## Engage NY Eureka Math Precalculus Module 3 Lesson 5 Answer Key

### Eureka Math Precalculus Module 3 Lesson 5 Example Answer Key

Example 1.

Look at the alternating sums of the rows of Pascalâ€™s triangle. An alternating sum alternately subtracts and then adds values. For example, the alternating sum of Row 2 would be 1 – 2 + 1, and the alternating sum of Row 3 would be

1 – 3 + 3 – 1.

a. Compute the alternating sum for each row of the triangle shown.

Answer:

The sums are all zero.

b. Use the binomial theorem to explain why each alternating sum of a row in Pascalâ€™s triangle is 0.

Answer:

The binomial theorem states that (u + v)^{n} = u^{n} + A_{1} u^{(n – 1)} v + A_{2} u^{(n – 2)} v^{2} + A_{(n – 1)} uv^{(n – 1)} + n.

So, 0 = (1 + ( – 1))^{n} = 1^{n} + A_{1}(1^{(n – 1)})( – 1) + A_{2}(1^{(n – 2)})( – 1)^{2} + â‹¯ + A_{(n – 1)} (1)( – 1)^{(n – 1)} + ( – 1)^{n} = 1 – A_{1} + A_{2} + â‹¯ – A_{(n – 1)} + 1 for all even values of n or 1 – A_{1} + A_{2} + â‹¯ + A_{(n – 1)} – 1 for all odd values of n.

Example 2.

We know that the volume V(r) and surface area S(r) of a sphere of radius r are given by these formulas:

V(r) = \(\frac{4}{3}\) Ï€r^{3}

S(r) = 4Ï€r^{2}

Suppose we increase the radius of a sphere by 0.01 units from r to r + 0.01.

a. Use the binomial theorem to write an expression for the increase in volume.

Answer:

V(r + 0.01) – V(r) = \(\frac{4}{3}\) Ï€(r + 0.01)^{3} – \(\frac{4}{3}\) Ï€r^{3}

= \(\frac{4}{3}\) Ï€((r)^{3} + 3(r)^{2} (0.01) + 3r(0.01)^{2} + (0.01)^{3}) – \(\frac{4}{3}\) Ï€r^{3}

= \(\frac{4}{3}\) Ï€r^{3} + 0.04Ï€(r)^{2} + 0.0004Ï€r + 0.000â€†001Ï€ – \(\frac{4}{3}\) Ï€r^{3}

= 0.04Ï€(r)^{2} + 0.0004Ï€r + 0.000â€†001Ï€

b. Write an expression for the average rate of change of the volume as the radius increases from r to r + 0.01.

Answer:

Average rate of change = \(\frac{V(r + 0.01) – V(r)}{(r + 0.01) – r}\)

c. Simplify the expression in part (b) to compute the average rate of change of the volume of a sphere as the radius increases from r to r + 0.01.

Answer:

Average rate of change = \(\frac{0.04 \pi(r)^{2} + 0.0004 \pi r + 0.000001 \pi}{0.01}\) = 4Ï€r^{2} + 0.04Ï€r + 0.0001Ï€

d. What does the expression from part (c) resemble?

Answer:

Surface area of the sphere with radius r

e. Why does it make sense that the average rate of change should approximate the surface area? Think about the geometric figure formed by V(r + 0.01) – V(r). What does this represent?

Answer:

It is a shell of volume, a layer 0.01 units thick, covering the surface area of the inner sphere of radius r.

f. How could we approximate the volume of the shell using surface area? And the average rate of change for the volume?

Answer:

The volume is approximately S(r)â‹…(0.01); rate of change of volume is approximately \(\frac{S(r) \times 0.01}{0.01}\) = S(r), which is the surface area of the sphere, S(r).

### Eureka Math Precalculus Module 3 Lesson 5 Exercise Answer Key

Opening Exercise

Write the first six rows of Pascalâ€™s triangle. Then, use the triangle to find the coefficients of the terms with the powers of u and v shown, assuming that all expansions are in the form (u + v)^{n}. Explain how Pascalâ€™s triangle allows you to determine the coefficient.

