Calculus: Rates of Change - Furman Universitymath.furman.edu/~dcs/courses/math15/lectures/lecture-21.pdf · Calculus: Rates of Change Mathematics 15: Lecture 21 Dan Sloughter Furman

Calculus: Rates of ChangeMathematics 15: Lecture 21

Dan Sloughter

Furman University

November 8, 2006

Dan Sloughter (Furman University) Calculus: Rates of Change November 8, 2006 1 / 20

Newton and Leibniz

I Isaac Newton

I 1642 - 1727I Mathematician, astronomer, physicist, alchemist, Master of the Mint,

et cetera

I Gottfried Wilhelm Leibniz

I 1646 - 1716I Philosopher, mathematician, logicianI Conceptualized a mechanical computing device


Newton and Leibniz

I Isaac NewtonI 1642 - 1727

I Mathematician, astronomer, physicist, alchemist, Master of the Mint,et cetera




Newton and Leibniz

I Isaac NewtonI 1642 - 1727I Mathematician, astronomer, physicist, alchemist, Master of the Mint,

et cetera




Newton and Leibniz


et cetera




Newton and Leibniz


et cetera

I Gottfried Wilhelm LeibnizI 1646 - 1716

I Philosopher, mathematician, logicianI Conceptualized a mechanical computing device


Newton and Leibniz


et cetera

I Gottfried Wilhelm LeibnizI 1646 - 1716I Philosopher, mathematician, logician

I Conceptualized a mechanical computing device


Newton and Leibniz


et cetera

I Gottfried Wilhelm LeibnizI 1646 - 1716I Philosopher, mathematician, logicianI Conceptualized a mechanical computing device


The calculus

I Major break with the mathematics of the Greeks

I Geometric methods replaced with new techniquesI New forms of mathematical argument become acceptableI For one example, the Greek abhorrence of the infinitely small and the

infinitely large would be set aside, to be replaced by a mathematicswhich freely worked with infinities.

I Although the mathematicians of the 17th and 18th centuries never fullyunderstood the subtleties of the infinite, mathematics was set on a pathwhich would eventually have to face these problems and offer solutions.

I Beyond the change in direction for mathematics, the invention ofcalculus represents a milestone of human thought.

I This new mathematics would fuel a scientific, and then an industrial,revolution that would change the way people live, work, and think.


The calculus

I Major break with the mathematics of the GreeksI Geometric methods replaced with new techniques

I New forms of mathematical argument become acceptableI For one example, the Greek abhorrence of the infinitely small and the






The calculus

I Major break with the mathematics of the GreeksI Geometric methods replaced with new techniquesI New forms of mathematical argument become acceptable

I For one example, the Greek abhorrence of the infinitely small and theinfinitely large would be set aside, to be replaced by a mathematicswhich freely worked with infinities.





The calculus

I Major break with the mathematics of the GreeksI Geometric methods replaced with new techniquesI New forms of mathematical argument become acceptableI For one example, the Greek abhorrence of the infinitely small and the






The calculus







The calculus







The calculus







An example

I The problem of velocity: given a function which determines theposition of an object, find the velocity of the object.

I Example: Suppose we start with the knowledge that an object in freefall near the surface of the earth falls a distance x = 16t2 in t seconds.

I Average velocities:

I Since x = 256 feet at time t = 4 and x = 400 feet at time t = 5, theaverage velocity of the object from time t = 4 to time t = 5 is

400− 256

5− 4=

144

1= 144 feet/second.

I Since x = 324 feet at time t = 4.5, the average velocity of the objectfrom time t = 4 to time t = 4.5 is

324− 256

4.5− 4=

68

0.5= 136 feet/second.


An example





400− 256

5− 4=

144

1= 144 feet/second.


324− 256

4.5− 4=

68



An example





400− 256

5− 4=

144

1= 144 feet/second.


324− 256

4.5− 4=

68



An example



I Average velocities:I Since x = 256 feet at time t = 4 and x = 400 feet at time t = 5, the

average velocity of the object from time t = 4 to time t = 5 is

400− 256

5− 4=

144

1= 144 feet/second.


