The document provides step-by-step worked solutions to pre-calculus problems involving sketching graphs based on transformations of an original function. The first problem involves calculating the length of an arc, area of a sector, and angle of an arc given information about a crop circle. The second problem involves sketching the graphs of three transformations - a stretch, inverse, and reciprocal - of an original function given in a graph. Detailed explanations and mathematical working is shown for arriving at the solutions to each part of the two problems.

1. Developing Expert Voices
Pre Calculus 40S enriched
2007
Hi, my name is Sandy and I like purple(:
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2. Problem One
You’re a Skydiver and you’ve just jumped out from the jet plane and
you’re heading for the ground. You pull the string to the parachute
and start to examine a crop circle that a farmer has created.
Given that the centre of the crop circle is labeled O,
the diameter of the circle is 10km, angle BOG is
120° and that D is …
Answer the following in radians.
a) Determine the length of the arc that subtends
an angle of 240°.
b) Determine the area of that sector using the information given above.
c) Determine the angle at the centre of the circle if the arc subtended by
the angle is .
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3. Problem One – answers (a)
From the original question, the answers are supposed to be in radians.
The formula to change Degrees into Radians is:
Where D is Degrees and R is Radians.
Plug in the given information into the formula. It should then look like this:
Now you cross multiply:
240°(π) = R(180°)
Divide everything by 180° to leave R by itself and solve!
The degree signs cancel each other out, and also reduces to .
Therefore, 240° in RADIANS is .
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4. Problem One – answers (a)
We have the angle in radians, so we can use that to solve for the
arc length with this formula:
Where R is the angle in Radians, r is the radius and L is the Arc
Length.
Now we can substitute the given information and what we found
into the formula. The 5 in 2π(r) was found
by dividing the diameter
by 2.
*The radius is HALF of the
diameter in a circle.
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5. Problem One – answers (a)
To get rid of the fraction in a fraction you multiply the numerator
by the reciprocal of the denominator.
Multiply the left side and then cross multiply.
(4π)(10π) = L(6π)
Divide everything by (6π) to leave L by itself. Therefore…
The π’s reduce, leaving you with only 1. Also 4 x 10 = 40 all
divided by 6. Reducing that, it leaves you with
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6. Problem One – answers (a)
A second way is to change an angle from degree to radians is:
Since we’ve been through this once already, it’s should be pretty
straight forward on how to acquire the correct answer. You
should get the same answer you got before which was:
So if you didn’t, you know you did something wrong. But don’t
quit there, go back and try again. ϑ
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7. Problem One – answers (b)
The answer again must be in radians so this is the formula you use to find the
area of a sector:
Where Θ is the angle in radians, S is the area of the sector and r is the radius of
the circle.
Plug in the information that we already know.
Again, we have to multiply the numerator by the reciprocal of the
denominator.
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8. Problem One – answers (b)
Next, we simplify it and then it’s time to cross multiply.
Afterwards, we divide both sides by (6π) so that S is by itself.
We now simplify. The π’s reduce, just like before, leaving you with only one
again, and 100 ÷ 6 reduces to
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9. Problem One – answers (b)
Another way to solve this part of the question is to use a different
formula:
Plugging in all the information we come up with this:
And from this point, it is also pretty straight forward on how to
acquire the answer. Also, you should arrive at the same
answer, if not… try again!
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10. Problem One – answers (c)
This is very similar to the first part of this question. However,
instead of solving for the ARC, we’re solving for the ANGLE.
Where R is the angle in radians, L the Arc Length and r is the
radius.
Now we can substitute the information we know into the formula.
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11. Problem One – answers (c)
Knowing the routine, we multiply the numerator by the reciprocal
of the denominator again. We simplify, then cross multiply.
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12. Problem One – answers (c)
Once again, divide both sides by 9π so that R will be by itself.
Now simplify. Remembering that the π’s reduce you’re left
with…
Therefore the angle subtended by the arc is .
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13. Problem Two
Given the graph of f(x) below, sketch the following three graphs.
a) b) c)
Please note
that for tests or
examinations,
add arrows
and label your
axis.
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14. Problem Two – answers (a)
The first thing you should remember before starting is:
STRETCHES BEFORE TRANSLATIONS
The basic formula for a question like this is: Af(B(x-C))+D
Where A is a STRETCH (the y-coordinates are multiplied by
A), B is a STRETCH (the x-coordinates are multiplied by /
the reciprocal), C is a TRANSLATION (the x-
coordinate is moved horizontally) and D is a
TRANSLATION (the y-coordinate is moved vertically).
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15. Problem Two – answers (a)
Looking at the graph, we can figure out all of the coordinates of
each point.
We’ll arrive at the numbers in this order remembering that the x-
coordinate always comes before the y-coordinate.
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16. Problem Two – answers (a)
We can take those coordinates and apply what we’re given.
Af(B(x-C))+D
Remember that A stretches the y-coordinate and B stretches the
x-coordinate.
First point: A(-6,-3).
A((-6)(3)),(-3)(-2))
The 3 came from the reciprocal in the original question given
above.
You’re then left with a new coordinate: A(-18,6).
BUT you’re not finished yet.
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17. Problem Two – answers (a)
Now you find the Translation part.
Af(B(x-C))+D
Knowing that C shifts the x-coordinate and D shifts the y-
coordinate.
C is (-4) (even though it says x+4 in the question) because in the
formula, it’s (x-C). So in order for C to be positive in the
question, 4 must be a negative number.
(x+4) ◊ (x-(-4))
One negative, multiplied by another negative, always gives you a
positive number.