Answer:

a. u^{2} v^{4}

Answer:

15; this is in Row 6 (sum of exponents is 6) and the fifth term since the power of u started at 6 and has decreased to 2, which takes 5 steps.

b. u^{3} v^{2}

Answer:

10; this is the third term in Row 5.

c. u^{2} v^{2}

Answer:

6; this is the third term in Row 4.

d. v^{10}

Answer:

1; this is the last term in Row 10 since there is no u.

Exercises 1â€“2

Exercise 1.

Consider Rows 0â€“6 of Pascalâ€™s triangle.

a. Find the sum of each row.

Answer:

SUM

1

2

4

8

16

32

64

b. What pattern do you notice in the sums computed?

Answer:

The sum of the coefficients in the nth row appears to be 2^{n}.

c. Use the binomial theorem to explain this pattern.

Answer:

2^{n} = (1 + 1)^{n}

The binomial theorem states that (u + v)^{n} = u^{n} + A_{1} u^{n – 1} v + A_{2} u^{n – 2} v^{2} + â‹¯ + A_{n – 1} uv^{n – 1} + v^{n}.

So,

(1 + 1)^{n} = 1^{n} + A_{1} (1^{n – 1})(1) + A_{2} (1^{n – 2})(1)^{2} + â‹¯ + A_{n – 1} (1)(1)^{n – 1} + 1^{n} = 1 + A_{1} + A_{2} + â‹¯ + A_{n – 1} + 1, which is the sum of the coefficients of the nth row of Pascalâ€™s triangle.

Exercise 2.

Consider the expression 11^{n}.

a. Calculate 11^{n}, where n = 0, 1, 2, 3, 4.

Answer:

11^{0} = 1

11^{1} = 11

11^{2} = 121

11^{3} = 1331

11^{4} = 14641

b. What pattern do you notice in the successive powers?

Answer:

The digits of 11^{n} correspond to the coefficients in the n^{th} row of Pascalâ€™s triangle.

c. Use the binomial theorem to demonstrate why this pattern arises.

Answer:

11^{n} = (10 + 1)^{n} = 10^{n} + A_{1} (10)^{n – 1} (1) + A_{2} (10)^{n – 2} (1)^{2} + â‹¯ + A_{n – 1} (10)(1)^{n – 1} + 1^{n}

= 10^{n} + A_{1} (10)^{n – 1} + A_{2} (10)^{n – 2} + â‹¯ + A_{n – 1} (10) + 1

d. Use a calculator to find the value of 11^{5}. Explain whether this value represents what would be expected based on the pattern seen in lower powers of 11.

Answer:

If we continued the pattern seen in 11^{n}, where n = 0, 1, 2, 3, 4, we would expect 11^{5} to comprise the digits of the fifth row in Pascalâ€™s triangle. In other words, we could conjecture that 11^{5} = 1|5|10|10|5|1. Because we cannot represent a 10 as a single digit, the number on the calculator would be 161,051.

### Eureka Math Precalculus Module 3 Lesson 5 Problem Set Answer Key

Question 1.

Consider the binomial (2u – 3v)^{6}.

a. Find the term that contains v^{4}.

Answer:

15(2u)^{2} (3v)^{4} = 4860u^{2} v^{4}

b. Find the term that contains u^{3}.

Answer:

20(2u)^{3} (3v)^{3} = 4320u^{3} v^{3}

c. Find the third term.

Answer:

15(2u)^{4} (3v)^{2} = 2160u^{4} v^{2}

Question 2.

Consider the binomial (u^{2} – v^{3})^{6}.

a. Find the term that contains v^{6}.

Answer:

v^{6} = (v^{3})^{2}, 15(u^{2})^{4} (v^{3})^{2} = 15u^8 v^{6}

b. Find the term that contains u^{6}.

