. Speed is defined to be theabsolute value of the velocity.
The second derivatived2s
dt2also has its physical interpretation. It
is, of course, the first derivative of the velocity. Thus, it measuresthe rate of change of the velocity with respect to time, which isknown as acceleration. If acceleration is denoted by a, then
a =dv
dt=
d2s
dt2.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Rectilinear Motion
Example
An object moves along a coordinate line so that its position ssatisfies s = 2t2 − 12t + 8. Determine the velocity of the objectwhen t = 1 and when t = 6. When is the velocity 0? When is itpositive?
There is a technical distinction between velocity and speed.Velocity has a sign associated with it. Speed is defined to be theabsolute value of the velocity.
The second derivatived2s
dt2also has its physical interpretation. It
is, of course, the first derivative of the velocity. Thus, it measuresthe rate of change of the velocity with respect to time, which isknown as acceleration. If acceleration is denoted by a, then
a =dv
dt=
d2s
dt2.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Rectilinear Motion
Example
An object moves along a coordinate line so that its position ssatisfies s = 2t2 − 12t + 8. Determine the velocity of the objectwhen t = 1 and when t = 6. When is the velocity 0? When is itpositive?
There is a technical distinction between velocity and speed.Velocity has a sign associated with it. Speed is defined to be theabsolute value of the velocity.
The second derivatived2s
dt2also has its physical interpretation. It
is, of course, the first derivative of the velocity. Thus, it measuresthe rate of change of the velocity with respect to time, which isknown as acceleration. If acceleration is denoted by a, then
a =dv
dt=
d2s
dt2.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Rectilinear Motion
When a > 0, v is increasing and when a < 0, v is decreasing.When a = 0, then v is not changing. Because the speed of theparticle at t seconds is |v |, we then have the following:
1 If v ≥ 0 and a > 0, then the speed is increasing.
2 If v ≥ 0 and a < 0, then the speed is decreasing.
3 If v ≤ 0 and a > 0, then the speed is decreasing.
4 If v ≤ 0 and a < 0, then the speed is increasing.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Rectilinear Motion
Example
A point moves along a horizontal coordinate line in such a waythat its position at time t is specified by
s = t3 − 12t2 + 36t − 30
(a) When is the velocity 0?
(b) When is the velocity positive?
(c) When is the point moving backward (that is, to the left)?
(d) When is the acceleration positive?
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
The Derivative as a Rate of Change
If the functional relationship between y and x is given y = f (x)and if x changes from x1 to x1 + ∆x , then the value of y changesfrom f (x1) to f (x1 + ∆x). Here, ∆y = f (x1 + ∆x)− f (x1), whenthe change in x is ∆x . The average rate of change of y per unitchange in x , as x changes from x1 to x1 + ∆x is given by
∆y
∆x=
f (x1 + ∆x)− f (x1)
∆x.
If the limit of this quotient exists, then this limit is what weconsider as the instantaneous rate of change of y per unit changein x at x1.
Definition
If y = f (x), the instantaneous rate of change of y per unit changein x at x1 is f ′(x1) if it exists.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
The Derivative as a Rate of Change
If the functional relationship between y and x is given y = f (x)and if x changes from x1 to x1 + ∆x , then the value of y changesfrom f (x1) to f (x1 + ∆x). Here, ∆y = f (x1 + ∆x)− f (x1), whenthe change in x is ∆x . The average rate of change of y per unitchange in x , as x changes from x1 to x1 + ∆x is given by
∆y
∆x=
f (x1 + ∆x)− f (x1)
∆x.
If the limit of this quotient exists, then this limit is what weconsider as the instantaneous rate of change of y per unit changein x at x1.
Definition
If y = f (x), the instantaneous rate of change of y per unit changein x at x1 is f ′(x1) if it exists.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
The Derivative as a Rate of Change
Example
Let V (x) cubic centimeters be the volume of a cube having anedge of x centimeters, measured to four significant digits. On acalculator compute the average rate of change of V (x) withrespect to x as x changes from (a) 3.000 to 3.200; (b) 3.000 to3.100; (c) 3.000 to 3.010; (d) 3.000 to 3.001; (e) What is theinstantaneous rate of change of V (x) with respect to x when x is3.000?
The average rate of change of V (x) with respect to x as x changesfrom x1 to x1 + ∆x is
V (x1 + ∆x)− V (x1)
∆x.
