Applications of the Derivative

```By Michele Bunch, Kimberly Duane and Mark Duane
Extreme Values
 Critical values – any value along f(x) where f’(x) = 0
OR f’(x) is undefined
 Relative max/min → over ENTIRE FUNCTION
 Note: NOT ALL CV’S ARE RELATIVE EXTREMA!
 Absolute max/min → over an INTERVAL
 If f(x) is continuous on a closed interval, extrema could be
at CV’s OR at endpoints – check them all!
 Key words
 If it asks WHERE the max/min is, it wants the x value
 If it asks WHAT it is, then it wants the value
Find the absolute max and min on the interval [-3,2]
Critical
Values
Because x=4 is not in the
interval, you don’t have to
test it IN THIS CASE!!
Absolute max: x=2
Absolute min: x=-3
Endpoints
st
1
Derivative Test
This is a test that uses the first derivative to help you find
relative maxes and mins.
1. Find critical values and create a number line of f’(x) with these
values.
1
4
x = 1 is a local max; x = 4 is a local
min
2. Use f’(x) to determine if f’(x) is positive or negative.
-- If f’(x) is positive, f(x) is increasing
-- If f’(x) is negative, f(x) is decreasing
3. If f’(x) changes from positive to negative, the CV is a local max.
If f’(x) changes from negative to positive, the CV is a local min.
nd
2
Derivative Test
Still a test for determining local max and
min!
1. Take the 2nd derivative, set it equal to zero, and solve.
5/2
2. Place these values on a number line for f”(x) and use f”(x) to
determine if f”(x) is positive or negative.
3. For the critical values:
-- If f”(CV) is positive, then the CV is a local min
-- If f”(CV) is negative, then the CV is a local max
Concavity
Concave Up
Looks like
this:
 f’(x) is increasing
 f”(x) is positive
 This means the slopes are
always increasing
Concave Down
Looks like
this:
 f’(x) is decreasing
 f”(x) is negative
 This means the slopes are
always decreasing
f(x), f’(x), and f”(x)
 The graph of the derivative is
essentially the slope of the
tangent line as it travels along
the curve.
 For every point on f(x) that
has a horizontal tangent line,
f’(x) will cross the x axis at
that same point.
 If the slope of the tangent
line is positive and increasing
for f(x), f’(x) will be positive
(and vice versa)
 f’(x) and f”(x) have the same
relationship as f(x) and f”(x)
Mean Value Theorem
Given a differentiable function over an interval, there is
at least one point on the curve where the derivative
is equal to the average derivative of the entire
interval
Examples
Find all values of “c” that satisfy the MVT.
1)
[-7,1]
f(-7)=-2
f(1)=0
f’(c)=2/8=1/4
2) y=x⁻¹
f(1)=1
f(4)=1/2
f’(c)=-1/6
[1,4]
Optimization
Using related formulas to find the maximum or minimum
solution to a problem
Steps:
1)Sketch the problem
2)Write equations (target and constraints)
3)Combine equations
4)Determine domain
5)Find 1st derivative and critical points
6)Find maximum/minimum
Example One
Two pens are to be constructed using a total of 900 feet of
fencing. One pen is to be a square X by X and the other
is to be a rectangle with one side twice as long as the
other (X by 2X). Determine the dimensions of the pens so
that the enclosed areas are as large as possible.
x
x
2x
x
So the dimensions with the
maximum area are 105.9 ft
by 105.9 ft and 79.4 ft by
158.8 ft.
Example Two
Given the curve y=x² in the first quadrant and a vertical
line x=3, determine the inscribed rectangle of
maximum area which has a right side on line x=3.
(x,y)
x²
3-x
The rectangle is 2 by
4, so the area of the
rectangle is 8.
Related Rates
Finding the rate at which a quantity changes by comparing
to a known rate of change
Usually used in relation to time
Steps:
1) Draw a picture
2) State wanted and known rates
3) Write formula to compare rates
4) Take the derivative
5) Plug in the known rate
6) Solve for unknown rate
Example One
A 20 foot ladder is leaning against a building. The ladder is
sliding down the wall at a constant rate of 2 ft/sec. At
what rate is the angle between the ladder and the ground
changing when the top of the ladder is 12 feet from the
ground?
h
20
ɵ
x
The angle is
changing at a
rate of -1/8
second.
Example Two
Water is flowing into and inverted cone at the rate of 5
in³/sec. If the cone has an altitude of 4 inches and a base
radius of 3 inches, how fast is the water level rising when
the water is 2 inches deep?
3
4
h
The water level is
rising at 20/(9π)
inches per second.
Particle Motion



Particle motion is described by a function and its
derivatives.
When given a function that describes a particles
linear motion, the first derivative describes the
particle's velocity and the absolute value of the first
derivative gives speed.
The second derivative gives the particles acceleration.
Example One
Given the following equation that gives the position of a
particle, find A) the velocity equation, B) the velocity
at time 15 minutes, and C) the speed at 15 minutes.
A)
B)f’(15)=-8.5 m/s
C)8.5 m/s
Example Two
 The velocity (in centimeters per second) of a blood
molecule flowing through a capillary of radius 0.008
cm is given by the equation below. Find the
acceleration when r = 0.004
The acceleration
at r=0.004 is 0.00008
centimeters per
second squared.
Linear Approximation


Linear approximation is using the slope of a tangent
line to approximate a value that is close to the point of
tangency.
Δ f(x)=f’(a)-x (yes, the delta is upside down). Δf(x) =
f(a + Δx) – f(a).
Example One
 The cube root of 27 is 3. How much larger is the cube
root of 27.2?






The derivative of f(x) is
Δf '(x)= (1/27)(0.2)
Δf '(x)= 1/135
So the cube root of 27.2 is 1/135 more than the cube
root of 27
Estimated value =3.00740
Actual value=3.00739
Example Two
•
The price of a bus pass between Albuquerque and Los
Alamos is set at x dollars, a bus company takes in a
monthly revenue of
in thousands of
dollars. Estimate the change in revenue from \$50 to \$53.
•
•
•
•
∆R'(50)=.5
.5*∆x = .5*3
= 1.5
So by increasing the tickets by three dollars the
company's revenue goes up by \$1500 to 51500. The actual
value for \$53 is \$514100.
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