Answer:

u^{6} = (u^{2})^{3}, 20(u^{2})^{3}(v^{3})^{3} = 20u^{6} v^{9}

c. Find the fifth term.

Answer:

15(u^{2})^{2} (v^{3})^{4} = 15u^{4} v^{12}

Question 3.

Find the sum of all coefficients in the following binomial expansion.

a. (2u + v)^{10}

Answer:

3^{10} = 59049

b. (2u – v)^{10}

Answer:

1

c. (2u – 3v)^{11}

Answer:

– 1

d. (u – 3v)^{11}

Answer:

– 2^{11}

e. (1 + i)^{10}

Answer:

((1 + i)^{2})^{5} = (1 + 2i – 1)^{5} = (2i)^{5} = 32i

f. (1 – i)^{10}

Answer:

((1 – i)^{2})^{5} = (1 – 2i – 1)^{5} = ( – 2i)^{5} = – 32i

g. (1 + i)^{200}

Answer:

((1 + i)^{2})^{100} = (1 + 2i – 1)^{100} = (2i)^{100} = 2^{100}

h. (1 + i)^{201}

Answer:

(1 + i) (1 + i)^{200} = (1 + i)((1 + i)^{2})^{100} = (1 + i)2^{100}

Question 4.

Expand the binomial (1 + \(\sqrt{2}\)i)^{6}.

Answer:

1 + 6(1)^{5} (\(\sqrt{2}\)i) + 15(1)^{4} (\(\sqrt{2}\)i)^{2} + 20(1)^{3}(\(\sqrt{2}\)i)^{3} + 15(1)^{2} (\(\sqrt{2}\)i)^{4} + 6(1) (\(\sqrt{2}\)i)^{5} + (\(\sqrt{2}\)i)^{6}

1 + 6\(\sqrt{2}\)i – 30 – 40\(\sqrt{2}\)i + 60 + 24\(\sqrt{2}\)i – 8 = 23 – 10\(\sqrt{2}\)i

Question 5.

Show that (2 + \(\sqrt{2}\)i)^{20} + (2 – \(\sqrt{2}\)i)^{20} is an integer.

Answer:

Letâ€™s get the first few terms for both to see whether it has any patterns that we can simplify.

(2 + \(\sqrt{2}\)i)^{2} = 2^{20} + A_{1} (2)^{19} (\(\sqrt{2}\)i)^{1} + A_{2} (2)^{18}(\(\sqrt{2}\)i)^{2} + A_{3} (2)^{17} (\(\sqrt{2}\)i)^{3} + A_{4} (2)^{18}(\(\sqrt{2}\)i)^{4} + A_{5} (2)^{18}(\(\sqrt{2}\)i)^{5} + â‹¯ – A_{17} (2)^{3} (\(\sqrt{2}\)i)^{17} + A_{18} (2)^{2} (\(\sqrt{2}\)i)^{18}Â + A_{19} (2)^{1} (\(\sqrt{2}\)i)^{19} + (\(\sqrt{2}\)i)^{20}

(2 – \(\sqrt{2}\)i)^{2} = 2^{20} – A_{1} (2)^{19} (\(\sqrt{2}\)i)^{1} + A_{2} (2)^{18}(\(\sqrt{2}\)i)^{2} – A_{3} (2)^{17} (\(\sqrt{2}\)i)^{3} + A_{4} (2)^{18}(\(\sqrt{2}\)i)^{4} + A_{5} (2)^{18}(\(\sqrt{2}\)i)^{5} + â‹¯ – A_{17} (2)^{3} (\(\sqrt{2}\)i)^{17} + A_{18} (2)^{2} (\(\sqrt{2}\)i)^{18}Â – A_{19} (2)^{1} (\(\sqrt{2}\)i)^{19} + (\(\sqrt{2}\)i)^{20}

All the even terms are canceled, and the remaining terms would involve multiplying powers of 2 with even powers of \(\sqrt{2}\). So, the result would be the sum of some positive and negative integers, which is 116,967,424.