324− 256

4.5− 4=

68



An example



I Average velocities:I Since x = 256 feet at time t = 4 and x = 400 feet at time t = 5, the

average velocity of the object from time t = 4 to time t = 5 is

400− 256

5− 4=

144

1= 144 feet/second.


324− 256

4.5− 4=

68



An example (cont’d)


I Since x = 268.96 at time t = 4.1, the average velocity of the objectfrom time t = 4 to time t = 4.1 is

268.96− 256

4.1− 4=

12.96

0.1= 129.6 feet/second.

I Continuing in this manner, we create the following table:Initial Time Final Time Average Velocity

4.0000 5.0000 144.00004.0000 4.5000 136.00004.0000 4.1000 129.60004.0000 4.0100 128.16004.0000 4.0010 128.01604.0000 4.0001 128.0016



I Average velocities:I Since x = 268.96 at time t = 4.1, the average velocity of the object

from time t = 4 to time t = 4.1 is

268.96− 256

4.1− 4=

12.96



4.0000 5.0000 144.00004.0000 4.5000 136.00004.0000 4.1000 129.60004.0000 4.0100 128.16004.0000 4.0010 128.01604.0000 4.0001 128.0016



I Average velocities:I Since x = 268.96 at time t = 4.1, the average velocity of the object

from time t = 4 to time t = 4.1 is

268.96− 256

4.1− 4=

12.96



4.0000 5.0000 144.00004.0000 4.5000 136.00004.0000 4.1000 129.60004.0000 4.0100 128.16004.0000 4.0010 128.01604.0000 4.0001 128.0016



I But what if we want to know the instantaneous velocity of the objectexactly at time t = 4?

I No matter how small we take the time interval, the procedure wehave outlined above will always give an average velocity, never aninstantaneous velocity.

I Note: we cannot compute the velocity from time t = 4 to time t = 4because that would give us the indeterminate expression:

256− 256

4− 4=

0

0.

I Yet it appears that the average velocities are approaching 128 feet persecond as the time interval shrinks toward 4.






256− 256

4− 4=

0

0.







256− 256

4− 4=

0

0.







256− 256

4− 4=

0

0.




I We can justify our conjecture as follows:

I Suppose we first compute the average velocity of our object over thetime interval from t = 4 to t = 4 + ∆t, where we think of ∆t as asmall, but arbitrary, increment in time.

I Now

x = 16(4 + ∆t)2 = 16(16 + 8∆t + ∆t2) = 256 + 128∆t + 16∆t2

at time t = 4 + ∆t and x = 256 at time t = 4I So the average velocity from time t = 4 to time t = 4 + ∆t is

(256 + 128∆t + 16∆t2)− 256

(4 + ∆t)− 4=

128∆t + 16∆t2

∆t

=∆t(128 + 16∆t)

∆t= 128 + 16∆t.



I We can justify our conjecture as follows:I Suppose we first compute the average velocity of our object over the

time interval from t = 4 to t = 4 + ∆t, where we think of ∆t as asmall, but arbitrary, increment in time.

I Now

x = 16(4 + ∆t)2 = 16(16 + 8∆t + ∆t2) = 256 + 128∆t + 16∆t2


(256 + 128∆t + 16∆t2)− 256

(4 + ∆t)− 4=

128∆t + 16∆t2

∆t

=∆t(128 + 16∆t)

∆t= 128 + 16∆t.





I Now

x = 16(4 + ∆t)2 = 16(16 + 8∆t + ∆t2) = 256 + 128∆t + 16∆t2

at time t = 4 + ∆t and x = 256 at time t = 4

I So the average velocity from time t = 4 to time t = 4 + ∆t is

(256 + 128∆t + 16∆t2)− 256

(4 + ∆t)− 4=

128∆t + 16∆t2

∆t

=∆t(128 + 16∆t)

∆t= 128 + 16∆t.





I Now

x = 16(4 + ∆t)2 = 16(16 + 8∆t + ∆t2) = 256 + 128∆t + 16∆t2


(256 + 128∆t + 16∆t2)− 256

(4 + ∆t)− 4=

128∆t + 16∆t2

∆t

=∆t(128 + 16∆t)

∆t= 128 + 16∆t.



