few_shots: 
Example 1:
Concept/skill: 
Understand that by similarity, side ratios in right triangles are properties of the angles in the triangle, leading to definitions of trigonometric ratios for acute angles.

Problem/activity:
Find the general solution for:
$${\mathrm{cos}\left({\pi\over2}+x\right)=\mathrm{tan}{\pi\over4}}$$


Answer: 
No. This problem does not directly teach how side ratios in right triangles lead to definitions of trigonometric ratios. Instead, it involves solving a trigonometric function.

Example 2:
Concept/skill: 
Solve quadratic equations in one variable.

Problem/activity:
Task
Solve the quadratic equation
$$ x^2 = (2x - 9)^2 $$
using as many different methods as possible.


Answer: 
Yes. This task shows one quadratic equation with one variable, x.

Example 3:
Concept/skill: 
Construct a tangent line from a point outside a given circle to the circle.

Problem/activity:
Task
Suppose $A$ is an angle with vertex $P$, as pictured below:
[IMG]
Draw a circle with center $P$ and with  radius $r \gt 0$.  Explain why the circle
meets each ray of angle $A$ in a single point.  Label these points $Q$ and $R$ respectively.
Draw circles with centers $Q$ and $R$ respectively and radius $r$.  These circles meet at $P$ and a second point to be labelled $B$.
Show that ray $\overrightarrow{PB}$  bisects angle $P$.


Answer: 
No. This task does not involve drawing lines that are tangent to circles. The only line that may be drawn is the ray between points P and B.

Now, determine whether the following problem or activity can enable students to understand the concepts or skills in the provided description.