Thus,
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
The Derivative as a Rate of Change
1 x1 = 3.000, ∆x = 0.200
V (3.200)− V (3.000)
0.200=
(3.200)3 − (3.000)3
0.200= 28.84.
2 x1 = 3.000, ∆x = 0.100⇒V (x1 + ∆x)− V (x1)
∆x= 27.91
3 x1 = 3.000, ∆x = 0.010⇒V (x1 + ∆x)− V (x1)
∆x= 27.09
4 x1 = 3.000, ∆x = 0.001⇒V (x1 + ∆x)− V (x1)
∆x= 27.01
5 The instantaneous rate of change when x = 3.000 isV ′(3.000).
V (x) = x3 ⇒ V ′(x) = 3x2 ⇒ V ′(3) = 3(32) = 27.
We conclude that when the length of the cube is 3 cm, theinstantaneous rate of change of the volume is 27 cm3 percentimeter change in the length of the edge.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
The Derivative as a Rate of Change
1 x1 = 3.000, ∆x = 0.200
V (3.200)− V (3.000)
0.200=
(3.200)3 − (3.000)3
0.200= 28.84.
2 x1 = 3.000, ∆x = 0.100⇒V (x1 + ∆x)− V (x1)
∆x= 27.91
3 x1 = 3.000, ∆x = 0.010⇒V (x1 + ∆x)− V (x1)
∆x= 27.09
4 x1 = 3.000, ∆x = 0.001⇒V (x1 + ∆x)− V (x1)
∆x= 27.01
5 The instantaneous rate of change when x = 3.000 isV ′(3.000).
V (x) = x3 ⇒ V ′(x) = 3x2 ⇒ V ′(3) = 3(32) = 27.
We conclude that when the length of the cube is 3 cm, theinstantaneous rate of change of the volume is 27 cm3 percentimeter change in the length of the edge.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Related Rates
A problem in related rates is one involving rates of change ofrelated variables. In real-world applications related rates, thevaraibles have a specific relatiionship for values of t where t is ameasure of time. This realationship is usually expressed in the formof an equation which represents a mathematical model of thesituation.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Related Rates
Suggested steps in solving a related rate problem.
(1) READ and UNDERSTAND the problem.
(2) Draw a figure or and illustration of the conditions in theproblem.
(3) Identify all relevant variables, including those whose rates aregiven and those whose rates are to be found.
(4) Express all given rates and rates to be found as derivatives.
(5) Find an equation connecting the identified variables.
(6) Implicitly differentiate the equation relating the differentvariables using the chain rule where appropriate. (Usually, thevariables are dependent of time and hence, the implicitdifferentiation is with respect to time.)
(7) Substitute the known values and solve for the inknown rate.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Related Rates
Suggested steps in solving a related rate problem.
(1) READ and UNDERSTAND the problem.
(2) Draw a figure or and illustration of the conditions in theproblem.
(3) Identify all relevant variables, including those whose rates aregiven and those whose rates are to be found.
(4) Express all given rates and rates to be found as derivatives.
(5) Find an equation connecting the identified variables.
(6) Implicitly differentiate the equation relating the differentvariables using the chain rule where appropriate. (Usually, thevariables are dependent of time and hence, the implicitdifferentiation is with respect to time.)
(7) Substitute the known values and solve for the inknown rate.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Related Rates
Suggested steps in solving a related rate problem.
(1) READ and UNDERSTAND the problem.
(2) Draw a figure or and illustration of the conditions in theproblem.
(3) Identify all relevant variables, including those whose rates aregiven and those whose rates are to be found.
(4) Express all given rates and rates to be found as derivatives.
(5) Find an equation connecting the identified variables.
(6) Implicitly differentiate the equation relating the differentvariables using the chain rule where appropriate. (Usually, thevariables are dependent of time and hence, the implicitdifferentiation is with respect to time.)
(7) Substitute the known values and solve for the inknown rate.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Related Rates
Suggested steps in solving a related rate problem.
(1) READ and UNDERSTAND the problem.
(2) Draw a figure or and illustration of the conditions in theproblem.
(3) Identify all relevant variables, including those whose rates aregiven and those whose rates are to be found.
(4) Express all given rates and rates to be found as derivatives.
(5) Find an equation connecting the identified variables.
(6) Implicitly differentiate the equation relating the differentvariables using the chain rule where appropriate. (Usually, thevariables are dependent of time and hence, the implicitdifferentiation is with respect to time.)
