Applications of the Derivative

Edward ParkJuan Viera-DimarcoMichael Blanchard

1st Derivative Test

• Take the derivative of f• Every point where f’ = 0 or does not exist is a

critical point on f• Create a number line for f’ to represent this,

mark all critical points on the line• Test points on either side of each critical point• Everywhere the graph of f’ is positive, f is

increasing, and everywhere the graph of f’ is negative, f is decreasing

Practice

f(x) = x2 – 2x + 4f’(x) = 2x – 20 = 2(x – 1)X = 1

X = 1

Test points on either side of x=1f’(0) = 2(0) - 2 = -2, which is negativef’(2) = 2(2) – 2 = 2, which is positive

f’(x)- +

f(x)

Wherever f’(x) changes from + to – f(x) has a maxWherever f’(x) changes from – to + f(x) has a minTherefore, in this equation, there is a min at x=1

Use the first derivative test to find critical points and tell if they are a max or min

2nd Derivative Test• Take the derivative of f’ • Every point where f’’ = 0 or does not exist is a

point of inflection on f• Test points on either side of the point where f’’ =

0 or DNE• Make sure f’’ changes sign, if it doesn’t then

there is no POI• Everywhere the graph of f’’ is positive, f is

concave up• Everywhere the graph of f’’ is negative, f is

concave down

continued

• This test can be used to determine weather a critical point is a max or min

• If f’’ is positive, the point is a min• If f’’ is negative, the point is a max

Practicefind the local max, min and the intervals where f(x) is concave

up and down using the second derivative testf(x) = x3 – 12x2 + 45xf’(x) = 3x2 – 24x + 450 = 3(x2 – 8x + 15)0 = 3(x – 3)(x – 5)x = 3, 5f’’(x) = 6x – 240 = 6(x – 4)X = 4

f’(x)

f’’(x)

f(x)

f(x)

+

f’(0) = 45f’(4) = 3(16) – 96 + 45 = -3f’(6) = 3(36) – 144 + 45 = 9

- +

f’’(0) = -24f’’(5) = 6(5) – 24 = 6

- +

f(x) is concave down on (-∞, 4), concave down on (4,∞)f(x) has a local max at x=3 and a local min at x=5

x=4

X=3 X=5

Relationships between the graphs of f, f’, and f’’

• Where f’ crosses the x-axis, f is a max or min• Where f’’ crosses the x-axis, f’ is a max or min and f

changes inflection• When f’’ is above the x-axis, f is concave up and f’

is increasing• When f’’ is below the x-axis, f is concave down and

f’ is decreasing

PracticeIdentify which graph is f, which one is f’, and

which one is f’’

a)

b)

c)

f’

f

f’’

MVT (official)

• Assume that f is continuous on the closed interval [a,b] and differentiable on (a,b). Then there exists at least one value c in (a,b) such that f’(c) = [f(b)-f(a)]/[b-a]

• Blah blah blah

MVT (sparknotes version)

• There is a point c between a and b for every function where the slope of the point at c is equal to the slope of the line connecting a and b.

• For example......

Practice• Find all values of “c” that satisfy the MVT for f(x)=(x-

1)^(1/2) [1,3]f(1) = 0f(3) = 2^(1/2)m = 2^(1/2)/(3-1) = 2^(1/2)/2f’(c) = .5(c-1)^(-1/2)2^(1/2)/2 = .5(c-1)^(-1/2)(2^(1/2))^(-2) = [(c-1)^(-1/2)]^(-2)½ = c-13/2 = c

Rolle’s Theorem

• A special case of the MVT where f(a) = f(b).• If f(a) = f(b), then the slope is 0.• This means that f’(c) is either a max or a min.

Optimization1. Sketch2. Write equations

a) target: what you are trying to optimizeb) Constraints: limits on the variable

3. Combine equations: write a function in one variable4. Determine the “feasible domain” of the equation from step

3. Domain gives the “endpoints”.5. Find the 1st derivative and determine critical points where

the function may be maximized or minimized.6. Find the max/min of the target function using endpoints

and critical points7. Write your answer using a complete sentence.

Practice• Farmer Edward wants to put 40 ft of fence along the side of his barn for

his cattle. Find the dimensions and the maximum area of this new addition.

Barnx

x

y

A = xy

2x + y = 40

y = 40-2x

A = x(40-2x)

A = 40x – 2x^2

A’ = 0 = 40 – 4x

40 = 4x; x = 10

2(10) + y = 40

y = 20

The maximum area is 200 ft^2

BarnBarnx

Barn yx

Barn

x

yx

Barn yx

Barnx

Related Rates• In related rate problems, the goal is to

calculate an unknown rate of change in terms of other rates of change that are known

• For example, how fast does the top of a ladder standing against a wall move if the bottom is pulled away from the wall at constant speed?