shot_info: fl_problem_001894_neg_0

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im_task_000870_pos_0

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im_task_001160_neg_0

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standard_description: Write a function that describes a relationship between two quantities.
problem_activity: In Class Launch
Use after Unit 2, Lesson 3
Elicit students’ prior knowledge about the solar system and what it looks like. Sketch a picture like this of the sun and a few planets with their orbital paths.
[IMG]
Using student input and making corrections if necessary, write a list of the planets in order from nearest to farthest away from the sun (Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune). Ask students to estimate how long it takes each planet to go around the sun, and write the estimates next to each planet.
Tell students that there is a relationship between a planet’s distance from the sun and its orbital period (the time it takes to complete one revolution around the sun). In this activity, they will figure out how to describe that relationship using polynomials.
Teacher Instructions
Although students saw how to find equations to fit data in an earlier course, this is the first time they fit polynomial equations to data. If students need a hint about how to find a polynomial equation that fits the data better, it may help them to know that polynomials of higher degree can fit data points more closely. One good strategy is to start with a polynomial of degree 1 or 2 and then add more terms. For example, students could start with an equation of the form
\(y=nx^2\)
and then move to
\(y=nx^2+mx\)
and
\(y=nx^2+mx+k\)
. If the maximum amount of terms (1 greater than the degree) is reached and the polynomial is still not a good fit, then the degree can be increased.
Graphing technology can be a very helpful tool for finding models in this way, and students should be encouraged to use it. Graphing calculators can automatically find polynomials of specified forms that are the closest possible fit to a set of data. Desmos also has this feature. Calculators may report the value of
\(R^2\)
for the functions they produce.
\(R^2\)
measures how closely a function fits a set of data points. Tell students that the closer to 1
\(R^2\)
is, the better the function fits the data.
To conclude the task, invite students to share their models for the relationship between the orbital period and distance, and for the relationship between the orbital period and the square root of the distance. The square root should have a much simpler relationship to the orbital period: if we cube the square root of the distance and multiply it by some constant, we get the period. This is exactly what Kepler found. Share with students the statement of Kepler’s third law, which is: “The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit”. What this means is that the square of the orbital period is some constant times the cube of the distance. This can be expressed as
\(p^2 = n \boldcdot d^3\)
. This can be rearranged into something of the form
\(p= k (d^{1/2})^3\)
, which is the relationship that students have been investigating.
Once Kepler saw this relationship, he could divide the square of any planet’s period by the cube of its distance in order to find out what the proportionality constant is. Students can try this for themselves and verify that all planets have the same proportionality constant. They may also want to compare what happens with Jupiter’s moons—do they have the same proportionality constant as the planets? Newton’s laws of gravity later explained why this relationship between distance and period is true.
Student-Facing Statement
In the early 1600s, Johannes Kepler (1571–1630) studied the motions of the planets to find a good mathematical model for them. In 1619, he published his third law of planetary motion, which says how the orbital periods of the planets are related to their distances from the sun. In Kepler’s time, Uranus and Neptune had not been discovered, but here is the data for all 8 planets:
| planet  | distance (millions of km) | period (days) |
|---------|---------------------------|---------------|
| Mercury |           57.9            |     88.0      |
|  Venus  |           108.2           |     224.7     |
|  Earth  |           149.6           |     365.2     |
|  Mars   |           227.9           |     687.0     |
| Jupiter |           778.6           |     4,331     |
| Saturn  |          1,433.5          |    10,747     |
| Uranus  |          2,872.5          |    30,589     |
| Neptune |          4,495.1          |    59,800     |
Plot the distance (
\(x\)
) and period (
\(y\)
) of each planet, and find a polynomial model that fits the data as well as possible. You may have to experiment with both the degree of your polynomial function and the number of terms.
Make another plot that uses the square root of each distance instead, and find a polynomial model that fits those points as well as possible.
Which model do you think is best?
Jupiter has a lot of moons. Look up the periods and distances of some of them, and use that data to make a polynomial model of the relationship between period and distance for Jupiter’s moons.
Use your model to predict the period of another one of Jupiter’s moons using the radius of its orbit.
How good was the prediction? What are some possible sources of error?
Teacher Instructions
Although students saw how to find equations to fit data in an earlier course, this is the first time they fit polynomial equations to data. If students need a hint about how to find a polynomial equation that fits the data better, it may help them to know that polynomials of higher degree can fit data points more closely. One good strategy is to start with a polynomial of degree 1 or 2 and then add more terms. For example, students could start with an equation of the form
\(y=nx^2\)
and then move to
\(y=nx^2+mx\)
and
\(y=nx^2+mx+k\)
. If the maximum amount of terms (1 greater than the degree) is reached and the polynomial is still not a good fit, then the degree can be increased.
Graphing technology can be a very helpful tool for finding models in this way, and students should be encouraged to use it. Graphing calculators can automatically find polynomials of specified forms that are the closest possible fit to a set of data. Desmos also has this feature. Calculators may report the value of
\(R^2\)
for the functions they produce.
\(R^2\)
measures how closely a function fits a set of data points. Tell students that the closer to 1
\(R^2\)
is, the better the function fits the data.
To conclude the task, invite students to share their models for the relationship between the orbital period and distance, and for the relationship between the orbital period and the square root of the distance. The square root should have a much simpler relationship to the orbital period: if we cube the square root of the distance and multiply it by some constant, we get the period. This is exactly what Kepler found. Share with students the statement of Kepler’s third law, which is: “The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit”. What this means is that the square of the orbital period is some constant times the cube of the distance. This can be expressed as
\(p^2 = n \boldcdot d^3\)
. This can be rearranged into something of the form
\(p= k (d^{1/2})^3\)
, which is the relationship that students have been investigating.
Once Kepler saw this relationship, he could divide the square of any planet’s period by the cube of its distance in order to find out what the proportionality constant is. Students can try this for themselves and verify that all planets have the same proportionality constant. They may also want to compare what happens with Jupiter’s moons—do they have the same proportionality constant as the planets? Newton’s laws of gravity later explained why this relationship between distance and period is true.
Student-Facing Statement
In the early 1600s, Johannes Kepler (1571–1630) studied the motions of the planets to find a good mathematical model for them. In 1619, he published his third law of planetary motion, which says how the orbital periods of the planets are related to their distances from the sun. In Kepler’s time, Uranus and Neptune had not been discovered, but here is the data for all 8 planets:
| planet  | distance (millions of km) | period (days) |
|---------|---------------------------|---------------|
| Mercury |           57.9            |     88.0      |
|  Venus  |           108.2           |     224.7     |
|  Earth  |           149.6           |     365.2     |
|  Mars   |           227.9           |     687.0     |
| Jupiter |           778.6           |     4,331     |
| Saturn  |          1,433.5          |    10,747     |
| Uranus  |          2,872.5          |    30,589     |
| Neptune |          4,495.1          |    59,800     |
Plot the distance (
\(x\)
) and period (
\(y\)
) of each planet, and find a polynomial model that fits the data as well as possible. You may have to experiment with both the degree of your polynomial function and the number of terms.
Make another plot that uses the square root of each distance instead, and find a polynomial model that fits those points as well as possible.
Which model do you think is best?
Jupiter has a lot of moons. Here are the periods and distances of the Galilean moons, which were discovered in 1610:
|   moon   | distance (thousands of km) | period (days) |
|----------|----------------------------|---------------|
|    Io    |           421.8            |     1.77      |
|  Europa  |           671.1            |     3.55      |
| Ganymede |           1070.4           |     7.16      |
| Callisto |           1882.7           |     16.69     |
Use the data to make a polynomial model of the relationship between period and distance for Jupiter’s moons.
Another moon of Jupiter, number LXXI, was discovered in 2018. Its distance from Jupiter is 11,483 thousand km. According to your model, what should its period be?
The actual period of the moon is 252.0 days. How close was the prediction? What are some possible sources of error?


aligns: yes