Motion & Derivative Applications

PDM Standard 9.B

9.B Proficiency Scale

The Derivative at a Point

Chapter 9 Section 2

Calculating Instantaneous Velocity

Consider the example from the last lesson:

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A projectile is propelled into the air from ground level with an initial velocity of 800 ft/sec. If Earth’s gravity is considered to be constant, it’s height (in feet) after t seconds is given by the function

h(t) = 800t - 16t²

Calculating Instantaneous Velocity

To determine the velocity of the projectile at a particular instant of time, we calculated the difference quotient (the equation for the slope) and used it to find the average velocity over smaller and smaller intervals.

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Example:

• Average velocity over interval from 5 to 6 seconds is 624 ft/sec
• Average velocity over interval from 5 to 5.5 seconds is 632 ft/sec
• Average velocity over interval from 4.9 to 5 second is 641.6 ft/sec

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Calculating Instantaneous Velocity

This process of calculating average velocities over smaller and smaller intervals could be continued forever, and the numbers obtained would come closer and closer to the projectile’s velocity at exactly 5 seconds after launch.

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For this reason, we define the term instantaneous velocity to mean the limit approached by these numbers as the lengths of the time intervals approach zero.

Definition

Suppose an object is moving so that at each time t it is at position f(t), Then the instantaneous velocity at time t is

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Provided the limit exists and is finite

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Calculating instantaneous velocity example

For the projectile modeled by the equation h(t) = 800t - 16t², what is the instantaneous velocity at time t = 5?

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Recall that the difference quotient was 800 - 32t - 16𝚫t

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Calculating instantaneous velocity example

For the projectile modeled by the equation h(t) = 800t - 16t², what is the instantaneous velocity at time t = 5?

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Picturing Instantaneous Velocity

Just as the idea of average velocity has a geometric interpretation in terms of slopes of secant lines, the idea of instantaneous velocity alo has a geometric interpretation as a tangent line.

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The slopes of the secant lines get closer and closer to the slope of the tangent line.

Pictured Example

We can see from the graph that as the intervals decrease, the slope of the secant line gets closer and closer to the slope of the tangent line.

The Derivative at a Point

In the discussion above, the quantity

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appears both in the definition of instantaneous rate of change and in the slope of the tangent line. This quantity arises naturally in many other settings, so it has a special name. It is called the derivative of f at x.

The Derivative at a Point

The derivative of a real function of f at x, denoted

is given by:

Provided this limit exists and is finite

Finding the Derivative at a Point

Let f(x) = x² + 4 for all real numbers x. Find f’(x), then find f’(3), the derivative of f at x = 3, and explain its meaning.

Finding the Derivative at a Point

Let f(x) = x² + 4 for all real numbers x.

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Let’s begin by finding f’(x) using the definition:

Finding the Derivative at a Point

Let f(x) = x² + 4 for all real numbers x.

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Let’s begin by finding f’(x) using the definition:

Finding the Derivative at a Point

Let f(x) = x² + 4 for all real numbers x.

Now let’s find f’(3) using our difference quotient

2x + 𝚫x = 2(3) + 𝚫x = 6 + 𝚫x

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Recall that to find the instantaneous velocity we look for the limit of the difference quotient as 𝚫x approaches zero:

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Finding the Derivative at a Point

Let’s think about the meaning. This means that when

x = 3, the instantaneous rate of change of the function f(x) = x² + 4 is 6. Which also means that at this point in time, the function is increasing 6 times faster in the y-direction than in the x-direction.

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We can also interpret this as the tangent line going through the function at (3, f(3)) will have a slope of 6.

Finding the Derivative at a Point

To check your answer, we can graph the original function

f(x) = x² + 4 and sketch the tangent line at point (3, f(3)). We can see that the slope of this line seems approximately equal to 6.

Instantaneous Rates of Change are Derivatives

Velocity is only one of many kinds of rates of change. It applies when x represents time and f(x) represents the position of an object in one dimension such as height or distance.

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If x and y are any quantities related by an equation f(x), then, for a particular value of x, the instantaneous rate of change of f at x is defined to be the derivative of f at x.