I Justifying the conjecture (cont’d):

I Note: we can now generate the values in the above table by using thevalues 1, 0.5, 0.1, 0.01, 0.01, 0.001, and 0.0001 for ∆t.

I Moreover, it should now be clear that as ∆t approaches 0, the averagevelocities will approach 128.

I In modern terminology, we say that 128 is the limit of the averagevelocities as ∆t approaches 0.

I Notation: we would write

lim∆t→0

(128 + 16∆t) = 128.



I Justifying the conjecture (cont’d):I Note: we can now generate the values in the above table by using the

values 1, 0.5, 0.1, 0.01, 0.01, 0.001, and 0.0001 for ∆t.

I Moreover, it should now be clear that as ∆t approaches 0, the averagevelocities will approach 128.



lim∆t→0

(128 + 16∆t) = 128.




values 1, 0.5, 0.1, 0.01, 0.01, 0.001, and 0.0001 for ∆t.I Moreover, it should now be clear that as ∆t approaches 0, the average

velocities will approach 128.



lim∆t→0

(128 + 16∆t) = 128.





velocities will approach 128.I In modern terminology, we say that 128 is the limit of the average

velocities as ∆t approaches 0.


lim∆t→0

(128 + 16∆t) = 128.





velocities will approach 128.I In modern terminology, we say that 128 is the limit of the average

velocities as ∆t approaches 0.I Notation: we would write

lim∆t→0

(128 + 16∆t) = 128.



I Note: we might try to find the instantaneous velocity of the object atan arbitrary time t.

I To do this, we would first find ∆x , the change in position of theobject from time t to time t + ∆t:

∆x = 16(t + ∆t)2 − 16t2 = 16(t2 + 2t∆t + ∆t2)− 16t2

= 32t∆t + 16∆t2 = ∆t(32t + 16∆t).

I Then the average velocity of the object from time t to time t + h is

∆x

∆t=

∆t(32t + 16∆t)

∆t= 32t + 16∆t.

I Thus the instantaneous velocity of the object is

v = lim∆t→0

(32t + 16∆t) = 32t.





∆x = 16(t + ∆t)2 − 16t2 = 16(t2 + 2t∆t + ∆t2)− 16t2

= 32t∆t + 16∆t2 = ∆t(32t + 16∆t).


∆x

∆t=

∆t(32t + 16∆t)

∆t= 32t + 16∆t.


v = lim∆t→0

(32t + 16∆t) = 32t.





∆x = 16(t + ∆t)2 − 16t2 = 16(t2 + 2t∆t + ∆t2)− 16t2

= 32t∆t + 16∆t2 = ∆t(32t + 16∆t).


∆x

∆t=

∆t(32t + 16∆t)

∆t= 32t + 16∆t.


v = lim∆t→0

(32t + 16∆t) = 32t.





∆x = 16(t + ∆t)2 − 16t2 = 16(t2 + 2t∆t + ∆t2)− 16t2

= 32t∆t + 16∆t2 = ∆t(32t + 16∆t).


∆x

∆t=

∆t(32t + 16∆t)

∆t= 32t + 16∆t.


v = lim∆t→0

(32t + 16∆t) = 32t.



I For example, at time t = 4, v = 128 feet per second, and at timet = 2, v = 64 feet per second.

I Notation: Leibniz would write

v =dx

dt,

and Newton would writev = x .



I For example, at time t = 4, v = 128 feet per second, and at timet = 2, v = 64 feet per second.

I Notation: Leibniz would write

v =dx

dt,

and Newton would writev = x .


Rates of change

I More general problem: finding the instantaneous rate of change ofone variable with respect to another.

I If y is a function of x and ∆y is the change in y as x changes from xto x + ∆x , then

∆y

∆x

is the average rate of change of y with respect to x .

I Thendy

dx= lim

∆x→0

∆y

∆x

is the instantaneous rate of change of y with respect to x .

I We call dydx the derivative of y with respect to x .

I Newton’s word for derivative was fluxion.


