Rate of change calculus problems

Rate of change calculus problems DEFAULT

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As with the first problem we first need to take the derivative of the function.

\[A'\left( t \right) = {t^4} - {t^3} - {t^2} = 15{t^2}\left( {9{t^2} - 12t - 26} \right)\]

Next, we need to determine where the function isn’t changing. This is at,

\[\begin{align*}t & = 0\\ t & = \frac{{12 \pm \sqrt { - 4\left( 9 \right)\left( { - 26} \right)} }}{{18}} = \frac{{12 \pm \sqrt {} }}{{18}} = \frac{{12 \pm 6\sqrt {30} }}{{18}} = \frac{{2 \pm \sqrt {30} }}{3} = - ,\,\,\,\,\end{align*}\]

So, the function is not changing at three values of \(t\). Finally, to determine where the function is increasing or decreasing we need to determine where the derivative is positive or negative. Recall that if the derivative is positive then the function must be increasing and if the derivative is negative then the function must be decreasing. The following number line gives this information.

RateOfChange_Ex2

So, from this number line we can see that we have the following increasing and decreasing information.

\[{\mbox{Increasing :}}\,\, - \infty < t < - ,\,\,\, < t < \infty \,\,\,\,\,\,\,\,{\mbox{Decreasing :}} - < t < 0,\,\,\,0 < t < \]

Sours: https://tutorial.math.lamar.edu/classes/calci/RateOfChange.aspx
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Solve Rate of Change Problems in Calculus

Problem 1

A rectangular water tank (see figure below) is being filled at the constant rate of 20 liters / second. The base of the tank has dimensions w = 1 meter and L = 2 meters. What is the rate of change of the height of water in the tank?(express the answer in cm / sec).
tangent lines to the graph of y = x<sup>3</sup> - 3x

Solution to Problem 1:
  • The volume V of water in the tank is given by.
    V = w*L*H
  • We know the rate of change of the volume dV/dt = 20 liter /sec. We need to find the rate of change of the height H of water dH/dt. V and H are functions of time. We can differentiate both side of the above formula to obtain
    dV/dt = W*L*dH/dt
  • note W and L do not change with time and are therefore considered as constants in the above operation of differentiation.
  • We now find a formula for dH/dt as follows.
    dH/dt = dV/dt / W*L
  • We need to convert liters into cubic cm and meters into cm as follows
    1 litter = 1 cubic decimeter
    = cubic centimeters
    = cm 3
    and 1 meter = centimeter.
  • We now evaluate the rate of change of the height H of water.
    dH/dt = dV/dt / W*L
    = ( 20* cm 3 / sec ) / ( cm * cm)
    = 1 cm / sec.


Problem 2

An airplane is flying in a straight direction and at a constant height of meters (see figure below). The angle of elevation of the airplane from a fixed point of observation is a. The speed of the airplane is km / hr. What is the rate of change of angle a when it is 25 degrees? (Express the answer in degrees / second and round to one decimal place).
tangent lines to the graph of y = x<sup>3</sup> - 3x Pin it!Share on Facebook

Solution to Problem 2:
  • The airplane is flying horizontally at the rate of dx/dt = km/hr. We need a relationship between angle a and distance x. From trigonometry, we can write
    tan a = h/x
  • angle a and distance x are both functions of time t. Differentiate both sides of the above formula with respect to t.
    d(tan a)/dt = d(h/x)/dt
  • We now use the chain rule to further expand the terms in the above formula
    d(tan a)/dt = (sec 2 a) da/dt
    d(h/x)/dt = h*(-1 / x 2) dx/dt.
    (note: height h is constant)
  • Substitute the above into the original formula to obtain
    (sec 2 a) da/dt = h*(-1 / x 2) dx/dt
  • The above can be written as
    da/dt = [ h*(-1 / x 2) dx/dt ] / (sec 2 a)
  • We now use the first formula to find x in terms of a and h follows
    x = h / tan a
  • Substitute the above into the formula for da/dt and simplify
    da/dt = [ h*(- tan 2a / h 2) dx/dt ] / (sec 2 a)
    = [ (- tan 2a / h) dx/dt ] / (sec 2 a)
    = (- sin 2a / h) dx/dt
  • Use the values for a, h and dx/dt to approximate da/dt with the right conversion of units: 1km = m and 1 hour = sec.
    da/dt = [- sin 2(25 deg)/ m]*[ m/ sec]
    = radians/sec
    = * [ degrees / Pi radians] /sec
    = degrees/sec