Question 6.

We know (u + v)^{2} = u^{2} + 2uv + v^{2} = u^{2} + v^{2} + 2uv. Use this pattern to predict what the expanded form of each expression would be. Then, expand the expression, and compare your results.

a. (u + v + w)^{2}

Answer:

u^{2} + v^{2} + w^{2} + 2uv + 2uw + 2vw

b. (a + b + c + d)^{2}

Answer:

a^{2} + b^{2} + c^{2} + d^{2} + 2ab + 2ac + 2ad + 2bc + 2bd + 2cd

Question 7.

Look at the powers of 101 up to the fourth power on a calculator. Explain what you see. Predict the value of 101^{5}, and then find the answer on a calculator. Are they the same?

Answer:

It is similar to Pascalâ€™s triangle, but zeros are inserted between the numbers in each row. The answers are not the same; the expanded form of 1|0|5|0|10|0|10|0|5|0|1 is 10, 510, 100, 501; however, if we could represent a 10 as a single digit in our base – ten arithmetic, then both answers would be the same.

Question 8.

Can Pascalâ€™s triangle be applied to (\(\frac{1}{u}\) + \(\frac{1}{v}\))^{n} given u,vâ‰ 0?

Answer:

Yes. The answer is (\(\frac{1}{u}\) + \(\frac{1}{v}\))^{n} = \(\frac{1}{u^{n}}\) + \(\frac{A_{1}}{u^{(n – 1)} v}\)) + \(\frac{A_{2}}{u^{(n – 2)} v^{2}}\) + â‹¯ + \(\frac{A_{n – 2}}{u^{2} v^{(n – 2)}}\) + \(\frac{A_{n – 1}}{u v^{(n – 1)}}\) + \(\frac{1}{v^{n}}\) .

Question 9.

The volume and surface area of a sphere are given by V = \(\frac{4}{3}\) Ï€r^{3} and S = 4Ï€r^{2}. Suppose we increase the radius of a sphere by 0.001 units from r to r + 0.001.

a. Use the binomial theorem to write an expression for the increase in volume V(r + 0.001) – V(r) as the sum of three terms.

Answer:

V(r + 0.001) – V(r) = \(\frac{4}{3}\) Ï€(r + 0.001)^{3} – \(\frac{4}{3}\) Ï€r^{3}

= \(\frac{4}{3}\) Ï€[(r)^{3} + 3(r)^{2} (0.001) + 3r(0.001)^{2} + (0.001)^{3} ] – \(\frac{4}{3}\) Ï€r^{3}

= \(\frac{4}{3}\) Ï€r^{3} + 0.004Ï€(r)^{2} + 0.000â€†004Ï€r + \(\frac{0.000000004 \pi}{3}\) – \(\frac{4}{3}\) Ï€r^{3}

= 0.004Ï€(r)^{2} + 0.000â€†004Ï€r + \(\frac{0.000000004 \pi}{3}\)

b. Write an expression for the average rate of change of the volume as the radius increases from r to r + 0.001.

Answer:

Average rate of change: \(\frac{V(r + 0.001) – V(r)}{(r + 0.001) – r}\)

c. Simplify the expression in part (b) to compute the average rate of change of the volume of a sphere as the radius increases from r to r + 0.001.

Answer:

Average rate of change: \(\frac{0.004 \pi(r)^{2} + 0.000004 \pi r + \frac{0.000000004 \pi}{3}}{0.001}\) = 4Ï€r^{2} + 0.004Ï€r + 0.000â€†004Ï€

d. What does the expression from part (c) resemble?

Answer:

Surface area of the sphere with radius r

e. Why does it make sense that the average rate of change should approximate the surface area? Think about the geometric figure formed by V(r + 0.001) – V(r). What does this represent?

Answer:

It is a shell of the volume, a layer 0.001 units thick, covering the surface area of the inner sphere of radius r.

f. How could we approximate the volume of the shell using surface area? And the average rate of change for the volume?