Finding the Equation of the Tangent Line

Let’s think about the last example: f(x) = x² + 4

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We found that f’(3) = 6. This means that the slope when x = 3 is 6. To find the equation of the tangent line, let’s find the point that we are going through.

f(3) = 3² + 4 = 13 so we are going through the point (3, 13)

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We can use point-slope form to write the equation:

y - 13 = 6(x - 3)

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Finding the Equation of the Tangent Line

Find the equation of the tangent line for f(x) = 2x² + 3 at x = 5 given that f’(5) = 20.

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Finding the Equation of the Tangent Line

Find the equation of the tangent line for f(x) = 2x² + 3 at x = 5 given that f’(5) = 20.

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f’(5) = 20 means the slope is 20

f(5) = 2(5)² + 3 = 53

Therefore we are looking for the equation of the line with a slope of 20 going through the point (5, 53)

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y - 53 = 20(x - 5)

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Homework

Khan Academy:

• Derivative as the slope of the curve
• The derivative & tangent line equations
• Derivative as a limit

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Need more help? Check this out

The Derivative Function

Chapter 9 Section 3

The Derivative Function (Tables & Graphs)

Recall that the derivative of a curve correlates to the slope at specified points.

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Let’s fill in the second column by plugging in values of x into function f(x)

The Derivative Function (Tables & Graphs)

Recall that the derivative of a curve correlates to the slope at specified points.

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Now, let’s graph the points for f(x) on the graph so we can estimate the slope at each point

The Derivative Function (Tables & Graphs)

Now, let’s graph the points for f(x) so we can estimate the slope at each point.

The Derivative Function (Tables & Graphs)

Use graphing software, and you can zoom in on each point to find the slope of points very close to each other.

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Ex: to find slope at x = -4

Zoom in and find points close by

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7.41 - 8.61 = -1.2 = -6

-3.9 - (-4.1) .2

So we can say f’(-4) = -6

The Derivative Function (Tables & Graphs)

Recall that the derivative of a curve correlates to the slope at specified points.

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Use this strategy to fill in the “Slope at x” column of the table

The Derivative Function (Tables & Graphs)

Recall that the derivative of a curve correlates to the slope at specified points.

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The Derivative Function (Tables & Graphs)

Now let’s use this table to sketch a graph of both f(x) and f’(x) on the same set of axes.

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The Derivative Function

(Tables & Graphs)

Now let’s use this table to sketch a graph of both f(x) and f’(x) on the same set of axes.

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Definition

Suppose that f is a function that has a derivative f’(x) at each point x in the domain of f. Then the function f’(x) for all x in the domain of f is called the derivative function of f.

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You may recall from the last lesson that we determined that the difference quotient of f(x) is the derivative of f(x):

Definition

We also determined that to find instantaneous velocity we can look at the limit of f’(x) as 𝚫x approaches zero:

Definition

Let’s now use this formula to find the derivative of f(x) = x² + 2x

Definition

f(x) = x² + 2x

Definition

The Derivative Function (Tables & Graphs)

Does our derivative function f’(x) = 2x + 2 match our graph and table?

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Finding the Formula for Velocity

Suppose the height (ft) of a projectile t seconds after launch is given by h(t) = 800t - 16t². Find a formula for its velocity after launch.

Finding the Formula for Velocity

Suppose the height (ft) of a projectile t seconds after launch is given by h(t) = 800t - 16t². Find a formula for its velocity after launch.

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Since we are looking for a formula for instantaneous velocity, we will use the formula:

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Finding the Formula for Velocity

Suppose the height (ft) of a projectile t seconds after launch is given by h(t) = 800t - 16t². Find a formula for its velocity after launch.

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Finding the Formula for Velocity

Since the formula for its velocity after launch is

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We could use this to determine the velocity of the object at any point in time.

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For example, after 30 seconds the velocity is

800 - 32(30) = -160 ft/sec, which means that the projectile is moving downward at a speed of 160 ft/sec.