(7) Substitute the known values and solve for the inknown rate.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Related Rates
Suggested steps in solving a related rate problem.
(1) READ and UNDERSTAND the problem.
(2) Draw a figure or and illustration of the conditions in theproblem.
(3) Identify all relevant variables, including those whose rates aregiven and those whose rates are to be found.
(4) Express all given rates and rates to be found as derivatives.
(5) Find an equation connecting the identified variables.
(6) Implicitly differentiate the equation relating the differentvariables using the chain rule where appropriate. (Usually, thevariables are dependent of time and hence, the implicitdifferentiation is with respect to time.)
(7) Substitute the known values and solve for the inknown rate.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Related Rates
Suggested steps in solving a related rate problem.
(1) READ and UNDERSTAND the problem.
(2) Draw a figure or and illustration of the conditions in theproblem.
(3) Identify all relevant variables, including those whose rates aregiven and those whose rates are to be found.
(4) Express all given rates and rates to be found as derivatives.
(5) Find an equation connecting the identified variables.
(6) Implicitly differentiate the equation relating the differentvariables using the chain rule where appropriate. (Usually, thevariables are dependent of time and hence, the implicitdifferentiation is with respect to time.)
(7) Substitute the known values and solve for the inknown rate.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Related Rates
Suggested steps in solving a related rate problem.
(1) READ and UNDERSTAND the problem.
(2) Draw a figure or and illustration of the conditions in theproblem.
(3) Identify all relevant variables, including those whose rates aregiven and those whose rates are to be found.
(4) Express all given rates and rates to be found as derivatives.
(5) Find an equation connecting the identified variables.
(6) Implicitly differentiate the equation relating the differentvariables using the chain rule where appropriate. (Usually, thevariables are dependent of time and hence, the implicitdifferentiation is with respect to time.)
(7) Substitute the known values and solve for the inknown rate.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Related Rates
Suggested steps in solving a related rate problem.
(1) READ and UNDERSTAND the problem.
(2) Draw a figure or and illustration of the conditions in theproblem.
(3) Identify all relevant variables, including those whose rates aregiven and those whose rates are to be found.
(4) Express all given rates and rates to be found as derivatives.
(5) Find an equation connecting the identified variables.
(6) Implicitly differentiate the equation relating the differentvariables using the chain rule where appropriate. (Usually, thevariables are dependent of time and hence, the implicitdifferentiation is with respect to time.)
(7) Substitute the known values and solve for the inknown rate.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 1
Example
A 26-foot ladder is placed against a wall. If the top of the ladder issliding down the wall at 2 ft per second, at what rate is thebottom of the ladder moving away from the wall when the bottomof the ladder is 10 feet away from the wall?
Illustration Solution
x = x(t) : distance of thebottom of the ladder at timet.
y = y(t) : distance of thetop of the ladder at time t.
x2 + y2 = 262 : Relationshipof the variables.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 1
Example
A 26-foot ladder is placed against a wall. If the top of the ladder issliding down the wall at 2 ft per second, at what rate is thebottom of the ladder moving away from the wall when the bottomof the ladder is 10 feet away from the wall?
Illustration Solution
x = x(t) : distance of thebottom of the ladder at timet.
y = y(t) : distance of thetop of the ladder at time t.
x2 + y2 = 262 : Relationshipof the variables.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 1
Example
A 26-foot ladder is placed against a wall. If the top of the ladder issliding down the wall at 2 ft per second, at what rate is thebottom of the ladder moving away from the wall when the bottomof the ladder is 10 feet away from the wall?
Illustration
Solution
x = x(t) : distance of thebottom of the ladder at timet.
y = y(t) : distance of thetop of the ladder at time t.
x2 + y2 = 262 : Relationshipof the variables.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 1
Example
A 26-foot ladder is placed against a wall. If the top of the ladder issliding down the wall at 2 ft per second, at what rate is thebottom of the ladder moving away from the wall when the bottomof the ladder is 10 feet away from the wall?
Illustration Solution
x = x(t) : distance of thebottom of the ladder at timet.
y = y(t) : distance of thetop of the ladder at time t.
x2 + y2 = 262 : Relationshipof the variables.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 1
Example
A 26-foot ladder is placed against a wall. If the top of the ladder issliding down the wall at 2 ft per second, at what rate is thebottom of the ladder moving away from the wall when the bottomof the ladder is 10 feet away from the wall?
