## Engage NY Eureka Math Algebra 2 Module 3 Lesson 6 Answer Key

### Eureka Math Algebra 2 Module 3 Lesson 6 Exercise Answer Key

Exercises 1 – 3:

Exercise 1.

Assume that there is initially 1 cm of water in the tank, and the height of the water doubles every 10 seconds. Write an equation that could be used to calculate the height H(t) of the water in the tank at any time t.

Answer:

The height of the water at time t seconds con be modeled by H(t) = 2^{t/10}

Exercise 2.

How would the equation in Exercise 1 change if…

a. the initial depth of water in the tank was 2 cm?

Answer:

H(t) = 2 . 2^{t/10}

b. the initial depth of water in the tank was cm?

Answer:

H(t) = \(\frac{1}{2}\) . 2^{t/10}

c. the initial depth of water in the tank was 10 cm?

Answer:

H(t) = 10 . 2^{t/10}

d. the initial depth of water in the tank was A cm, for some positive real number A?

Answer:

H(t) = A . 2^{t/10}

Exercise 3.

How would the equation in Exercise 2, part (d), change if…

a. the height tripled every ten seconds?

Answer:

H(t) = A . 3^{t/10}

b. the height doubled every five seconds?

Answer:

H(t) = A . 2^{t/5}

c. the height quadrupled every second?

Answer:

H(t) = A . 4^{t}

d. the height halved every ten seconds?

Answer:

H(t) = A . (0.5^{t/10})

Example 1.

Consider two identical water tanks, each of which begins with a height of water 1 cm and fills with water at a different rate. Which equations can be used to calculate the height of water in each tank at time t? Use H_{1} for tank 1 and H_{2} for tank 2.

Answer:

a. If both tanks start filling at the same time, which one fills first?

Answer:

Tank 2 fills first because the level is rising more quickly.

b. We want to know the average rate of change of the height of the water in these tanks over an interval that starts at a fixed time T as they are filling up. What is the formula for the average rate of change of a function f on an interval [a, b]?

Answer:

\(\frac{f(b)-f(a)}{b-a}\)

c. What is the formula for the average rate of change of the function H_{1} on an interval [a, b]?

Answer:

\(\frac{\boldsymbol{H}_{1}(\boldsymbol{b})-\boldsymbol{H}_{1}(\boldsymbol{a})}{\boldsymbol{b}-\boldsymbol{a}}\)

d. Let’s calculate the average rate of change of the function H_{1} on the interval [T, T + 0. 1], which is an interval one-tenth of a second long starting at an unknown time T.

Answer:

Exercises 4 – 8:

Exercise 4.

For the second tank, calculate the average change in the height, H_{2}, from time T seconds to T + 0. 1 second. Express the answer as a number times the value of the original function at time T. Explain the meaning of these findings.

Answer:

On average, over the time interval [T, T + 0. 1], the water in tank 2 rises at a rate of approximately 1.16123H_{2}(T) centimeters per second.

Exercise 5.

For each tank, calculate the average change in height from time T seconds to T + 0.001 second. Express the answer as a number times the value of the original function at time T. Explain the meaning of these findings.

Answer:

Tank 1:

On overage, over the time interval [T, T + 0.01], the water in tank 1 rises at a rate of approximately 0.693387H_{1}(T) centimeter per second.

Tank 2:

Over the time interval [T, T + 0.001], the water in tank 2 rises at an average rate of approximately 1. 09922H_{2}(T) centimeters per second.

Exercise 6.

In Exercise 5, the average rate of change of the height of the water in tank 1 on the interval [T, T + 0.001] can be described by the expression c_{1} 2T, and the average rate of change of the height of the water in tank 2 on the interval [T, T + 0. 001] can be described by the expression c_{2} 3T• What are approximate values of c_{1} and c_{2}?

Answer:

c_{1} ≈ 0.69339 and c_{2} ≈ 1. 09922.

Exercise 7.

As an experiment, let’s look for a value of b so that if the height of the water can be described by H(t) = b^{t}, then the expression for the average rate of change on the interval [T, T + 0. 001] is 1 . H(T).

a. Write out the expression for the average rate of change of H(t) = b^{t} on the interval [T, T + 0.001].