Problem 3

If two resistors with resistances R1 and R2 are connected in parallel as shown in the figure below, their electrical behavior is equivalent to a resistor of resistance R such that
1 / R = 1 / R1 + 1 / R2

If R1 changes with time at a rate r = dR1/dt and R2 is constant, express the rate of change dR / dt of the resistance of R in terms of dR1/dt, R1 and R2.
tangent lines to the graph of y = x<sup>3</sup> - 3x

Solution to Problem 3:
  • We start by differentiating, with respect to time, both sides of the given formula for resistance R, noting that R2 is constant and d(1/R2)/dt = 0
    (-1/R 2)dR/dt = (-1/R1 2)dR1/dt
  • Arrange the above to obtain
    dR/dt = (R/R1) 2dR1/dt
  • From the formula 1 / R = 1 / R1 + 1 / R2, we can write
    R = R1*R2 / (R1 + R2)
  • Substitute R in the formula for dR/dt and simplify
    dR/dt = (R1*R2 / R1*(R1 + R2)) 2dR1/dt
    = (R2 / (R1 + R2)) 2dR1/dt

Exercises

1 - Find a formula for the rate of change dV/dt of the volume of a balloon being inflated such that it radius R increases at a rate equal to dR/dt.
2 - Find a formula for the rate of change dA/dt of the area A of a square whose side x centimeters changes at a rate equal to 2 cm/sec.
3 - Two cars start moving from the same point in two directions that makes 90 degrees at the constant speeds of s1 and s2. Find a formula for the rate of change of the distance D between the two cars.

Solutions to the Above Exercises

1 -    dV/dt = 4*Pi*R 2dR/dt
2 -    dA/dt = 4x cm 2/sec
3 -    dD/dt = sqrt( s1 2 + s2 2 )
More references on calculus problems
Sours: https://www.analyzemath.com/calculus/Problems/rate_change.html
Related Rates - Conical Tank, Ladder Angle \u0026 Shadow Problem, Circle \u0026 Sphere - Calculus

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Change calculus of problems rate

Derivatives as Rates of Change

3. Derivatives

Learning Objectives

  • Determine a new value of a quantity from the old value and the amount of change.
  • Calculate the average rate of change and explain how it differs from the instantaneous rate of change.
  • Apply rates of change to displacement, velocity, and acceleration of an object moving along a straight line.
  • Predict the future population from the present value and the population growth rate.
  • Use derivatives to calculate marginal cost and revenue in a business situation.

In this section we look at some applications of the derivative by focusing on the interpretation of the derivative as the rate of change of a function. These applications include acceleration and velocity in physics, population growth rates in biology, and marginal functions in economics.

Another use for the derivative is to analyze motion along a line. We have described velocity as the rate of change of position. If we take the derivative of the velocity, we can find the acceleration, or the rate of change of velocity. It is also important to introduce the idea of speed, which is the magnitude of velocity. Thus, we can state the following mathematical definitions.

Definition

Let &#;(&#;) be a function giving the position of an object at time &#;.

The velocity of the object at time &#; is given by &#;(&#;)=&#;^&#;\&#;&#;&#;&#;&#;&#;(&#;).

The speed of the object at time &#; is given by &#;&#;(&#;)&#;.

The acceleration of the object at &#; is given by a(&#;)=&#;^&#;\&#;&#;&#;&#;&#;&#;(&#;)=&#;''(&#;).

A particle moves along a coordinate axis. Its position at time &#; is given by &#;(&#;)=&#;^2-5&#;+1. Is the particle moving from right to left or from left to right at time &#;=3?

In addition to analyzing velocity, speed, acceleration, and position, we can use derivatives to analyze various types of populations, including those as diverse as bacteria colonies and cities. We can use a current population, together with a growth rate, to estimate the size of a population in the future. The population growth rate is the rate of change of a population and consequently can be represented by the derivative of the size of the population.

Definition

If P(&#;) is the number of entities present in a population, then the population growth rate of P(&#;) is defined to be P^&#;\&#;&#;&#;&#;&#;&#;(&#;).

Estimating a Population

The population of a city is tripling every 5 years. If its current population is 10,, what will be its approximate population 2 years from now?

