## Engage NY Eureka Math Precalculus Module 1 Lesson 23 Answer Key

### Eureka Math Precalculus Module 1 Lesson 23 Exercise Answer Key

Opening Exercise

a. Describe the geometric effect of performing the transformation \(\left(\begin{array}{l}

x \\

y

\end{array}\right)\)→R\(\left(\begin{array}{l}

x \\

y

\end{array}\right)\).

Answer:

Applying R rotates each point in the plane about the origin through \(\frac{\pi}{3}\) radians in a counterclockwise direction.

b. Plot the point \(\left(\begin{array}{l}

1 \\

0

\end{array}\right)\), and then find R\(\left(\begin{array}{l}

1 \\

0

\end{array}\right)\) and plot it.

Answer:

c. If we want to show that R has been applied twice to (1,0), we can write R^{2}\(\left(\begin{array}{l}

1 \\

0

\end{array}\right)\), which represents R\(\left(\begin{array}{l}

1 \\

0

\end{array}\right)\). Find R^{2}\(\left(\begin{array}{l}

1 \\

0

\end{array}\right)\) and plot it. Then, find R^{3}\(\left(\begin{array}{l}

1 \\

0

\end{array}\right)\)=R(R(R\(\left(\begin{array}{l}

1 \\

0

\end{array}\right)\))), and plot it.

Answer:

d. Describe the matrix transformation \(\left(\begin{array}{l}

x \\

y

\end{array}\right)\)→R^{2}\(\left(\begin{array}{l}

x \\

y

\end{array}\right)\) using a single matrix.

Answer:

R^{2}\(\left(\begin{array}{l}

x \\

y

\end{array}\right)\) is the transformation that rotates points through 2∙\(\frac{\pi}{3}\) radians, so a formula for R^{2}\(\left(\begin{array}{l}

x \\

y

\end{array}\right)\) is

Exercises

Exercise 1.

a. Suppose f(t) represents the position of a moving object that starts at (1,0). How long does it take for this object to return to its starting point? When the argument of the trigonometric function changes from t to 2t, what effect does this have?

Answer:

The object will return to (1,0) when 2t=2π. Thus, it will take t=π seconds for this to happen. Changing the argument from t to 2t causes the object to move twice as fast.

b. If the position is given instead by g(t), how long would it take the object to return to its starting point? When the argument of the trigonometric function changes from t to \(\frac{t}{2}\), what effect does this have?

Answer:

The object will return to (1,0) when \(\frac{t}{2}\)=2π. Thus, it will take t=4π seconds for this to happen.

Changing the argument from t to \(\frac{t}{2}\) causes the object to move half as fast.

Exercise 2.

a. Draw the path that P=G(t) traces out as t varies within the interval 0≤t≤1.

b. Where will the object be at t=3 seconds?

Answer:

G(3)=(1,0)

c. How long will it take the object to reach (0,-1)?

Answer:

These coordinates represent (cos(π),sin(π)), so (cos(\(\frac{π}{2}\)∙2)”,” sin(\(\frac{π}{2}\)∙2) ); the object reaches this location at t=2 seconds. G(2)=(0,-1), so it will take 2 seconds to reach that location.

Exercise 3.

a. Draw the path that P=H(t) traces out as t varies within the interval 0≤t≤2.

b. Where will the object be at t=1 second?

Answer:

H(1)=(-4,1)

c. How long will it take the object to return to its starting point?

Answer:

H(4)=(1,4), so it will take 4 seconds to return to its starting point.

Exercise 4.

Suppose you want to write a program that takes the point (3,5) and rotates it about the origin to the point (-3,-5) over a 1-second interval. Write a function P=f(t) that encodes this rotation.

Answer:

Let f(t)=. We have f(0)=(3,5) and f(1)=(-3,-5), as required.

Exercise 5.

If instead you wanted the rotation to take place over a 1.5-second interval, how would your function change?

Answer:

Let f(t)=. We have f(0)=(3,5) and f(1.5)=(-3,-5), as required.

### Eureka Math Precalculus Module 1 Lesson 23 Problem Set Answer Key

Question 1.

Let . Find the following.

a.

Answer:

b. How many transformations do you need to take so that the image returns to where it started?

Answer:

It rotates by \(\frac{π}{4}\) radians for each transformation; therefore, it takes 8 times to get to 2π, which is where it started.

c. Describe the matrix transformation \(\left(\begin{array}{l}

x \\

y

\end{array}\right)\)→R^{2}\(\left(\begin{array}{l}

x \\

y

\end{array}\right)\) and R^{n}\(\left(\begin{array}{l}

x \\

y

\end{array}\right)\) using a single matrix.

Answer:

R^{2}=\(\left(\begin{array}{l}

x \\

y

\end{array}\right)\) is the transformation that rotates the point through 2×\(\frac{π}{4}\) radian, so a formula for R^{2}\(\left(\begin{array}{l}

x \\

y

\end{array}\right)\) is

Question 2.

For f(t)=, it takes 2π to transform the object at \(\left(\begin{array}{l}

1 \\

1

\end{array}\right)\) back to where it starts. How long does it take the following functions to return to their starting point?

a.

Answer:

3t=2π,t=\(\frac{2π}{3}\)

b.

Answer:

\(\frac{t}{3}\)=2π,t=6π

c.