Illustration Solution
x = x(t) : distance of thebottom of the ladder at timet.
y = y(t) : distance of thetop of the ladder at time t.
x2 + y2 = 262 : Relationshipof the variables.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 1
Example
A 26-foot ladder is placed against a wall. If the top of the ladder issliding down the wall at 2 ft per second, at what rate is thebottom of the ladder moving away from the wall when the bottomof the ladder is 10 feet away from the wall?
Illustration Solution
x = x(t) : distance of thebottom of the ladder at timet.
y = y(t) : distance of thetop of the ladder at time t.
x2 + y2 = 262 : Relationshipof the variables.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 1
Example
A 26-foot ladder is placed against a wall. If the top of the ladder issliding down the wall at 2 ft per second, at what rate is thebottom of the ladder moving away from the wall when the bottomof the ladder is 10 feet away from the wall?
Illustration Solution
x = x(t) : distance of thebottom of the ladder at timet.
y = y(t) : distance of thetop of the ladder at time t.
x2 + y2 = 262 : Relationshipof the variables.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 1
Given informations in the problem are as follows:
dy
dt= −2 : since y is decreasing at a constant rate of 2 feet
per second.
When x = 10 the value of y = 24. This can be computedusing the Pythgorean theorem.dx
dt: unknown rate of change.
Differentiating the equation with respect to t gives us,
2xdx
dt+ 2y
dy
dt= 0.
Solving for the unknown rate gives usdx
dt= 4.8 feet per second.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 1
Given informations in the problem are as follows:
dy
dt= −2 : since y is decreasing at a constant rate of 2 feet
per second.
When x = 10 the value of y = 24. This can be computedusing the Pythgorean theorem.dx
dt: unknown rate of change.
Differentiating the equation with respect to t gives us,
2xdx
dt+ 2y
dy
dt= 0.
Solving for the unknown rate gives usdx
dt= 4.8 feet per second.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 1
Given informations in the problem are as follows:
dy
dt= −2 : since y is decreasing at a constant rate of 2 feet
per second.
When x = 10 the value of y = 24. This can be computedusing the Pythgorean theorem.dx
dt: unknown rate of change.
Differentiating the equation with respect to t gives us,
2xdx
dt+ 2y
dy
dt= 0.
Solving for the unknown rate gives usdx
dt= 4.8 feet per second.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 1
Given informations in the problem are as follows:
dy
dt= −2 : since y is decreasing at a constant rate of 2 feet
per second.
When x = 10 the value of y = 24. This can be computedusing the Pythgorean theorem.
dx
dt: unknown rate of change.
Differentiating the equation with respect to t gives us,
2xdx
dt+ 2y
dy
dt= 0.
Solving for the unknown rate gives usdx
dt= 4.8 feet per second.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 1
Given informations in the problem are as follows:
dy
dt= −2 : since y is decreasing at a constant rate of 2 feet
per second.
When x = 10 the value of y = 24. This can be computedusing the Pythgorean theorem.dx
dt: unknown rate of change.
Differentiating the equation with respect to t gives us,
2xdx
dt+ 2y
dy
dt= 0.
Solving for the unknown rate gives usdx
dt= 4.8 feet per second.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 1
Given informations in the problem are as follows:
dy
dt= −2 : since y is decreasing at a constant rate of 2 feet
per second.
When x = 10 the value of y = 24. This can be computedusing the Pythgorean theorem.dx
dt: unknown rate of change.
Differentiating the equation with respect to t gives us,
2xdx
dt+ 2y
dy
dt= 0.
Solving for the unknown rate gives usdx
dt= 4.8 feet per second.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 1
Given informations in the problem are as follows:
dy
dt= −2 : since y is decreasing at a constant rate of 2 feet
per second.
When x = 10 the value of y = 24. This can be computedusing the Pythgorean theorem.dx
dt: unknown rate of change.
Differentiating the equation with respect to t gives us,
2xdx
dt+ 2y
dy
dt= 0.
Solving for the unknown rate gives usdx
dt= 4.8 feet per second.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 2
Example
Air is being pumped into a spherical balloon such that its radiusincreases at a rate of 0.75 inches per minute. Find the rate ofchange of its volume when the radius is 5 inches.
V = V (t) : Volume of the spherical ballon at time t.
r = r(t) : radius of the spherical ballon at time t.
dr
dt: rate of change of the radius with respect to t.
V =4
3πr3 : relationship of the variables.