Solution

Let P(&#;) be the population (in thousands) &#; years from now. Thus, we know that P(0)=10 and based on the information, we anticipate P(5)=30. Now estimate P^&#;\&#;&#;&#;&#;&#;&#;(0), the current growth rate, using

P^&#;\&#;&#;&#;&#;&#;&#;(0)\a&#;&#;&#;&#;&#; \&#;&#;ac&#;P(5)-P(0)&#;&#;5-0&#;=\&#;&#;ac&#;30-10&#;&#;5&#;=4.

By applying (Figure) to P(&#;), we can estimate the population 2 years from now by writing

P(2)\a&#;&#;&#;&#;&#; P(0)+(2)P^&#;\&#;&#;&#;&#;&#;&#;(0)\a&#;&#;&#;&#;&#; 10+2(4)=18;

thus, in 2 years the population will be approximately 18,

The current population of a mosquito colony is known to be 3,; that is, P(0)=3,000. If P^&#;\&#;&#;&#;&#;&#;&#;(0)=100, estimate the size of the population in 3 days, where &#; is measured in days.

In addition to analyzing motion along a line and population growth, derivatives are useful in analyzing changes in cost, revenue, and profit. The concept of a marginal function is common in the fields of business and economics and implies the use of derivatives. The marginal cost is the derivative of the cost function. The marginal revenue is the derivative of the revenue function. The marginal profit is the derivative of the profit function, which is based on the cost function and the revenue function.

We can roughly approximate

MC(&#;)=C^&#;\&#;&#;&#;&#;&#;&#;(&#;)=\&#;&#;&#;&#;&#;&#;&#;&#;&#;&#;\&#;&#; 0&#;&#;\&#;&#;&#;&#;\&#;&#;ac&#;C(&#;+&#;)-C(&#;)&#;&#;&#;&#;

by choosing an appropriate value for &#;. Since &#; represents objects, a reasonable and small value for &#; is 1. Thus, by substituting &#;=1, we get the approximation MC(&#;)=C^&#;\&#;&#;&#;&#;&#;&#;(&#;)\a&#;&#;&#;&#;&#; C(&#;+1)-C(&#;). Consequently, C^&#;\&#;&#;&#;&#;&#;&#;(&#;) for a given value of &#; can be thought of as the change in cost associated with producing one additional item. In a similar way, MR(&#;)=R^&#;\&#;&#;&#;&#;&#;&#;(&#;) approximates the revenue obtained by selling one additional item, and MP(&#;)=P^&#;\&#;&#;&#;&#;&#;&#;(&#;) approximates the profit obtained by producing and selling one additional item.

Applying Marginal Revenue

Assume that the number of barbeque dinners that can be sold, &#;, can be related to the price charged, &#;, by the equation &#;(&#;)=9-0.03&#;, \, 0\&#;&#; &#;\&#;&#; 300.

In this case, the revenue in dollars obtained by selling &#; barbeque dinners is given by

R(&#;)=&#;&#;(&#;)=&#;(9-0.03&#;)=-0.03&#;^2+9&#;for 0\&#;&#; &#;\&#;&#; 300.

Use the marginal revenue function to estimate the revenue obtained from selling the st barbeque dinner. Compare this to the actual revenue obtained from the sale of this dinner.

Solution

First, find the marginal revenue function: MR(&#;)=R^&#;\&#;&#;&#;&#;&#;&#;(&#;)=-0.06&#;+9.

Next, use R^&#;\&#;&#;&#;&#;&#;&#;(100) to approximate R(101)-R(100), the revenue obtained from the sale of the st dinner. Since R^&#;\&#;&#;&#;&#;&#;&#;(100)=3, the revenue obtained from the sale of the st dinner is approximately $3.

The actual revenue obtained from the sale of the st dinner is

R(101)-R(100)=602.97-600=2.97, or \$2.97.

The marginal revenue is a fairly good estimate in this case and has the advantage of being easy to compute.

Suppose that the profit obtained from the sale of &#; fish-fry dinners is given by P(&#;)=-0.03&#;^2+8&#;-50. Use the marginal profit function to estimate the profit from the sale of the st fish-fry dinner.

Key Concepts

  • Using &#;(a+&#;)\a&#;&#;&#;&#;&#; &#;(a)+&#;^&#;\&#;&#;&#;&#;&#;&#;(a)&#;, it is possible to estimate &#;(a+&#;) given &#;^&#;\&#;&#;&#;&#;&#;&#;(a) and &#;(a).
  • The rate of change of position is velocity, and the rate of change of velocity is acceleration. Speed is the absolute value, or magnitude, of velocity.
  • The population growth rate and the present population can be used to predict the size of a future population.
  • Marginal cost, marginal revenue, and marginal profit functions can be used to predict, respectively, the cost of producing one more item, the revenue obtained by selling one more item, and the profit obtained by producing and selling one more item.

