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| 8 Theory slides |
| 8 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Here are a few recommended readings before getting started with this lesson.
After becoming a subscriber of a streaming platform, Magdalena became interested in a particular crime series. While watching an episode, she found out that after an accident, police can determine the speed of a car before the driver started braking.
The speed of a car can be estimated by using the following formula.Magdalena shared her streaming account with Kriz so that they can watch a new nature documentary. The documentary is about the work of a Swiss agricultural biologist, Max Kleiber. He researched how the mass of an animal influences the animal's metabolic rate.
Metabolic Rate |
The rate at which a person or an animal burns calories to maintain its body weight. |
Mammal | Mass |
---|---|
Mouse | 20 g |
Kriz | 81 kg |
Horse | 625 kg |
Mammal | Mass (kg) | Metabolic Rate (kcal/day) |
---|---|---|
Mouse | 0.02 | ≈3.72 |
Kriz | 81 | 1890 |
Horse | 625 | 8750 |
Multiply
LHS/70=RHS/70
LHS4=RHS4
(ab)m=ambm
(na)n=a
3LHS=3RHS
3a⋅b=3a⋅3b
3a3=a
na=an1
(am)n=(an)m
an1=na
Calculate root
Calculate power and product
LHS⋅32=RHS⋅32
32⋅32a=a
Identity Property of Multiplication
LHS−v=RHS−v
Rearrange equation
LHS2=RHS2
(a)2=a
Distribute 64
(a−b)2=a2−2ab+b2
LHS−v2=RHS−v2
(ab)m=ambm
Calculate power and product
LHS−64s=RHS−64s
LHS/(-64)=RHS/(-64)
Distribute -641
t=1
1a=1
Identity Property of Multiplication
Add terms
Next, the second equation y=2t−16 will be graphed in the same coordinate plane by using a table of values. Remember to substitute t≥8 so that the square root can be calculated right away.
t | 2t−16 | y=2t−16 |
---|---|---|
8 | 2(8)−16 | 0=0 |
10 | 2(10)−16 | 4=2 |
16 | 2(16)−16 | 16=4 |
26 | 2(26)−16 | 36=6 |
40 | 2(40)−16 | 64=8 |
58 | 2(58)−16 | 100=10 |
All these points will be plotted and connected with a smooth curve.
x | 1.113+x | T=1.113+x |
---|---|---|
-3 | 1.113+(-3) | 0 |
-2 | 1.113+(-2) | 1.11 |
1 | 1.113+1 | 2.22 |
6 | 1.113+6 | 3.33 |
13 | 1.113+13 | 4.44 |
Next, the points from the table will be plotted on a coordinate plane and connected with a smooth curve.
Since the square root of a greater number is also greater, the function will increase to infinity. Additionally, the square root of a non-negative number is also non-negative. Consequently, the range of the function consists of numbers greater than or equal to 0. The obtained domain and range correspond to option C.
up.
In addition to watching TV, Magdalena also loves programming and developing simple applications for personal use in her spare time.
Option | Statement |
---|---|
A | The average increase in profit of the first company will be greater over the next 10 years. |
B | The average increase in profit of the second company will be greater over the next 10 years. |
C | On average, the profit of the companies increases at the same rate. |
D | On average, the annual profit of the second company is about 0.175 millions of dollars. |
Keep in mind that the order of the transformations may differ. For example, the horizontal translation could be the last transformation.
t | 213t−2+2 | P1(t) |
---|---|---|
2 | 2132−2+2 | 2 |
10 | 21310−2+2 | 3 |
18 | 21318−2+2 | ≈3.26 |
Now, substitute values of t and P1(t) into the formula for the average rate of change over both intervals.
Interval [t1,t2] | t2−t1P1(t2)−P1(t1) | Substitute and Evaluate |
---|---|---|
[2,10] | 10−2P1(10)−P1(2) | 10−23−2=0.125 |
[10,18] | 18−10P1(18)−P1(10) | 18−103.26−3≈0.033 |
The average rate of change R1 is greater than R2. Also, both values are positive. This means that after 10 years, the profit of the first company will increase but more slowly.
t | 213t−2+2 | P1(t) |
---|---|---|
0 | 2130−2+2 | ≈1.37 |
10 | 21310−2+2 | 3 |
Next, approximate the values of P2(0) and P2(10) from the graph of the profit P2.
It was found that P2(0) is about 1.25 and P2(10) is equal to 3. Now, both average rates of change over the interval [0,10] can be determined. Use a calculator if needed.
Profit Function | x2−x1f(x2)−f(x1) | Substitute and Evaluate |
---|---|---|
P1 | 10−0P1(10)−P1(0) | 10−03−1.37=0.163 |
P2 | 10−0P2(10)−P2(0) | 10−03−1.25=0.175 |
The average rate of change of the annual profit of the second company is greater over the next 10 years. This means that the average increase in profit of the second company is greater. Keep in mind that the average rate of change measures the average change in profit, not the actual profit.