Therefore: A((-18-4),(6+5))
And the new coordinate IS: A(-22,11)
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18. Problem Two – answers (a)
Now there are 3 more points to do. Just for practice, try them
yourself first and once your finished.. Go to the next slide and
check if your answers are correct. ϑ
The remaining points are:
B(-3,-3)
C(1,1)
D(4,1)
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19. Problem Two – answers (a)
Since we’ve done this once before, I’m going to go through it a
bit faster.
B(-3,-3)
Stretches first! ◊ B((-3)(3),(-3)(-2)) ◊ B(-9,6)
Now Translations! ◊ B(((-9)-4),(6+5)) ◊ (-13,11)
C(1,1)
Stretches first! ◊ C((1)(3),(1)(-2)) ◊ C(3,-2)
Now Translations! ◊ C((3-4),((-2)+5)) ◊ C(-1,3)
D(4,1)
Stretches first! ◊ D((4)(3),(1)(-2)) ◊ D(12,-2)
Now Translations! ◊ D(((12)-4),(-2+5)) ◊ D(8,3)
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20. Problem Two – answers (a)
Now that we’ve calculated all the new coordinates, it’s time to plot those
points onto a graph and connect the dots. After you’re finished, it should
look like this:
*New Graph: Is red.
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21. Problem Two – answers (b)
We’re looking for the INVERSE of the function.
Since we already have the coordinates, all we have to do is
switch the y-coordinate and the x-coordinate.
A(-6,-3) Now, SWITCH THEM. A(-3,-6)
Easy right? Okay. So lets do the others now.
B(-3,-3) ◊ B(-3,-3)
C(1,1) ◊ C(1,1)
D(4,1) ◊ D(1,4)
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22. Problem Two – answers (b)
Now just like the other one, plot the points, connect the dots and
you have your solution!
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23. Problem Two – answers (c)
It’s asking for the reciprocal of the original function.
1. Look for the “invariant points.” ((-1,-1) and (1,1))
- These points are on both the original function and the reciprocal function because
the reciprocal of 1 is 1.
2. Examine the straight horizontal lines. One of them is y=(-3) and the other is y=(1).
On the graph, you find the reciprocal and graph it Therefore y=(1/3) and y=(1).
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24. Problem Two – answers (c)
Now use the “Biggering, smallering” game. If a number
increases, its reciprocal decreases and vice versa. Once it’s
finished, it should look like this:
Don’t forget to add an
asymptote where the
graph has zero(s) or
where x = 0.
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25. Problem Three
Prove:
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26. Problem Three - answers
First thing to always do is to draw the “Great Wall of China.”
Expand the problem.
- We know that COT is the reciprocal of tan
and instead of tan, you can use sin and cos.
We also changed 1 to sinΘ/sinΘ because
that equals one. So we don’t necessarily add
anything to the original question.
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27. Problem Three - answers
Here we simply just multiplied everything
out.
Just like we were doing before, to get rid of the
double fraction we multiplied the numerator by
the reciprocal of the denominator.
The sinΘ’s reduce!
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28. Problem Three - answers
You then end up with this after you reduce!
You’re probably wondering how the heck does
cos2Θ equal 1-sin2Θ? Well, knowing our
identities: sin2Θ + cos2Θ = 1
If we rearrange the order, we can see that
cos2Θ = 1-sin2Θ
If you notice, 1-sin2Θ is a difference of squares.
SO, it is easily factorable.
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29. Problem Three - answers
The 1-sinΘ’s reduce.
And you’re left with 1+sinΘ!
*When You’re
Q.E.D
finished solving, you
must always put
Q.E.D to indicate that
you are finished.
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30. Problem Four
At this very moment, there are 1135 students that attend Daniel
McIntyre High School. 2 years ago there was only 960
students that attended the High School.
a) At what rate is the student body population increasing at?
b) Assuming that the rate continues to increase at this rate, how
much longer will it take for the student body population to be
1500
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31. Problem Four – answers(a)
We are looking for the rate, or the model, that the student body
population is increasing. Here’s the formula that we use to
solve this question:
P = P0(Model)t
Where P is the population at the end of the time period, P0 is the
population at the beginning of the time period, Model is the, in
this case, rate at which the population increases and t is the
change in time.
So using the information we know, we can use it to plug it into
the formula.
1135 = 960(Model)2
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32. Problem Four – answers(a)
1135 = 960(Model)2
Divide both sides by 960 to leave (Model)2 by itself and reduce:
You can use “ln” to solve or you there’s an alternate way. First
way is to use ln. Take the ln of both sides:
Multiply by ½ on both sides to remove the 2 from the right.
Calculate the left side out.
0.0837 = ln(Model)
Therefore e0.0837 = Model. And the rate is (0.0837)(100) = 8.37%
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33. Problem Four – answers(a)
1135 = 960(Model)2
The other way is to do it like this:
These reduce.
And the result is 1.0873 = Model.
To get the actual model you minus one, because when the
population increases, it keeps the original amount (1) and from
there increases (+ 0.0873%).
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34. Problem Four – answers(b)
We’re asked to find the time it takes for the population to
increase from 1135 to 1500.
1500 = 1135(1.0873)t
Divide both sides by 1135 and reduce:
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35. Problem Four – answers(b)
Now you can use ln to solve for t:
Divide both sides by ln(1.0873).
Then solve for t! and t is 3.331428783 ◊ 3.3314 years.
So it will take 3.3314 years for the student body population to
grow to 1500.
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36. Toodle-Loo Kangaroo!
=)
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