For the following exercises, the given functions represent the position of a particle traveling along a horizontal line.

  1. Find the velocity and acceleration functions.
  2. Determine the time intervals when the object is slowing down or speeding up.

1. &#;(&#;)=2&#;^3-3&#;^2-12&#;+8

2. &#;(&#;)=2&#;^3-15&#;^2+36&#;-10

Solution

a. &#;(&#;)=6&#;^2-30&#;+36, \, a(&#;)=12&#;-30
b. Speeds up: (2,2.5)\c&#;&#; (3,\&#;&#;&#;&#;&#;); Slows down: (0,2)\c&#;&#; (2.5,3)

3. &#;(&#;)=\&#;&#;ac&#;&#;&#;&#;1+&#;^2&#;

4. A rocket is fired vertically upward from the ground. The distance &#; in feet that the rocket travels from the ground after &#; seconds is given by &#;(&#;)=-16&#;^2+560&#;.

  1. Find the velocity of the rocket 3 seconds after being fired.
  2. Find the acceleration of the rocket 3 seconds after being fired.

Solution

a. 464 \, \&#;&#;&#;&#;&#;&#;&#;/&#;&#;^2
b. -32 \, \&#;&#;&#;&#;&#;&#;&#;/&#;&#;^2

5. A ball is thrown downward with a speed of 8 ft/s from the top of a foot-tall building. After &#; seconds, its height above the ground is given by &#;(&#;)=-16&#;^2-8&#;+64.

  1. Determine how long it takes for the ball to hit the ground.
  2. Determine the velocity of the ball when it hits the ground.

7. The position of a hummingbird flying along a straight line in &#; seconds is given by &#;(&#;)=3&#;^3-7&#; meters.

  1. Determine the velocity of the bird at &#;=1 sec.
  2. Determine the acceleration of the bird at &#;=1 sec.
  3. Determine the acceleration of the bird when the velocity equals 0.

9. The position function &#;(&#;)=&#;^3-8&#; gives the position in miles of a freight train where east is the positive direction and &#; is measured in hours.

  1. Determine the direction the train is traveling when &#;(&#;)=0.
  2. Determine the direction the train is traveling when a(&#;)=0.
  3. Determine the time intervals when the train is slowing down or speeding up.

 The cost function, in dollars, of a company that manufactures food processors is given by C(&#;)=200+\&#;&#;ac&#;7&#;&#;&#;&#;+\&#;&#;ac&#;&#;^2&#;&#;7&#;, where &#; is the number of food processors manufactured.

  1. Find the marginal cost function.
  2. Find the marginal cost of manufacturing 12 food processors.
  3. Find the actual cost of manufacturing the thirteenth food processor.

[T] A profit is earned when revenue exceeds cost. Suppose the profit function for a skateboard manufacturer is given by P(&#;)=30&#;-0.3&#;^2-250, where &#; is the number of skateboards sold.

  1. Find the exact profit from the sale of the thirtieth skateboard.
  2. Find the marginal profit function and use it to estimate the profit from the sale of the thirtieth skateboard.

 The centripetal force of an object of mass &#; is given by F(&#;)=\&#;&#;ac&#;&#;&#;^2&#;&#;&#;&#;, where &#; is the speed of rotation and &#; is the distance from the center of rotation.

  1. Find the rate of change of centripetal force with respect to the distance from the center of rotation.
  2. Find the rate of change of centripetal force of an object with mass kilograms, velocity of m/s, and a distance from the center of rotation of meters.

The following questions concern the population (in millions) of London by decade in the 19th century, which is listed in the following table.

Years since Population (millions)
1
11
21
31
41
51
61
71
81
91

[T]

  1. Using a calculator or a computer program, find the best-fit linear function to measure the population.
  2. Find the derivative of the equation in (a) and explain its physical meaning.
  3. Find the second derivative of the equation and explain its physical meaning.

Solution

a. P(&#;)=0.03983+0.4280
b. P^&#;\&#;&#;&#;&#;&#;&#;(&#;)=0.03983. The population is increasing.
c. P''(&#;)=0. The rate at which the population is increasing is constant.