Answer:

\(\frac{\boldsymbol{H}_{b}(\boldsymbol{T}+\mathbf{0 . 0 0 1})-\boldsymbol{H}_{b}(\boldsymbol{T})}{\mathbf{0 . 0 0 1}}\)

b. Set your expression in part (a) equal to 1 H(T), and reduce to an expression involving a single b.

Answer:

c. Now we want to find the value of b that satisfies the equation you found in part (b), but we do not have a way to explicitly solve this equation. Look back at Exercise 6; which two consecutive integers have b between them?

Answer:

We are looking for the base of the exponent that produces a rate of change on a small interval near t that is 1. H(t). When that base is 2, the value of the rate is roughly 0.69H (t). When the base is 3, the value of the rate is roughly 1.1H. Since 0.69 < 1 < 1.1, the base we are looking for is somewhere between 2 and 3.

d. Use your calculator and a guess-and-check method to find an approximate value of b to 2 decimal places.

Answer:

Students may choose to use a table such as a table shown below. Make sure that students are maintaining enough decimal places of b^{0.001} to determine which value is closest to 0.001.

Then b ≈ 2.72.

Exercise 8.

Verify that for the value of b found in Exercise 7, \(\frac{H_{b}(T+0.001)-H_{b}(T)}{0.001}\) ≈ H_{b}(T), where H_{b}(T) = b^{T}

Answer:

When the height of the water increases by a factor of 2.72 units per second, the height at any time is equal to the rate of change of height at that time.

### Eureka Math Algebra 2 Module 3 Lesson 6 Problem Set Answer Key

Question 1.

The product 4 . 3 . 2 . 1 is called 4 factorial and is denoted by 4!. Then 10! = 10 . 9 . 8 . 7 . 6 . 5 . 4 . 3 . 2 . 1, and for any positive integer n, n! = n(n – 1)(n – 2) ……… 3 . 2 . 1.

a. Complete the following table of factorial values:

Answer:

b. Evaluate the sum \(1+\frac{1}{1 !}\).

Answer:

2

c. Evaluate the sum \(1+\frac{1}{1 !}+\frac{1}{2 !}\)

Answer:

2.5

d. Use a calculator to approximate the sum \(1+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}\) to 7 decimal places. Do not round the fractions before evaluating the sum.

Answer:

\(\frac{8}{3}\) ≈ 2.6666667

e. Use a calculator to approximate the sum \(1+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}\) to 7 decimal places. Do not round the fractions before evaluating the sum.

Answer:

\(\frac{65}{24}\) ≈ 2.7083333

f. Use a calculator to approximate sums of the form \(1+\frac{1}{1 !}+\frac{1}{2 !}+\cdots+\frac{1}{k !}\) to 7 decimal places for k = 5, 6, 7,8,9, 10. Do not round the fractions before evaluating the sums with a calculator.

Answer:

If k = 5, the sum is \(\frac{163}{60}\) ≈ 2.1766667.

If k = 6, the sum is \(\frac{1957}{720}\) ≈ 2.7180556.

1f k = 7,the sum is \(\frac{685}{252}\) ≈ 2. 7182540.

If k = 8, the sum is \(\frac{109 601}{40320}\) ≈ 2.7182788.

If k = 9, the sum is \(\frac{98 461}{36 288}\) ≈ 2.7182815.

If k = 10, the sum is \(\frac{9 864 101}{3 628 800}\) ≈ 2.7182818.

g. Make a conjecture about the sums \(1+\frac{1}{1 !}+\frac{1}{2 !}+\cdots+\frac{1}{k !}\) for positive integers k as k increases in size.

Answer:

It seems that as k gets larger, the sums \(1+\frac{1}{1 !}+\frac{1}{2 !}+\cdots+\frac{1}{k !}\) get closer to e.

h. Would calculating terms of this sequence ever yield an exact value of e? Why or why not?

Answer:

No. The number e is irrational, so it cannot be written as a quotient of integers. Any finite sum \(1+\frac{1}{1 !}+\frac{1}{2 !}+\cdots+\frac{1}{k !}\) can be expressed as a single rational number with denominator k!, so the sums are all rational numbers. However, the more terms that are calculated, the closer to e the sum becomes, so these sums provide better and better rational number approximations of e.

Question 2.

