Applications of the Derivative

By Michele Bunch, Kimberly Duane and Mark Duane

Extreme ValuesCritical values – any value along f(x) where

f’(x) = 0 OR f’(x) is undefined

These help you determine extreme values: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 wordsIf 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

Endpoints

Because x=4 is not in the interval, you don’t have to test it IN THIS CASE!!

Absolute max: x=2Absolute min: x=-3

1st Derivative TestThis 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.

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.

1 4x = 1 is a local max; x = 4 is a local min

2nd Derivative TestStill a test for determining local max and min!

1. Take the 2nd derivative, set it equal to zero, and solve.

2. Place these values on a number line for f”(x) and use f”(x) to determine if f”(x) is positive or negative.

5/2

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

ConcavityConcave Up Concave Down

f’(x) is increasingf”(x) is positiveThis means the slopes are

always increasing

f’(x) is decreasingf”(x) is negativeThis means the slopes

are always decreasing

Looks like this:

Looks like this:

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 TheoremGiven 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

ExamplesFind all values of “c” that satisfy the MVT.1) [-7,1]f(-7)=-2f(1)=0f’(c)=2/8=1/4

2) y=x ¹ [1,4]⁻f(1)=1f(4)=1/2f’(c)=-1/6

OptimizationUsing 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 equations4) Determine domain

5) Find 1st derivative and critical points6) Find maximum/minimum

7) Give answer in a sentence

Example OneTwo 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 TwoGiven 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)

3-xx²

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 rates3) Write formula to compare rates

4) Take the derivative5) Plug in the known rate6) Solve for unknown rate

Example OneA 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?

The angle is changing at a rate of -1/8 radians per second.

ɵx

h 20

Example TwoWater 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?

The water level is rising at 20/(9π) inches per second.

4h

3

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 OneGiven 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/sC)8.5 m/s

Example TwoThe 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 OneThe 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.