Answer:

The volume is approximately S(r)â‹…0.001; the rate of change of volume is approximately \(\frac{s(r) \times 0.001}{0.001}\), which is the surface area of the sphere, S(r).

g. Find the difference between the average rate of change of the volume and S(r) when r = 1.

Answer:

Average rate of change = 4Ï€(1)^{2} + 0.004Ï€(1) + 0.000â€†004Ï€ = 12.579

S(1) = 4Ï€(1)^{2} = 12.566

12.579 – 12.566 = 0.013

Question 10.

The area and circumference of a circle of radius r are given by A(r) = Ï€r^{2} and C(r) = 2Ï€r. Suppose we increase the radius of a sphere by 0.001 units from r to r + 0.001.

a. Use the binomial theorem to write an expression for the increase in area volume A(r + 0.001) – A(r) as a sum of three terms.

Answer:

A(r + 0.001) – V(r) = Ï€(r + 0.001)^{2} – Ï€r^{2}

= Ï€r^{2} + 0.002Ï€r + 0.000â€†001Ï€ – Ï€r^{2}

= 0.002Ï€r + 0.000â€†001Ï€

b. Write an expression for the average rate of change of the area as the radius increases from r to r + 0.001.

Answer:

Average rate of change: \(\frac{A(r + 0.001) – A(r)}{(r + 0.001) – r}\)

c. Simplify the expression in part (b) to compute the average rate of change of the area of a circle as the radius increases from r to r + 0.001.

Answer:

Average rate of change: \(\frac{0.002 \pi r + 0.000001 \pi}{0.001}\) = 2Ï€r + 0.001Ï€

d. What does the expression from part (c) resemble?

Answer:

Surface area of the circle with radius r

e. Why does it make sense that the average rate of change should approximate the area of a circle? Think about the geometric figure formed by A(r + 0.001) – A(r). What does this represent?

Answer:

It is a shell of the volume, a layer 0.01 units thick, covering the surface area of the inner circle of radius r.

f. How could we approximate the area of the shell using circumference? And the average rate of change for the area?

Answer:

The volume is approximately A(s)â‹…0.001; rate of change of volume is approximately \(\frac{A(s) \times 0.001}{0.001}\), which is the surface area of the circle, A(s).

g. Find the difference between the average rate of change of the area and C(r) when r = 1.

Answer:

Average rate of change = 2Ï€(1) + 0.001(1) = 6.284

C(1) = 2Ï€(1) = 6.283

6.284 – 6.283 = 0.001

### Eureka Math Precalculus Module 3 Lesson 5 Exit Ticket Answer Key

Question 1.

The area and circumference of a circle of radius r are given by

A(r) = Ï€r^{2}

C(r) = 2Ï€r

a. Show mathematically that the average rate of change of the area of the circle as the radius increases from r to r + 0.01 units is very close to the circumference of the circle.

Answer:

Average rate of change: (A(r + 0.01) – A(r))/((r + 0.01) – r)

A(r + 0.01) – A(r) = Ï€(r + 0.01)^{2} – Ï€r^{2}

= Ï€r^{2} + 2Ï€(r)(0.01) + Ï€(0.01)^{2} – Ï€r^{2}

= 0.02Ï€r + 0.0001Ï€

Average rate of change: \(\frac{0.02 \pi r + 0.0001 \pi}{0.01}\) = 2Ï€r + 0.01Ï€, which is approximately equal to C(r).

b. Explain why this makes sense geometrically.

Answer:

The difference A(r + 0.01) – A(r) represents the area of a ring of width 0.01 units, where the inner circle forming the ring is a circle with radius r. The area of the ring could be approximated by the expression

0.01â‹…2Ï€r. The average rate of change of the area is \(\frac{0.01 \cdot 2 \pi r}{0.01}\) = 2Ï€r, which is the circumference of the circle with radius r.