Consider the sequence given by a_{n} = (1 + \(\frac{1}{n}\))^{n} where n ≥ 1 is an integer.

a. Use your calculator to approximate the first 5 terms of this sequence to 7 decimal places.

Answer:

a_{1} = (1 + \(\frac{1}{1}\))^{1} = 2

a_{2} = (1 + \(\frac{1}{2}\))^{2} = 2.25

a_{3} = (1 + \(\frac{1}{3}\))^{3} ≈ 2.3703704

a_{4} = (1 + \(\frac{1}{4}\))^{4} ≈ 2.4414063

a_{5} = (1 + \(\frac{1}{5}\))^{5} = 2.4883200

b. Does it appear that this sequence settles near a particular value?

Answer:

No, the numbers get bigger, but we cannot tell if it keeps getting bigger or settles on or near a particular value.

c. Use a calculator to approximate the following terms of this sequence to 7 decimal places.

Answer:

i. a_{100} = 2.7081383

ii. a_{1000} = 2.7169239

iii. a_{10,000} = 2.7181459

iv. a_{100,000} = 2.7182682

v. a_{1,000,000} = 2.7182805

vi. a_{100,000} = 2.7182816

vii. a_{100,000,000} = 2.7182818

d. Does it appear that this sequence settles near a particular value?

Answer:

Yes, it appears that as n gets really large (at least 100,000,000), the terms a_{n} of the sequence settle near the value of e.

e. Compare the results of this exercise with the results of Problem 1. What do you observe?

Answer:

It took about 10 terms of the sum in Problem 1 to see that the sum settled at the value e, but it takes 100,000,000 terms of the sequence in this problem to see that the sum settles at the value e.

Question 3.

If x = 5a^{4} and a = 2e^{3}, express x in terms of e, and approximate to the nearest whole number.

Answer:

If x = 5a^{4} and a = 2e^{3}, then x = 5(2e^{3})^{4}. Rewriting the right side in an equivalent form gives x = 80e^{12} ≈ 13020383.

Question 4.

If a = 2b^{3} and b = –\(\frac{1}{2}\)e^{-2}, express a in terms of e, and approximate to four decimal places.

Answer:

If a = 2b^{3} and b = –\(\frac{1}{2}\)e^{-2}, then a = 2 (-\(\frac{1}{2}\) e^{-2})^{3}. Rewriting the right side in an equivalent form gives a = –\(\frac{1}{4}\)e^{-6} ≈ -0.0006.

Question 5.

If x = 3e^{4} and e = \(\frac{s}{2 x^{3}}\) show that s = 54e^{13}, and approximate sto the nearest whole number.

Answer:

Rewrite the equation e = \(\frac{s}{2 x^{3}}\) to isolate the variable s.

e = \(\frac{s}{2 x^{3}}\)

2x^{3}e = s

By the substitution property, if s = 2x^{3}e and x = 3e^{4}, then s = 2(3e^{4})^{3} e. Rewriting the right side in an equivalent form gives s = 2 . 27e^{12} . e = 54e^{13} ≈ 23890323.

Question 6.

The following graph shows the number of barrels of oil produced by the Glenn Pool well in Oklahoma from 1910 to 1916.

Source: Cutler, Willard W., Jr. Estimation of Underground Oil Reserves by Oil-Well Production Curves, U.S. Department of the Interior, 1924.

a. Estimate the average rate of change of the amount of oil produced by the well on the interval [1910, 1916], and explain what that number represents.

Answer:

Student responses will vary based on how they read the points on the graph. Over the interval [1910, 1916], the average rate of change is roughly

\(\frac{300-3200}{1916-1910}=-\frac{2900}{6}\) ≈ -483.33.

This says that the production of the well decreased by an average of about 483 barrels of oil each year between 1910 and 1916.

b. Estimate the average rate of change of the amount of oil produced by the well on the interval [1910, 1913], and explain what that number represents.

Answer:

Student responses will vary based on how they read the points on the graph. Over the interval [1910, 19131], the average rate of change is roughly

\(\frac{800-3200}{1913-1910}=-\frac{2400}{3}\) = -800.

This says that the production of the well decreased by an average of about 800 barrels of oil per year between 1910 and 1913.

c. Estimate the average rate of change of the amount of oil produced by the well on the interval [1913, 1916], and explain what that number represents.