Input | 18d | v1(d) | 163p | v2(p) |
---|---|---|---|---|
0 | 18(0) | 0 | 1630 | 0 |
216 | 18(216) | ≈62.35 | 163216 | 96 |
Now, both average rates of change over the interval [0,216] can be calculated by substituting the obtained values.
Function | x2−x1f(x2)−f(x1) | Substitute and Evaluate |
---|---|---|
v1 | 216−0v1(216)−v1(0) | 216−062.35−0≈0.29 |
v2 | 216−0v2(216)−v2(0) | 216−096−0≈0.44 |
It can be concluded that, on average, the increase in speed of the second model is greater than the increase in speed of the first model. Also, recall that the average rate of change can be thought of as the slope of the corresponding linear function.
On average, if the input of a function increases by one unit, the value of the function increases by the average rate of change, depending on the interval considered. |
This means that the speed of the car increases by about 0.29 for each meter of braking distance, or skid marks left on the ground by braking, and by about 0.44 miles for each unit of horsepower. Therefore, options A and D are correct.
d | 18d | v1(d) |
---|---|---|
0 | 18(0) | 0 |
2 | 18(2) | 6 |
4 | 18(4) | ≈8.5 |
6 | 18(6) | ≈10.4 |
8 | 18(8) | 12 |
10 | 18(10) | ≈13.4 |
Now, plot the points on a coordinate plane and connect them with a smooth curve.
up.Also, the function is not defined for negative values of d.
p | 163p | v2(p) |
---|---|---|
-8 | 163-8 | -32 |
-6 | 163-6 | ≈-29.1 |
-4 | 163-4 | ≈-25.4 |
-2 | 163-2 | ≈-20.2 |
0 | 1630 | 0 |
2 | 1632 | ≈20.2 |
4 | 1634 | ≈25.4 |
6 | 1636 | ≈29.1 |
8 | 1638 | 32 |
Similarly, plot the obtained points and connect them with a smooth curve.
Magdalena noticed that when a stone is thrown into a pond, it generates circular water waves. She then used her smartphone to record a thrown stone and calculated the speed of a wave by analyzing the footage frame by frame. One of the waves appeared to travel at a rate of about 25 centimeters per second.
For every second, let's visualize the circular waves generated by the thrown stone.
As time t increases, the radius of a circle increases at a rate of 25 centimeters per second. To find the radius r at any second, multiply the rate by that time. r=25t We can also think of this as calculating the distance in every direction given the speed of the wave. The area A of any circle is given by π times the square of its radius. A=π r^2 This equation represents how the area of the circle depends on the radius. To represent the relationship between area and time, we will bring the obtained equations together to perform a composition of functions. Both equations have the radius as a variable. We can substitute the first equation into the second one.
Start by noticing that the area is given in square feet while the formula obtained in Part A was calculated for centimeters. We can use a conversion factor to rewrite the given area in square centimeters. Recall that 1 foot is equal to 30.48 centimeters.
200 ft^2 = 200( 1ft*30.48cm/1ft )^2
⇕
200 ft^2 = 185 806.08 cm^2
Finally, let's substitute A=185 806.08 into the equation obtained in Part A and solve for t.
This means that it will take the wave about 9.73 seconds to hit the bounding stones of the pond.
To find the number of bacteria after 3 12 hours, we will substitute 3 12 for t into the given formula. Our next step will be to rewrite this mixed number as a fraction. Let's do it!
Next, we will rewrite the fraction in the exponent using a radical.
Therefore, after three and a half hours, the population of bacteria will be about 336 359.
To find the time, we will start by rewriting the given formula for a more general case with a population size of P_0. The exponential growth remains the same.
B(t) = P_0 * 2^(t2)
We want to find the time it takes for the initial population to increase tenfold. This means that we need to find t at which B(t) is equal to 10P_0.
Therefore, it will take about 6.64 hours for the population to increase tenfold.
Start by determining the domain of the square root in the given equation. Remember that for a square root to be a real number, its radicand must be nonnegative. x^2 + 4 ≥ 0 Because x^2 is always nonnegative and the sum of a nonnegative and a positive number is also nonnegative, the domain of the square root is all real numbers. To solve the equation, we can isolate the radical and square both sides of the equation.
Notice that we obtained a linear equation. Let's solve it for x.
We obtained that the solution to the given equation is x= 32. Let's now check whether this is an extraneous solution by substituting it into the original equation.
We confirmed that x= 32 is indeed a solution to the equation.
Let's first rewrite the equation as a function by collecting all terms on one side of the equation.
Next, set the right-hand side equal to f(x). Then, the zeros of the function will correspond to the solutions to the given equation. x-4 + sqrt(x^2+4) = 0 ⇓ f(x) = x-4 + sqrt(x^2+4) Now, we can graph the obtained function by using a graphing calculator. To do so, we first press the Y= button and type the function in one of the rows. Having written the function, we can push GRAPH to draw it.
The x-coordinate of the point at which the function intersects the x-axis is a solution to the original equation.