Solution:dV
dt= 75π cubic inches per minute.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 2
Example
Air is being pumped into a spherical balloon such that its radiusincreases at a rate of 0.75 inches per minute. Find the rate ofchange of its volume when the radius is 5 inches.
V = V (t) : Volume of the spherical ballon at time t.
r = r(t) : radius of the spherical ballon at time t.
dr
dt: rate of change of the radius with respect to t.
V =4
3πr3 : relationship of the variables.
Solution:dV
dt= 75π cubic inches per minute.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 2
Example
Air is being pumped into a spherical balloon such that its radiusincreases at a rate of 0.75 inches per minute. Find the rate ofchange of its volume when the radius is 5 inches.
V = V (t) : Volume of the spherical ballon at time t.
r = r(t) : radius of the spherical ballon at time t.
dr
dt: rate of change of the radius with respect to t.
V =4
3πr3 : relationship of the variables.
Solution:dV
dt= 75π cubic inches per minute.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 2
Example
Air is being pumped into a spherical balloon such that its radiusincreases at a rate of 0.75 inches per minute. Find the rate ofchange of its volume when the radius is 5 inches.
V = V (t) : Volume of the spherical ballon at time t.
r = r(t) : radius of the spherical ballon at time t.
dr
dt: rate of change of the radius with respect to t.
V =4
3πr3 : relationship of the variables.
Solution:dV
dt= 75π cubic inches per minute.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 2
Example
Air is being pumped into a spherical balloon such that its radiusincreases at a rate of 0.75 inches per minute. Find the rate ofchange of its volume when the radius is 5 inches.
V = V (t) : Volume of the spherical ballon at time t.
r = r(t) : radius of the spherical ballon at time t.
dr
dt: rate of change of the radius with respect to t.
V =4
3πr3 : relationship of the variables.
Solution:dV
dt= 75π cubic inches per minute.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 2
Example
Air is being pumped into a spherical balloon such that its radiusincreases at a rate of 0.75 inches per minute. Find the rate ofchange of its volume when the radius is 5 inches.
V = V (t) : Volume of the spherical ballon at time t.
r = r(t) : radius of the spherical ballon at time t.
dr
dt: rate of change of the radius with respect to t.
V =4
3πr3 : relationship of the variables.
Solution:dV
dt= 75π cubic inches per minute.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 2
Example
Air is being pumped into a spherical balloon such that its radiusincreases at a rate of 0.75 inches per minute. Find the rate ofchange of its volume when the radius is 5 inches.
V = V (t) : Volume of the spherical ballon at time t.
r = r(t) : radius of the spherical ballon at time t.
dr
dt: rate of change of the radius with respect to t.
V =4
3πr3 : relationship of the variables.
Solution:dV
dt= 75π cubic inches per minute.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 3
Example
A car is traveling north towards an intersection at a rate of 60 kphwhile a truck is traveling east away from the intersection at a rateof 50 kph. Find the rate of change of the distance between the carand the truck when the car is 3 kilometers south of the intersectionand the truck is 4 kilometers east of the intersection?
Illustration Solution
x = x(t) : distance travelledby the truck.
y = y(t) : distance travelledby the car.
z = z(t) : distance betweenthe car and the truck.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 3
Example
A car is traveling north towards an intersection at a rate of 60 kphwhile a truck is traveling east away from the intersection at a rateof 50 kph. Find the rate of change of the distance between the carand the truck when the car is 3 kilometers south of the intersectionand the truck is 4 kilometers east of the intersection?
Illustration Solution
x = x(t) : distance travelledby the truck.
y = y(t) : distance travelledby the car.
z = z(t) : distance betweenthe car and the truck.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 3
Example
A car is traveling north towards an intersection at a rate of 60 kphwhile a truck is traveling east away from the intersection at a rateof 50 kph. Find the rate of change of the distance between the carand the truck when the car is 3 kilometers south of the intersectionand the truck is 4 kilometers east of the intersection?
Illustration
Solution
x = x(t) : distance travelledby the truck.
y = y(t) : distance travelledby the car.
z = z(t) : distance betweenthe car and the truck.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 3
Example
A car is traveling north towards an intersection at a rate of 60 kphwhile a truck is traveling east away from the intersection at a rateof 50 kph. Find the rate of change of the distance between the carand the truck when the car is 3 kilometers south of the intersectionand the truck is 4 kilometers east of the intersection?