[T]

  1. Using a calculator or a computer program, find the best-fit quadratic curve through the data.
  2. Find the derivative of the equation and explain its physical meaning.
  3. Find the second derivative of the equation and explain its physical meaning.

For the following exercises, consider an astronaut on a large planet in another galaxy. To learn more about the composition of this planet, the astronaut drops an electronic sensor into a deep trench. The sensor transmits its vertical position every second in relation to the astronaut’s position. The summary of the falling sensor data is displayed in the following table.

Time after dropping (s)Position (m)
00
1−1
2−2
3−5
4−7
5−14

[T]

  1. Using a calculator or computer program, find the best-fit quadratic curve to the data.
  2. Find the derivative of the position function and explain its physical meaning.
  3. Find the second derivative of the position function and explain its physical meaning.

Solution

a. &#;(&#;)=-0.6071&#;^2+0.4357&#;-0.3571
b. &#;^&#;\&#;&#;&#;&#;&#;&#;(&#;)=-1.214&#;+0.4357. This is the velocity of the sensor.
c. &#;''(&#;)=-1.214. This is the acceleration of the sensor; it is a constant acceleration downward.

[T]

  1. Using a calculator or computer program, find the best-fit cubic curve to the data.
  2. Find the derivative of the position function and explain its physical meaning.
  3. Find the second derivative of the position function and explain its physical meaning.
  4. Using the result from (c), explain why a cubic function is not a good choice for this problem.

The following problems deal with the Holling type I, II, and III equations. These equations describe the ecological event of growth of a predator population given the amount of prey available for consumption.

[T] The populations of the snowshoe hare (in thousands) and the lynx (in hundreds) collected over 7 years from to are shown in the following table. The snowshoe hare is the primary prey of the lynx.

Population of snowshoe hare (thousands)Population of lynx (hundreds)
2010
5515
6555
9560
  1. Graph the data points and determine which Holling-type function fits the data best.
  2. Using the meanings of the parameters a and &#;, determine values for those parameters by examining a graph of the data. Recall that &#; measures what prey value results in the half-maximum of the predator value.
  3. Plot the resulting Holling-type I, II, and III functions on top of the data. Was the result from part a. correct?
Sours: https://opentextbc.ca/calculusv1openstax/chapter/derivatives-as-rates-of-change/
Introduction to Related Rates

Average Rate Of Change In Calculus
w/ Step-by-Step Examples!

How do you find the average rate of change in calculus?

Jenn (B.S., M.Ed.) of Calcworkshop® teaching average rate of change for calculus

Jenn, Founder Calcworkshop®, 15+ Years Experience (Licensed &#; Certified Teacher)

Great question!

And that&#;s exactly what you&#;ll going to learn in today&#;s lesson.

Let&#;s go!

I&#;m sure you&#;re familiar with some of the following phrases:

  • Miles Per Hour
  • Cost Per Minute
  • Plants Per Acre
  • Kilometers Per Gallon
  • Tuition Fees Per Semester
  • Meters Per Second

How To Find Average Rate Of Change

Whenever we wish to describe how quantities change over time is the basic idea for finding the average rate of change and is one of the cornerstone concepts in calculus.

So, what does it mean to find the average rate of change?

The average rate of change finds how fast a function is changing with respect to something else changing.

It is simply the process of calculating the rate at which the output(y-values) changes compared to its input(x-values).

How do you find the average rate of change?

We use the slope formula!

average rate of change formula

Average Rate Of Change Formula

To find the average rate of change, we divide the change in y(output) by the change in x(input). And visually, all we are doing is calculating the slope of the secant line passing between two points.

how to find the slope of a secant line passing through two points

How To Find The Slope Of A Secant Line Passing Through Two Points

Now for a linear function, the average rate of change (slope) is constant, but for a non-linear function, the average rate of change is not constant (i.e., changing).

Let&#;s practice finding the average rate of a function, f(x), over the specified interval given the table of values as seen below.

Practice Problem #1

find the average rate of change of the function over the given interval example

Find The Average Rate Of Change Of The Function Over The Given Interval

Practice Problem #2

how to find average rate of change over an interval example

How To Find Average Rate Of Change Over An Interval

See how easy it is?

All you have to do is calculate the slope to find the average rate of change!

Average Vs Instantaneous Rate Of Change

But now this leads us to a very important question.

What is the difference is between Instantaneous Rate of Change and Average Rate of Change?