• Or, how fast does water pour into a tank if the height rises at a constant rate?

• It is important to note that in both of these problems, the rate actually changes with time, and so it is necessary to determine a specific time (t=1, t=9) to find an exact rate

How do we solve related rates?Each related rates problem can be solved using 3 steps

1. Assign variables and restate the problem

2. Find an equation that relates the variables and the differentiate

3. Use the data to find the unknown derivative

If possible, draw a diagram. It can help in many situations.

For example: A 17ft ladder is placed next to a wall. The bottom is 6 feet from the wall at t=0 and slides away from the wall at a rate of 1 ft/sec. What is the velocity of the top of the ladder at time t=2?

To solve this, we use the three steps:

1. Assign variables and restate the problem.

Let x be the distance from the bottom of the ladder to the wall

Let y be the height of the ladder’s top

The problem becomes: Compute at t=2 given that = 1ft/s and x(0) = 10ft

17ft

x

y

2. Find an equation that relates the variables and differentiate.

We need a relation between x and y. We use the Pythagorean Theorem, since we know that the hypotenuse of the right triangle is 17.

Since we need we differentiate with respect to t.

Now we solve for and plug in = 1

3. Use the data to find the unknown derivative.

At t=2, the ladder has moved at a rate of 1ft/sec away from a starting position of 6 feet, so x(2)=8. By the Pythagorean Theorem, y(2)=15. So, we plug it in to our equation.

Therefore, the answer is ft/sec

Another Practice ProblemBen is riding his bike starting at a point 8 miles directly south of his house. At t=0, Ben begins traveling east at 15mi/hr. At what rate is the distance between Ben and his house increasing at t=24 min?

1. Assign variables and restate the problem.

Let b be the distance that Ben has traveled, and let D be the distance between Ben and the house. Then the problem becomes:

Compute at t=24min given =15mi/hr

2. Find an equation that relates the variables and differentiate.

Use the Pythagorean Theorem and differentiate.

Solve for and plug in values.

3. Use the data to find the unknown derivative.

At t=24min=0.4hours, b=15*0.4=6 miles

By the Pythagorean Theorem, D= 10 miles. So, when plugging values in,

Therefore, the answer is 9 mi/hr.

D

b

8mi

Some Helpful Hints

• Rate means a derivative with respect to time• Anything that does not change should be plugged

in before taking the derivaative• “How fast” – the answer will be positive• “What rate” – negative or positive answer• Watch your units!• In related rates involving geometry, remember

the Pythagorean Theorem and similar triangles

Particle Motion• Imagine a particle moving in space. How is it moving? Is it speeding up or slowing

down? Is it going forwards or backwards? How fast is it moving?• By studying particle motion and position, velocity, and acceleration, we can

answer those questions. Some definitions:• Position is the place where the particle is at a certain time. It is represented by s(t)• Velocity is the rate of change of position. It is how fast the particle is moving and

in what direction. It is calculated with the equation and is represented by v(t)• Acceleration is the rate of change of velocity. It is how fast the particle speeds up

or slows down. It is calculated with the equation and is represented by a(t) • Remember that velocity is the derivative of position, and acceleration is the

derivative of velocity• Consequently, velocity is the antiderivative of acceleration, and position is the

antiderivative of velocity

Problems Concerning Particle Motion

• In many problems about motion, you are given the information about the position, velocity, or acceleration of an object and are asked to find a specific part of the motion

• To do these problems, you need to use your initial information to make an equation for position, velocity, or acceleration. Then, either differentiate or integrate the equation to a new equation that you want, and find the wanted piece of information.

For example,A large truck enters the off-ramp of a freeway at t=0. Its position after t seconds is for .(a) How fast is the truck going at the moment it enters the off-ramp?(b) Is the truck speeding up or slowing down?

Since velocity is the derivative of position, (a) Since the truck enters the off-ramp at t=0, the velocity is v(0)=84 ft/sec(b) If you look at the graph of the velocity function, you notice that the graph drops downwards. So, the truck is slowing down. Alternatively, you could take the derivative of velocity and get a(t)=-6t, which is negative. So, the truck slows down.

Practice• In many motion problems, equations are not just given to you; they have to be derived from

the given information. For example:

A rock is dropped from a height of 300m. What is its velocity when it hits the ground?

To solve this problem, we first need to come up with an equation . The acceleration due to gravity is -9.8 meters per second squared, so a(t)=-9.8

To find the velocity, we take the antiderivative and use the initial condition v(0)=0 (since at the very start of the drop, the rock is at rest)

Now find the position the same way; take the antiderivative and use the initial condition

We are trying to find out the velocity when it hits the ground, so we need the time at which it hits the ground

The velocity when the rock hits the ground is

The rock hits the ground at a velocity of -76.681 meters per second.

Bibliography

• The book• Mrs. Miller’s Notes