Illustration Solution
x = x(t) : distance travelledby the truck.
y = y(t) : distance travelledby the car.
z = z(t) : distance betweenthe car and the truck.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 3
Example
A car is traveling north towards an intersection at a rate of 60 kphwhile a truck is traveling east away from the intersection at a rateof 50 kph. Find the rate of change of the distance between the carand the truck when the car is 3 kilometers south of the intersectionand the truck is 4 kilometers east of the intersection?
Illustration Solution
x = x(t) : distance travelledby the truck.
y = y(t) : distance travelledby the car.
z = z(t) : distance betweenthe car and the truck.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 3
Example
A car is traveling north towards an intersection at a rate of 60 kphwhile a truck is traveling east away from the intersection at a rateof 50 kph. Find the rate of change of the distance between the carand the truck when the car is 3 kilometers south of the intersectionand the truck is 4 kilometers east of the intersection?
Illustration Solution
x = x(t) : distance travelledby the truck.
y = y(t) : distance travelledby the car.
z = z(t) : distance betweenthe car and the truck.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 3
Example
A car is traveling north towards an intersection at a rate of 60 kphwhile a truck is traveling east away from the intersection at a rateof 50 kph. Find the rate of change of the distance between the carand the truck when the car is 3 kilometers south of the intersectionand the truck is 4 kilometers east of the intersection?
Illustration Solution
x = x(t) : distance travelledby the truck.
y = y(t) : distance travelledby the car.
z = z(t) : distance betweenthe car and the truck.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 3
Solution
x2 + y2 = z2 : Relationship of the variables.dx
dt= 50kph rate of change for the truck.
dy
dt= −60kph rate of change for the car.
Answer:dz
dt= 4 kph. The distance between the car and the truck
is increasing at a rate of 4 kilometers per hour at that time.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 3
Solution
x2 + y2 = z2 : Relationship of the variables.dx
dt= 50kph rate of change for the truck.
dy
dt= −60kph rate of change for the car.
Answer:dz
dt= 4 kph. The distance between the car and the truck
is increasing at a rate of 4 kilometers per hour at that time.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 3
Solution
x2 + y2 = z2 : Relationship of the variables.
dx
dt= 50kph rate of change for the truck.
dy
dt= −60kph rate of change for the car.
Answer:dz
dt= 4 kph. The distance between the car and the truck
is increasing at a rate of 4 kilometers per hour at that time.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 3
Solution
x2 + y2 = z2 : Relationship of the variables.dx
dt= 50kph rate of change for the truck.
dy
dt= −60kph rate of change for the car.
Answer:dz
dt= 4 kph. The distance between the car and the truck
is increasing at a rate of 4 kilometers per hour at that time.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 3
Solution
x2 + y2 = z2 : Relationship of the variables.dx
dt= 50kph rate of change for the truck.
dy
dt= −60kph rate of change for the car.
Answer:dz
dt= 4 kph. The distance between the car and the truck
is increasing at a rate of 4 kilometers per hour at that time.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 3
Solution
x2 + y2 = z2 : Relationship of the variables.dx
dt= 50kph rate of change for the truck.
dy
dt= −60kph rate of change for the car.
Answer:dz
dt= 4 kph. The distance between the car and the truck
is increasing at a rate of 4 kilometers per hour at that time.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 4
Example
Suppose that two motorboats leave from the same point at thesame time. If one travels nort at 15 miles per hour and the othertravels east at 20 miles per hour, how fast will the distancebetween them be changing after 2 hours?
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 4
Example
Suppose that two motorboats leave from the same point at thesame time. If one travels nort at 15 miles per hour and the othertravels east at 20 miles per hour, how fast will the distancebetween them be changing after 2 hours?
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 4
Example
Suppose that two motorboats leave from the same point at thesame time. If one travels nort at 15 miles per hour and the othertravels east at 20 miles per hour, how fast will the distancebetween them be changing after 2 hours?
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 5
Example
A conical tank is 10 ft across the top (diamter is 10ft) and 12 feetdeep. If water is flowing into the tank at a rate of 10 cubic ft perminute, find the rate of change of teh depth of the water when thewater is 8 feet deep?
Height of the tank 12 ft.
Rate at which the water flows into the tank:dV
dt= 10ft3/min
Rate of change of the depth of the water when the water is 8
ft deep:dh
dt=?
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 5
Example
A conical tank is 10 ft across the top (diamter is 10ft) and 12 feetdeep. If water is flowing into the tank at a rate of 10 cubic ft perminute, find the rate of change of teh depth of the water when thewater is 8 feet deep?