Rates of change



∆y

∆x


I Thendy

dx= lim

∆x→0

∆y

∆x





Rates of change



∆y

∆x


I Thendy

dx= lim

∆x→0

∆y

∆x





Rates of change



∆y

∆x


I Thendy

dx= lim

∆x→0

∆y

∆x





Rates of change



∆y

∆x


I Thendy

dx= lim

∆x→0

∆y

∆x





Interpretation

I The interpretation of an instantaneous rate of change depends onwhat the variables represent in the particular situation.

I Example: If x is position and t is time, then dxdt is velocity.

I Example: If v is velocity and t is time, then dvdt is acceleration.


Interpretation





Interpretation





Interpretation (cont’d)

I Geometric interpretation:

I Recall: a straight line y = mx + b has slope m, which is the rate ofchange of y with respect to x .

I That is, for any two points on the line, the change in y divided by thechange in x is always m.

I Put another way, if y = mx + b, then

∆y = m(x + ∆x) + b− (mx + b) = mx + m∆x + b−mx − b = m∆x ,

so∆y

∆x=

m∆x

∆x= m.

I Thusdy

dx= lim

∆x→0m = m.



I Geometric interpretation:I Recall: a straight line y = mx + b has slope m, which is the rate of

change of y with respect to x .

I That is, for any two points on the line, the change in y divided by thechange in x is always m.



so∆y

∆x=

m∆x

∆x= m.

I Thusdy

dx= lim

∆x→0m = m.




change of y with respect to x .I That is, for any two points on the line, the change in y divided by the

change in x is always m.



so∆y

∆x=

m∆x

∆x= m.

I Thusdy

dx= lim

∆x→0m = m.





change in x is always m.I Put another way, if y = mx + b, then


so∆y

∆x=

m∆x

∆x= m.

I Thusdy

dx= lim

∆x→0m = m.





change in x is always m.I Put another way, if y = mx + b, then


so∆y

∆x=

m∆x

∆x= m.

I Thusdy

dx= lim

∆x→0m = m.



I Geometric interpretation (cont’d):

I For a curve given by an equation relating y with x , dydx represents the

slope of the curve at the point (x , y) on the curve.I This is also the slope of the line tangent to the curve at (x , y).I Note: unlike the slope of a line, in general the slope of a curve will

change from point to point on the curve.



I Geometric interpretation (cont’d):I For a curve given by an equation relating y with x , dy

dx represents theslope of the curve at the point (x , y) on the curve.

I This is also the slope of the line tangent to the curve at (x , y).I Note: unlike the slope of a line, in general the slope of a curve will






I This is also the slope of the line tangent to the curve at (x , y).

I Note: unlike the slope of a line, in general the slope of a curve willchange from point to point on the curve.





I This is also the slope of the line tangent to the curve at (x , y).I Note: unlike the slope of a line, in general the slope of a curve will



Example

I For the parabola y = x2,

∆y = (x + ∆x)2 − x2 = x2 + 2x∆x + ∆x2 − x2 = ∆x(2x + ∆x),

so∆y

∆x=

∆x(2x + ∆x)

∆x= 2x + ∆x .

I Hencedy

dx= lim

∆x→0(2x + ∆x) = 2x .

I For example: The curve y = x2 has slope 0 at x = 0, slope 4 atx = 2, and slope −2 at x = −1.


Example


∆y = (x + ∆x)2 − x2 = x2 + 2x∆x + ∆x2 − x2 = ∆x(2x + ∆x),

so∆y

∆x=

∆x(2x + ∆x)

∆x= 2x + ∆x .

I Hencedy

dx= lim

∆x→0(2x + ∆x) = 2x .



Example


∆y = (x + ∆x)2 − x2 = x2 + 2x∆x + ∆x2 − x2 = ∆x(2x + ∆x),

so∆y

∆x=

∆x(2x + ∆x)

∆x= 2x + ∆x .

I Hencedy

dx= lim

∆x→0(2x + ∆x) = 2x .



Example (cont’d)

I More generally, if k is a constant and y = kx2, then

dy

dx= 2kx .

I In particular, if x = 16t2, the distance a falling ball moves in tseconds, then

dx

dt= 32t,

the velocity as we saw before.