Answer:

Student responses will vary based on how they read the points on the graph. Over the interval [1913, 1916], the average rate of change is roughly

\(\frac{300-800}{1916-1913}=-\frac{500}{3}\) ≈ -166.67

This says that the production of the well decreased by on an average of about 166.67 barrels of oil per year between 1913 and 1916.

d. Compare your results for the rates of change in oil production in the first half and the second half of the time period in question in parts (b) and (c). What do those numbers say about the production of oil from the well?

Answer:

The production dropped much more rapidly in the first three years than it did in the second three years. Looking at the graph, it looks like the oil in the well might be running out, so less and less can be extracted each year.

e. Notice that the average rate of change of the amount of oil produced by the well on any interval starting and ending in two consecutive years is always negative. Explain what that means in the context of oil production.

Answer:

Because the average rate of change of oil production over a one-year period is always negative, the well is producing less oil each year than it did the year before.

Question 7.

The following table lists the number of hybrid electric vehicles (HEVs) sold in the United States between 1999 and 2013.

Source: U.S. Department of Energy, Alternative Fuels and Advanced Vehicle Data Center, 2013.

a. During which one-year interval is the average rate of change of the number of HEVs sold the largest? Explain how you know.

Answer:

The average rate of change of the number of HEVs sold is largest during [2011,2012] because of the number of HEVssold increases by the largest amount between those two years.

b. Calculate the average rate of change of the number of HEVs sold on the interval [2003,2004], and explain what that number represents.

Answer:

On the interval [2003,2004], the average rate of change in sales of HEVs is \(\frac{84199-47600}{2004-2003}\) which is 36,599. This means that during this one-year period, HEVs were selling at a rate of 36, 599 vehicles per year.

c. Calculate the average rate of change of the number of HEVs sold on the interval [2003, 2008], and explain what that number represents.

Answer:

On the interval [2003, 2008], the average rate of change in sales of HEVs is \(\frac{312,386-47,600}{2008-2003}\), which is 52,957.2. This means that during this five-year period, HEVs were selling at an average rate of 52,957 vehicles per year.

d. What does it mean if the average rate of change of the number of HEVs sold is negative?

Answer:

If the average rate of change of the vehicles sold is negative, then the sales are declining. This means that fewer cards were sold than in the previous year.

Extension:

Question 8.

The formula for the area of a circle of radius r can be expressed as a function A(r) = πr^{2}.

a. Find the average rate of change of the area of a circle on the interval [4, 5].

Answer:

\(\frac{A(5)-A(4)}{5-4}=\frac{25 \pi-16 \pi}{1}\) = 9π

b. Find the average rate of change of the area of a circle on the interval [4, 4.1].

Answer:

\(\frac{A(4.1)-A(4)}{4.1-4}=\frac{16.81 \pi-16 \pi}{0.1}\) = 8.1π

c. Find the average rate of change of the area of a circle on the interval [4, 4.01].

Answer:

\(\frac{A(4.01)-A(4)}{4.01-4}=\frac{16.0801 \pi-16 \pi}{0.01}\) = 8.01π

d. Find the average rate of change of the area of a circle on the interval [4, 4.001].

Answer:

\(\frac{A(4.001)-A(4)}{4.001-4}=\frac{16.008001 \pi-16 \pi}{0.001}\) = 8.001π

e. What is happening to the average rate of change of the area of the circle as the interval gets smaller and smaller?

Answer:

The average rate of change of the area of the circle appears to be getting close to 8π.

f. Find the average rate of change of the area of a circle on the interval [4, 4 + h] for some small positive number h.

Answer:

\(\frac{A(4+h)-A(4)}{(4+h)-4}\) = \(\frac{(4+h)^{2} \pi-16 \pi}{h}\)

= \(\frac{\left(16+8 h+h^{2}\right) \pi-16 \pi}{h}\)

= \(\frac{1}{h}\) (8h + h^{2})π

= \(\frac{1}{h}\) . h(8 + h)π

= (8 + h)π

g. What happens to the average rate of change of the area of the circle on the interval [4, 4 + h] as h → 0? Does this agree with your answer to part (d)? Should it agree with your answer to part (e)?