While both are used to find the slope, the average rate of change calculates the slope of the secant line using the slope formula from algebra. The instantaneous rate of change calculates the slope of the tangent line using derivatives.

secant line vs tangent line

Secant Line Vs Tangent Line

Using the graph above, we can see that the green secant line represents the average rate of change between points P and Q, and the orange tangent line designates the instantaneous rate of change at point P.

So, the other key difference is that the average rate of change finds the slope over an interval, whereas the instantaneous rate of change finds the slope at a particular point.

How To Find Instantaneous Rate Of Change

All we have to do is take the derivative of our function using our derivative rules and then plug in the given x-value into our derivative to calculate the slope at that exact point.

For example, let&#;s find the instantaneous rate of change for the following functions at the given point.

instantaneous rate of change calculus example

Instantaneous Rate Of Change Calculus &#; Example

Tips For Word Problems

But how do we know when to find the average rate of change or the instantaneous rate of change?

We will always use the slope formula when we see the word &#;average&#; or &#;mean&#; or &#;slope of the secant line.&#;

Otherwise, we will find the derivative or the instantaneous rate of change. For example, if you see any of the following statements, we will use derivatives:

  • Find the velocity of an object at a point.
  • Determine the instantaneous rate of change of a function.
  • Find the slope of the tangent to the graph of a function.
  • Calculate the marginal revenue for a given revenue function.

Harder Example

Alright, so now it&#;s time to look at an example where we are asked to find both the average rate of change and the instantaneous rate of change.

average and instantaneous rate of change of a function example

Average And Instantaneous Rate Of Change Of A Function &#; Example

Notice that for part (a), we used the slope formula to find the average rate of change over the interval. In contrast, for part (b), we used the power rule to find the derivative and substituted the desired x-value into the derivative to find the instantaneous rate of change.

Nothing to it!

Particle Motion

But why is any of this important?

Here&#;s why.

Because &#;slope&#; helps us to understand real-life situations like linear motion and physics.

The concept of Particle Motion, which is the expression of a function where its independent variable is time, t, enables us to make a powerful connection to the first derivative(velocity), second derivative(acceleration), and the position function(displacement).

The following notation is commonly used with particle motion.

displacement velocity acceleration notation calculus

Displacement Velocity Acceleration Notation Calculus

Ex) Position &#; Velocity &#; Acceleration

Let&#;s look at a question where we will use this notation to find either the average or instantaneous rate of change.

Suppose the position of a particle is given by \(x(t)=3 t^{3}+7 t\), and we are asked to find the instantaneous velocity, average velocity, instantaneous acceleration, and average acceleration, as indicated below.

a. Determine the instantaneous velocity at \(t=2\) seconds
\begin{equation}
\begin{array}{l}
x^{\prime}(t)=v(t)=9 t^{2}+7 \\
v(2)=9(2)^{2}+7=43
\end{array}
\end{equation}
Instantaneous Velocity: \(v(2)=43\)

b. Determine the average velocity between 1 and 3 seconds
\begin{equation}
A v g=\frac{x(4)-x(1)}{}=\frac{\left[3(4)^{3}+7(4)\right]-\left[3(1)^{3}+7(1)\right]}{}=\frac{}{3}=70
\end{equation}
Avgerage Velocity: \(\overline{v(t)}=70\)

c. Determine the instantaneous acceleration at \(t=2\) seconds
\begin{equation}
\begin{array}{l}
x^{\prime \prime}(t)=a(t)=18 t \\
a(2)=18(2)=36
\end{array}
\end{equation}
Instantaneous Acceleration: \(a(2)=36\)

d. Determine the average acceleration between 1 and 3 seconds
\begin{equation}
A v g=\frac{v(4)-v(1)}{}=\frac{x^{\prime}(4)-x^{\prime}(1)}{}=\frac{\left[9(4)^{2}+7\right]-\left[9(1)^{2}+7\right]}{}=\frac{}{3}=45
\end{equation}
Average Acceleration: \(\overline{a(t)}=45\)

Summary

Together we will learn how to calculate the average rate of change and instantaneous rate of change for a function, as well as apply our knowledge from our previous lesson on higher order derivatives to find the average velocity and acceleration and compare it with the instantaneous velocity and acceleration.

Let&#;s jump right in.

Video Tutorial w/ Full Lesson &#; Detailed Examples (Video)

calcworkshop jenn explaining average rate of change

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Sours: https://calcworkshop.com/derivatives/average-rate-of-change-calculus/

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