Height of the tank 12 ft.
Rate at which the water flows into the tank:dV
dt= 10ft3/min
Rate of change of the depth of the water when the water is 8
ft deep:dh
dt=?
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 5
Example
A conical tank is 10 ft across the top (diamter is 10ft) and 12 feetdeep. If water is flowing into the tank at a rate of 10 cubic ft perminute, find the rate of change of teh depth of the water when thewater is 8 feet deep?
Height of the tank 12 ft.
Rate at which the water flows into the tank:dV
dt= 10ft3/min
Rate of change of the depth of the water when the water is 8
ft deep:dh
dt=?
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 5
Example
A conical tank is 10 ft across the top (diamter is 10ft) and 12 feetdeep. If water is flowing into the tank at a rate of 10 cubic ft perminute, find the rate of change of teh depth of the water when thewater is 8 feet deep?
Height of the tank 12 ft.
Rate at which the water flows into the tank:dV
dt= 10ft3/min
Rate of change of the depth of the water when the water is 8
ft deep:dh
dt=?
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 5
Example
A conical tank is 10 ft across the top (diamter is 10ft) and 12 feetdeep. If water is flowing into the tank at a rate of 10 cubic ft perminute, find the rate of change of teh depth of the water when thewater is 8 feet deep?
Height of the tank 12 ft.
Rate at which the water flows into the tank:dV
dt= 10ft3/min
Rate of change of the depth of the water when the water is 8
ft deep:dh
dt=?
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 5
Example
A conical tank is 10 ft across the top (diamter is 10ft) and 12 feetdeep. If water is flowing into the tank at a rate of 10 cubic ft perminute, find the rate of change of teh depth of the water when thewater is 8 feet deep?
Height of the tank 12 ft.
Rate at which the water flows into the tank:dV
dt= 10ft3/min
Rate of change of the depth of the water when the water is 8
ft deep:dh
dt=?
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 5
Using the formula V =4
3πr2h and the relationship r =
5
12h
we have V =25
432πh3
Differentiating both sides of the equation with respect to time
gives usdV
dt=
75
432πh2
dh
dt.
Substitution of the known quantities gives usdh
dt=
9
10πft/min.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 5
Using the formula V =4
3πr2h and the relationship r =
5
12h
we have V =25
432πh3
Differentiating both sides of the equation with respect to time
gives usdV
dt=
75
432πh2
dh
dt.
Substitution of the known quantities gives usdh
dt=
9
10πft/min.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 5
Using the formula V =4
3πr2h and the relationship r =
5
12h
we have V =25
432πh3
Differentiating both sides of the equation with respect to time
gives usdV
dt=
75
432πh2
dh
dt.
Substitution of the known quantities gives usdh
dt=
9
10πft/min.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
Example 5
Using the formula V =4
3πr2h and the relationship r =
5
12h
we have V =25
432πh3
Differentiating both sides of the equation with respect to time
gives usdV
dt=
75
432πh2
dh
dt.
Substitution of the known quantities gives usdh
dt=
9
10πft/min.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
EXERCISES
1 A 15 foot ladder is resting against the wall. The bottom isinitially 10 feet away from the wall and is being pushedtowards the wall at a rate of ft/sec. How fast is the top of theladder moving up the wall 12 seconds after we start pushing?
2 A spherical snowball is being made so that its volume isincreasing at the rate of 8 ft3/min. Find the rate at which theradius is increasing when the snowball is 4ft in diameter.
3 Sand is being dropped at the rate of 10 m3/min onto aconical pile. If the height of the pile is always twice the baseradius, at what rate is height increasing when the pile is 8 mhigh?
4 A man 6 ft tall is walking toward a building at the rate of 5ft/sec. If there is a light one the ground 50 ft from thebuilding, how fast is the man’s shadow on the buildinggrowing shorter when he is 30 ft from the building?
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
EXERCISES
1 A trough is 12 ft long and its ends are in the form of invertedisosceles triangles having an altitude of 3 ft and base of 3 ft.Water is flowing into the trough at the rate of 2 ft3/min.How fast is the water level rising when the water is 1 ft deep?
2 A stone is dropped into a still pond. Concentric circularripples spread out, and the radius of disturbed region increasesat the rate of 16 cm/sec. At what rate does the area of thedisturbed region increase when its radius is 4 cm?