Answer:

As h → 0, 8 + h → 8, so as h gets smaller, the average rate of change approaches 8. This agrees with my response to part (e), and it should because as h → 0, the Interval [4, 4 + h] gets smaller.

h. Find the average rate of change of the area of a circle on the interval [r_{0}, r_{0} + h] for some positive number r_{0} and some small positive number h.

Answer:

i. What happens to the average rate of change of the area of the circle on the interval [r_{0}, r_{0} + h] as h → 0? Do you recognize the resulting formula?

Answer:

As h → 0, the expression for the average rate of change becomes 2πr_{0}, which is the circumference of the circle with radius r_{0}.

Question 9.

The formula for the volume of a sphere of radius r can be expressed as a function V(r) = \(\frac{4}{3}\)πr^{3}. As you work through these questions, you will see the pattern develop more clearly if you leave your answers in the form of a coefficient times π. Approximate the coefficient to five decimal places.

a. Find the average rate of change of the volume of a sphere on the interval [2, 3].

Answer:

\(\frac{V(3)-V(2)}{3-2}=\frac{\frac{4}{3} \cdot 27 \pi-\frac{4}{3} \cdot 8 \pi}{1}\) = \(\frac{4}{3}\) . 19π ≈ 25.33333π

b. Find the average rate of change of the volume of a sphere on the interval [2, 2. 1].

Answer:

\(\frac{V(2.1)-V(2)}{2.1-2}=\frac{\frac{4}{3} \pi\left(2.1^{3}-8\right)}{0.1}\) ≈ 16.81333π

c. Find the average rate of change of the volume of a sphere on the interval [2, 2. 01].

Answer:

\(\frac{V(2.01)-V(2)}{2.01-2}=\frac{\frac{4}{3} \pi\left(2.01^{3}-8\right)}{0.01}\) ≈ 16.08013π

d. Find the average rate of change of the volume of a sphere on the interval [2, 2.001].

Answer:

\(\frac{V(2.001)-V(2)}{2.001-2}=\frac{\frac{4}{3} \pi\left(2.001^{3}-8\right)}{0.001}\) ≈ 16.00800π

e. What is happening to the average rate of change of the volume of a sphere as the interval gets smaller and smaller?

Answer:

The average rate of change of the volume of the sphere appears to be getting close to 16π.

f. Find the average rate of change of the volume of a sphere on the interval [2, 2 + h] for some small positive number h.

Answer:

g. What happens to the average rate of change of the volume of a sphere on the interval [2, 2 + h] as h → 0? Does this agree with your answer to part (e)? Should it agree with your answer to part (e)?

Answer:

As h → 0, the value of the polynomial 12 + 6h + h^{2} approaches 12. Then the average rate of change approaches \(\frac{4 \pi}{3}\) . 12 = 16. This agrees with my response to part (e), and it should because as h → 0, the interval [2, 2 + h] gets smaller.

h. Find the average rate of change of the volume of a sphere on the interval [r_{0}, r_{0} + h] for some positive number r and some small positive number h.

Answer:

i. What happens to the average rate of change of the volume of a sphere on the interval [r_{0}, r_{0} + h] as h → 0? Do you recognize the resulting formula?

Answer:

As h → 0, the expression for the average rate of change becomes 4πr_{0}^{2}, which is the surface area of the sphere with radius r_{0}.

### Eureka Math Algebra 2 Module 3 Lesson 6 Exit Ticket Answer Key

Question 1.

Suppose that water is entering a cylindrical water tank so that the initial height of the water is 3 cm and the height of the water doubles every 30 seconds. Write an equation of the height of the water at time t seconds.

Answer:

H(t) = 3(\(2^{\frac{t}{30}}\))

Question 2.

Explain how the number e arose in our exploration of the average rate of change of the height of the water in the water tank.

Answer:

We first noticed that if the water level in the tank was doubling every second, then the average rate of change of the height of the water was roughly 0.69 times the height of the water at that time. And if the water level in the tank was tripling every second, then the average rate of change of the height of the water was roughly 1.1 times the height of the water at that time.

When we went looking for a base b so that the average rate of change of the height of the water was 1.0 times the height of the water at that time, we found that the base was roughly e. Calculating the average rate of change over shorter intervals gave a better approximation of e.