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Derivative Rules

The Derivative of a Linear Function: f(x) = mx + b

Is f’(x) = m (pertains to the slope)

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The Derivative of a Constant Function: f(x) = k

Is f’(x) = 0 (the slope of a constant function is zero)

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Finding a Formula for the Derivative of a Quadratic Function

As you’ve seen, our projectile motion problems tend to be modeled by a quadratic equation in the form:

f(x) = ax² + bx + c

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Is it possible for us to find a formula for the derivative of any quadratic function in this form?

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Let’s use the formula to find out:

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Finding a Formula for the Derivative of a Quadratic Function

y = ax² + bx + c

Finding a Formula for the Derivative of a Quadratic Function

Finding a Formula for the Derivative of a Quadratic Function

Therefore, we can say that the derivative of a quadratic function: f(x) = ax² + bx + c

Is f’(x) = 2ax + b for all real numbers x.

So the derivative of a quadratic function is a linear function

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When we look back at the last two examples we did, this formula applies:

h(t) = 800t - 16t² f(x) = x² + 2x

h’(t) = -32t + 800 f’(x) = 2x + 2

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Derivative Rules

The Derivative of a Quadratic Function: f(x) = ax² + bx + c

Is f’(x) = 2ax + b for all real numbers x.

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The Derivative of a Power Function: f(x) = axⁿ

Is f’(x) = nax^n-1 for all real numbers a and n

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Note: another way to describe a derivative is

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Ex:

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Derivative Rules

Find the derivative of each of the following functions using the derivative rules:

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Derivative Rules

Find the derivative of each of the following functions using the derivative rules:

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f’(x) = 14x

f’(x) = -4.5

f’(x) = 14x - 3

f’(x) = 30x⁵ - 12x²

Derivatives of Projectiles

If the function f(x) models a projectile’s position (distance or height) in terms of time, then the derivative of f(x) models the velocity of the object over time. The derivative of the velocity function represents the acceleration of the object over time.

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• If h(t) represents height of the object (ft) over time t (sec)
• Then h’(t) represents the object's velocity in ft/sec
• Slope of h(t)
• And h’’(t) represents the object's acceleration in ft/sec²
• Slope of h’(t)

Derivatives of Projectiles

A particle moves so that the distance s traveled in meters at time t seconds is given by the function s(t) = t² + 5t + 4

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• Find the average velocity between 3 and 4 seconds
• Find the instantaneous velocity of the particle at time t = 8
• What is the initial velocity of the particle?

Derivatives of Projectiles

A particle moves so that the distance s traveled in meters at time t seconds is given by the function s(t) = t² + 5t + 4

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Let’s begin by finding the derivative of the equation:

s’(t) = 2t + 5

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Now we can use this to answer the questions.

Derivatives of Projectiles

s(t) = t² + 5t + 4 s’(t) = 2t + 5

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• Find the average velocity between 3 and 4 seconds

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s’(3) = 2(3) + 5 = 11 13 + 11 = 24 = 12 m/sec

s’(4) = 2(4) + 5 = 13 2 2

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Derivatives of Projectiles

s(t) = t² + 5t + 4 s’(t) = 2t + 5

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• Find the instantaneous velocity of the particle at time t = 8

s’(8) = 2(8) + 5 = 21 m/sec

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• What is the initial velocity of the particle?

Initial velocity occurs at time t = 0

s’(0) = 2(0) + 5 = 5 m/sec

Homework

Khan Academy:

• Power Rule (Positive Integer Powers)
• Differentiate Polynomials

Homework

Acceleration and Deceleration

Chapter 9 Section 4

Acceleration and Deceleration Activity

Consider the function f(x) = x²

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Fill in the table of values for f(x) shown.

Acceleration and Deceleration Activity

Consider the function f(x) = x²

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Now let’s find the derivative f’(x) and use it to fill out the table for the slope of the tangent line (instantaneous velocity at time t)

Acceleration and Deceleration Activity

Consider the function f(x) = x²

f’(x) = 2x

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Now sketch a graph of f(x) as well as the graph of f’(x) on the same set of axes.

