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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.
Start by analyzing the units of the given values. Recall that liters are equivalent to cube decimeters. FlowQ:& 200 .dm^3 /min. Speedv:& 6403π .dm /s. Let's rewrite the flow in cube decimeters per seconds so that the parameters of water are measured in the same time interval. Use the conversion factor to rewrite minutes as seconds. 200 dm^3/min = 200 dm^3/60s = 10/3 dm^3/s Next, we will use the given formula to find the diameter of the hose. The hose allows a maximum flow at a speed of 6403π decimeters per second. Let's substitute the given values into the formula and solve for d.
Now that we have d^2 isolated, we need to take the square root of both sides of the equation in order to have d isolated on one side of the equation. Also, we will apply the n^\text{th} Roots of n^\text{th} Powers Property and Quotient Property of Radicals. Keep in mind that the diameter is always positive.
Therefore, the diameter d of the fire hose is 0.25 decimeters.
A mathematical pendulum is an idealized model of a simple pendulum. It consists of an object suspended from a fixed point with a string. The string cannot be stretched.
We want to find the time of a complete swing — one period — of a pendulum with the given length. Since we are given that g is equal to 32 feet per second square, we have everything we need to find the time of a complete swing of this pendulum. Let's substitute L= 89 and g= 32 into the given formula and evaluate.
It takes about 10.5 seconds for this pendulum to make a complete swing.
This time we are given the period of a pendulum. Based on that time, we need to find what the length the pendulum should be. To find its length, let's start by solving the given formula for L.
Now, we will substitute the given values of T and g to find the length of the pendulum.
The length of this pendulum would be about 729.5 feet, which is the height of, for example, Times Square Tower located in New York City.
We are given the speed of a car and the coefficient of friction between the tires and two types of roads. Coefficient of Friction rc Wet Roads: & f_1 = 0.4 Dry Roads: & f_2 = 0.7 We are asked to find the braking distance for both surfaces. To do so, we will first solve the given formula for d.
Next, consider both coefficients of friction one at a time. We will substitute the given values of v and f_1 to find the length of the skid marks d_1 on the wet surface.
Therefore, the braking distance on a wet road is 300 feet when the car is going 60 miles per hour. Now, calculate the braking distance d_2 on the dry surface by substituting f_2=0.4.
Finally, we can calculate the difference between the braking distances. d_1 - d_2 = 300 - 171 = 129 The difference is 129 feet. Notice that the braking distance nearly doubles on a wet road compared with a dry road.
Let's recall the formula for the braking distance that we found in Part A. d=v^2/30f Let's substitute the doubled speed 2v for the v into the formula and determine how it influences the braking distance.
Notice that now the braking distance is 4 times greater than the initial braking distance. In general, the length of the skid marks depends on the speed squared. d=v ^2/30f This means that if we increase the speed of the car by a factor of a, the length of the skid marks would increase a^2 times.
After riding one of the fastest roller coasters in the United States, Ignacio got interested in how the speed of a roller coaster like that can be determined.
We are asked to write a function that models the situation. To do so, we can use the formula for the speed of this coaster. Since we know that the initial speed is 10 feet per second and the gravity acceleration is about 9.8 meters per square second, we can substitute these values for v_0 and g into the formula, respectively.
Therefore, the speed v now depends on only d and be written as a function of d. v( d) = sqrt(100 + 19.6 d)
We are told that the roller coaster should have a speed of 35 meters per second at the bottom of the hill. Also, it is given that the height of the hill is 50 meters. This means that the vertical drop is also 50 meters. To find the initial speed of the coaster that will reach the desired velocity, start by solving the given formula for v_0.
Because the speed is always non-negative, we can simplify the absolute value. | v_0 | = sqrt(v^2-2gd) [0.6em] ⇓ [0.6em] v_0 = sqrt(v^2-2gd) Now, we can substitute v= 35, g= 9.8, and d= 50 into the obtained formula to calculate the initial speed.
Therefore, the initial speed of the roller coaster should be about 15.65 meters per second.