Example (cont’d)


dy

dx= 2kx .

I In particular, if x = 16t2, the distance a falling ball moves in tseconds, then

dx

dt= 32t,

the velocity as we saw before.


Example

I For the cubic y = x3,

∆y = (x + ∆x)3 − x3 = (x + ∆x)(x2 + 2x∆x + ∆x2)− x3

= x3 + 3x2∆x + 3x∆x2 + ∆x3 − x3

= ∆x(3x2 + 3x∆x + ∆x2).

I Thus

∆y

∆x=

∆x(3x2 + 3x∆x + ∆x2)

∆x= 3x2 + 3x∆x + ∆x2.

I Hencedy

dx= lim

∆x→0(3x2 + 3x∆x + ∆x2) = 3x2.

I For example, the curve y = x3 has slope 3 at x = −1, slope 0 atx = 0, and slope 3 at x = 1.


Example


∆y = (x + ∆x)3 − x3 = (x + ∆x)(x2 + 2x∆x + ∆x2)− x3

= x3 + 3x2∆x + 3x∆x2 + ∆x3 − x3

= ∆x(3x2 + 3x∆x + ∆x2).

I Thus

∆y

∆x=

∆x(3x2 + 3x∆x + ∆x2)

∆x= 3x2 + 3x∆x + ∆x2.

I Hencedy

dx= lim

∆x→0(3x2 + 3x∆x + ∆x2) = 3x2.



Example


∆y = (x + ∆x)3 − x3 = (x + ∆x)(x2 + 2x∆x + ∆x2)− x3

= x3 + 3x2∆x + 3x∆x2 + ∆x3 − x3

= ∆x(3x2 + 3x∆x + ∆x2).

I Thus

∆y

∆x=

∆x(3x2 + 3x∆x + ∆x2)

∆x= 3x2 + 3x∆x + ∆x2.

I Hencedy

dx= lim

∆x→0(3x2 + 3x∆x + ∆x2) = 3x2.



Example


∆y = (x + ∆x)3 − x3 = (x + ∆x)(x2 + 2x∆x + ∆x2)− x3

= x3 + 3x2∆x + 3x∆x2 + ∆x3 − x3

= ∆x(3x2 + 3x∆x + ∆x2).

I Thus

∆y

∆x=

∆x(3x2 + 3x∆x + ∆x2)

∆x= 3x2 + 3x∆x + ∆x2.

I Hencedy

dx= lim

∆x→0(3x2 + 3x∆x + ∆x2) = 3x2.



Example (cont’d)


dy

dx= 3kx2.


Problems

1. A rock dropped from a height on the planet Rigel 7 travels a distanceof x = 4t2 meters in t seconds.

a. Find the average velocity of the rock from time t = 0 to time t = 4.b. Find the average velocity of the rock from time t = 2 to time t = 4.c. Find the average velocity of the rock from time t = 2 to time t = 2.5.d. Find the average velocity of the rock from time t = 2 to time

t = 2 + ∆t, where ∆t 6= 0.e. Use the preceding result to find the instantaneous velocity of the rock at

time t = 2.f. Find ∆x , the change in x as t changes from t to t + ∆t, ∆t 6= 0.g. Use the preceding result to find v , the instantaneous velocity of the rock

at time t.h. Use the preceding result to find the instantaneous velocity of the rock at

times t = 1, t = 2, and t = 3.


Problems (cont’d)

2. For each of the following, find dydx by first finding ∆y and then

computing lim∆x→0∆y∆x .

a. y = 3xb. y = 5x − 6c. y = 4x2

d. y = 5x2 + 3xe. y = 3x2 − 4f. y = 5x3

3. For each of the following, find dydx using the formulas we developed in

class and then find the slope of the curve at the pointsx = −2,−1, 0, 1, 2.

a. y = 3x + 4b. y = 6x2

c. y = 4x3

d. y = −7x2


Documents

Calculus: Rates of Change - Furman Universitymath.furman.edu/~dcs/courses/math15/lectures/lecture-21.pdf · Calculus: Rates of Change Mathematics 15: Lecture 21 Dan Sloughter Furman