Acceleration and Deceleration Activity

Consider the function f(x) = x²

f’(x) = 2x

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We can see that when the graph is a quadratic, the derivative ends up being linear

Acceleration and Deceleration Activity

Consider the function

f(x) = -x² + 3

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Fill in the table of values for f(x) shown.

Acceleration and Deceleration Activity

Consider the function

f(x) = -x² + 3

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Now let’s find the derivative f’(x) and use it to fill out the table for the slope of the tangent line (instantaneous velocity at time t)

Acceleration and Deceleration Activity

Consider the function

f(x) = -x² + 3

f’(x) = -2x

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Now sketch a graph of f(x) as well as the graph of f’(x) on the same set of axes.

Acceleration and Deceleration Activity

Consider the function f(x) = -x² + 3

f’(x) = -2x

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Reflection Questions

Finish each sentence

• When the slope of the tangent lines are negative, the graph of the derivative…
• When the slope of the tangent lines are zero the graph of the derivative…
• When the slope of the tangent lines are positive, the graph of the derivative...

Reflection Questions

Finish each sentence

• When the slope of the tangent lines are negative, the graph of the derivative is below the x-axis (negative y)

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• When the slope of the tangent lines are zero (happens at maxima and minima) the graph of the derivative crosses the x-axis (at y = 0)

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• When the slope of the tangent lines are positive, the graph of the derivative is above the x-axis (positive y)

Projectile Problems

When we are looking at problems that measure the position of a projectile over time h(t), recall that

• The derivative h’(t) represents the velocity (speed) of the projectile.
• Units are ft/sec or m/sec
• The derivative of the velocity equation h’’(t) represents the acceleration of the projectile.
• Units are in ft/sec² or m/sec²

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Knowing this, what is the acceleration from #1 and #2?

Projectile Problems

• f(x) = x² original distance

f’(x) = 2x first derivative velocity

f’’(x) = 2 second derivative acceleration

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• f(x) = -x² + 3 original distance

f’(x) = -2x first derivative velocity

f’’(x) = -2 second derivative acceleration

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Introduction to Acceleration

Acceleration

So, acceleration is the slope of the tangent line of the velocity graph, or the derivative of the velocity function.

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So if the path of an object is modeled by f(x) = x², then the velocity is modeled by f’(x) = 2x and the acceleration is modeled by f’’(x) = 2

Acceleration

Let h(t) = -16t² + 40t + 2 be the height of a ball (in feet) tossed upwards at a time of t seconds.

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• Find v(t)
• Find the velocity of the ball when t = 1
• Find a(t)
• Find the acceleration of the ball when t = 1

Acceleration

Let h(t) = -16t² + 40t + 2 be the height of a ball (in feet) tossed upwards at a time of t seconds.

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• Find v(t)

v(t) = -32t + 40

• Find the velocity of the ball when t = 1

v(1) = -32(1) + 40 = 8 ft/sec

Acceleration

Let h(t) = -16t² + 40t + 2 be the height of a ball (in feet) tossed upwards at a time of t seconds.

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• Find a(t)

a(t) = -32 ft/sec²

• Find the acceleration of the ball when t = 1

The acceleration is -32 ft/sec²

Acceleration

Note: When acceleration is constant, we tend to not speak of instantaneous acceleration at time t, but simply the acceleration (since it doesn’t change)

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Since acceleration is the derivative of velocity, we can say that acceleration at a specific point in time is the slope of the tangent line at that point on the velocity graph.

Summarizing Projectile Applications

Recall that projectile equations are of the form

h(t) = -½gt² + v₀t + h₀ where g is the acceleration due to gravity (gravitational constant), v₀ is initial velocity, and h₀ is initial height.

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• The first derivative h’(t) = -gt + v₀ is linear and represents velocity
• And the second derivative h’’(t) = -g is just the acceleration due to gravity.

Acceleration - Deeper Dive

Read this article and try the embedded practice problems

Homework

Khan Academy:

• Calculating average acceleration from graphs
• Connecting acceleration and velocity graphs

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Remember… acceleration is just the slope of the velocity vs. time graph.

Homework