3 An automobile traveling at a rate of 30 ft/sec is approachingan intersection. When the automobile is 120 ft from theintersection, a truck traveling at the rate of 40 ft/sec crossesthe intersection. The automobile and the truck are on roadsthat are right angles to each other. How fast are theautomobile and the truck separating 2 sec after the truckleaves the intersection?
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
The Derivative as a Rate of Change
Example
A cylindrical tank has a vertical axis and is initially filled with 600gal of water. This tank takes 60 min to empty after a drain in itsbottom is opened. Suppose that the drain is opened at time t = 0.Suppose also that the volume V of water remaining in the tankafter t minutes is
V (t) =1
6(600− t)2 gal.
Find the instantaneous rate at which the water is flowing out ofthe tank at time t = 15 min and at time t = 45 min. Also, findthe average rate at which the water flows out of the tank duringthe first half hour from t = 15 to t = 45.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
The Derivative as a Rate of Change
Exercises (TC7, pp 160-167)
1 Let A(x) cm2 be the area of a square having a side of xcentimeters, measured to four significant digits. On yourcalculator compute the average rate of change of A(x) withrespect to x as x changes from (a) 4.000 to 4.600; (b) 4.000to 4.300; (c) 4.000 to 4.100; (d) 4.000 to 4.050; (e) What isthe instantaneous rate of change of A(x) with respect to xwhen x is 4.000?
2 The length of rectangle is 4 in. more than its width, and the4-in. difference is maintained as the rectangle increases insize. Let A(w) square inches be the area of the rectanglehaving a width of w inches, measured to four significantdigits. On your calculator compute the average rate of changeof A(w) with respect to w as w changes from (a) 3.000 to3.200; (b) 3.000 to 3.100; (c) 3.000 to 3.010; (d) 3.000 to3.001; (e) What is the instantaneous rate of change of A(w)with respect to w when w is 3.000?
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
The Derivative as a Rate of Change
Exercises (TC7, pp 160-167)
1 Let A(x) cm2 be the area of a square having a side of xcentimeters, measured to four significant digits. On yourcalculator compute the average rate of change of A(x) withrespect to x as x changes from (a) 4.000 to 4.600; (b) 4.000to 4.300; (c) 4.000 to 4.100; (d) 4.000 to 4.050; (e) What isthe instantaneous rate of change of A(x) with respect to xwhen x is 4.000?
2 The length of rectangle is 4 in. more than its width, and the4-in. difference is maintained as the rectangle increases insize. Let A(w) square inches be the area of the rectanglehaving a width of w inches, measured to four significantdigits. On your calculator compute the average rate of changeof A(w) with respect to w as w changes from (a) 3.000 to3.200; (b) 3.000 to 3.100; (c) 3.000 to 3.010; (d) 3.000 to3.001; (e) What is the instantaneous rate of change of A(w)with respect to w when w is 3.000?
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
The Derivative as a Rate of Change
3 Suppose a right-circular cylinder has a constant height of10.00 in. Let V cubic inches be the volume of theright-circular cylinder and r inches be the radius of its base.Find the average rate of change of V with respect to r as rchanges from (a) 5.00 to 5.40; (b) 5.00 to 5.10; (c) 5.00 to5.01; (d) What is the instantaneous rate of change of V withrespect to r when r is 5.00?
4 Let r inches be the radius of a circular metal plate of areaA(r) square inches and circumference C (r) inches. If heat isexpanding the plate, find (a) the instantaneous rate of changeof A(r) with respect to r , and (a) the instantaneous rate ofchange of C (r) with respect to r . (c) Compare your answersin parts (a) and (b) and explain how these rates of changediffer.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change
The Derivative as a Rate of Change
3 Suppose a right-circular cylinder has a constant height of10.00 in. Let V cubic inches be the volume of theright-circular cylinder and r inches be the radius of its base.Find the average rate of change of V with respect to r as rchanges from (a) 5.00 to 5.40; (b) 5.00 to 5.10; (c) 5.00 to5.01; (d) What is the instantaneous rate of change of V withrespect to r when r is 5.00?
4 Let r inches be the radius of a circular metal plate of areaA(r) square inches and circumference C (r) inches. If heat isexpanding the plate, find (a) the instantaneous rate of changeof A(r) with respect to r , and (a) the instantaneous rate ofchange of C (r) with respect to r . (c) Compare your answersin parts (a) and (b) and explain how these rates of changediffer.
Francis Joseph H. Campena Mathematics Department De La Salle UniverisityDerivative as a Rate of Change