Answer:

\(\frac{2t}{5}\)=2π,t=5π

Question 3.

Let F(t)=, where t is measured in radians. Find the following:

a. F(\(\frac{3π}{2}\)),F(\(\frac{7π}{2}\)), and the radius of the path

Answer:

The path of the point from 0≤t≤2π is a circle with a center at (0,0).

b. Show that the radius is always \(\sqrt{x^{2}+y^{2}}\) for the path of this transformation (t)=\(\left(\begin{array}{cc}

\cos (t) & -\sin (t) \\

\sin (t) & \cos (t)

\end{array}\right)\left(\begin{array}{l}

x \\

y

\end{array}\right)\).

Answer:

Question 4.

Let F(t)=, where t is a real number.

a. Draw the path that P=F(t) traces out as t varies within each of the following intervals:

i. 0≤t≤1

ii. 1≤t≤2

iii. 2≤t≤3

iv. 3≤t≤4

Answer:

b. Where will the object be located at t=2.5 seconds?

Answer:

c. How long does it take the object to reach \(\left(\begin{array}{c}

-8 \sqrt{6} \\

8 \sqrt{2}

\end{array}\right)\)?

Answer:

The point \(\left(\begin{array}{c}

-8 \sqrt{6} \\

8 \sqrt{2}

\end{array}\right)\) is in Quadrant II; the reference angle is \(\frac{πt}{2}\)=arctan\(\left(\frac{8 \sqrt{2}}{8 \sqrt{6}}\right)\)=\(\frac{π}{6}\),\(\frac{πt}{2}\)=\(\frac{π}{6}\),t=\(\frac{1}{3}\) second.

It takes 1.5 seconds to rotate the point to π; therefore, 1.5-\(\frac{1}{3}\)=\(\frac{7}{6}\) second.

Question 5.

Let F(t)=

a. Draw the path that P=F(t) traces out as t varies within the interval 0≤t≤1.

Answer:

b. How long does it take the object to reach (\(\sqrt{3}\),0)?

Answer:

The point (\(\sqrt{3}\),0) lies on the x-axis. Therefore, t=2 seconds to rotate to the point (-1,-\(\sqrt{3}\)).

c. How long does it take the object to return to its starting point?

Answer:

It takes 6 seconds.

Question 6.

Find the function that will rotate the point (4,2) about the origin to the point (-4,- 2) over the following time intervals.

a. Over a 1-second interval

Answer:

b. Over a 2-second interval

Answer:

c. Over a \(\frac{1}{3}\)-second interval

Answer:

d. How about rotating it back to where it starts over a 4/5-second interval?

Answer:

Question 7.

Summarize the geometric effect of the following function at the given point and the time interval.

a.

Answer:

At t=0, the point \(\left(\begin{array}{l}

4 \\

3

\end{array}\right)\) is dilated by a factor of 5 to \(\left(\begin{array}{l}

20 \\

15

\end{array}\right)\).

At t=1, the image \(\left(\begin{array}{l}

20 \\

15

\end{array}\right)\) then is rotated by \(\frac{π}{4}\) radians counterclockwise about the origin.

b.

Answer:

At t=0, the point \(\left(\begin{array}{l}

6 \\

2

\end{array}\right)\) is dilated by a factor of \(\frac{1}{2}\) to \(\left(\begin{array}{l}

3 \\

1

\end{array}\right)\).

At t=1, the image \(\left(\begin{array}{l}

3 \\

1

\end{array}\right)\) then is rotated by \(\frac{π}{6}\) radians counterclockwise about the origin.

Question 8.

In programming a computer video game, Grace coded the changing location of a rocket as follows:

At the time t second between t=0 seconds and t=4 seconds, the location \(\left(\begin{array}{l}

x \\

y

\end{array}\right)\) of the rocket is given by

At a time of t seconds between t=4 and t=8 seconds, the location of the rocket is given by

a. What is the location of the rocket at time t=0? What is its location at time t=8?

Answer:

b. Mason is worried that Grace may have made a mistake and the location of the rocket is unclear at time t=4 seconds. Explain why there is no inconsistency in the location of the rocket at this time.

Answer:

These are consistent.

c. What is the area of the region enclosed by the path of the rocket from time t=0 to t=8?

Answer:

The path traversed is a semicircle with a radius of 2; the area enclosed is A=\(\frac{\pi r^{2}}{2}\)=\(\frac{4π}{2}\)=2π square units.

### Eureka Math Precalculus Module 1 Lesson 23 Exit Ticket Answer Key

Write a function f(t) that incorporates the following actions. Make a drawing of the path the point follows during the time interval 0≤t≤3.

a. During the time interval 0≤t≤1, move the point (8,6) through \(\frac{π}{4}\) radians about the origin in a counterclockwise direction.

Answer:

b. During the time interval 1<t≤3, move the image along a straight line to (6,-8).

Answer:

The image is \(\left(\begin{array}{c}

\sqrt{2} \\

7 \sqrt{2}

\end{array}\right)\)→\(\left(\begin{array}{c}

6 \\

-8

\end{array}\right)\) in 2 seconds from 1<t≤3.

\(\sqrt{2}\)-kt=6

\(\sqrt{2}\)-2k=6

\(\sqrt{72}\)-mt=-8

7\(\sqrt{2}\)-2m=-8

m = \(\frac{7 \sqrt{2}+8}{2}\)