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Pre Calculus 40S enriched2007

My name is Sandy and I like the colour purple.

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Problem OneYou’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 is120° 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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Problem One – answers (a)

The best way to start this problem is to carefully look over the information that is given to you. Right away you can observe from the original question that the answers are supposed to be in radians. So, the first thing you should do is change 240° for question “a)” into radians.

The formula to change Degrees into Radians is:

Where D is Degrees and R is Radians.Since we know the degrees, which in this case is 240°, we can now plug it

into the equation and solve for R!It should then look like this:

From this point it’s a simple as cross multiplying to continue to solve for R. So…

240°(π) = R(180°)

You then 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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Now that we’ve discovered what 240° is in radians, we can use the formula to solve for the length of an arc which is:

Where R is Radians, r is the radius and L is the Arc Length.Just like when we were trying to change degrees into

radians, we substitute the numbers into the formula.

Just in case you’re lost on how I got 5, looking back to the original question we’re given that the diameter of the crop circle is 10km. Since we’re all so brilliant, we know that the radius is HALF the diameter. Therefore the radius is 10 ÷ 2 which equals 5.

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Before we cross multiply like we did earlier, we first have to find a way to get rid of the “Fraction in a Fraction.” What you have to do is multiply the numerator by the reciprocal of the denominator. Do you think you can do it?

After this we multiply it out and finally you can cross multiply.

(4π)(10π) = L(6π)

Once again, you 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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That’s only ONE way to do part “a)” to question one. If you remember back to the beginning when we changed degrees into radians, we used this formula:

But, this isn’t the only way we can change degrees to radians, we can also use this formula:

Do you see the difference?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.

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Problem One – answers (b)

Moving along, we’re now asked to use the information we’ve been given to solve for the area of the sector. Once again the answer must be in radians, so the formula to find the area of a sector that we are going to use is:

Where Θ is the angle in radians, S is the area of the sector and r is the radius of the circle.

Once again, we plug in the information that we already know.

From here, we once again have to multiply the numerator by the reciprocal of the denominator.

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Once we’ve done this, we simplify it and then it’s time to cross multiply.

After we’ve cross multiplied, 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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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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Problem One – answers (c)

This question is also very straight forward. Typically, it’s almost exactly the same as part “a)” to 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.

Knowing the routine, we multiply the numerator by the reciprocal of the denominator again. We simplify, then cross multiply.

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Problem One – answers (c)

From here we, 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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Problem TwoGiven the graph of f(x) below, sketch the following

three graphs.a) b) c)

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Please note that for tests or examinations, add arrows and label your axis.

Problem Two – answers (a)

With these type of questions, the first thing you should remember before starting is:

STRETCHES BEFORE TRANSLATIONSNow, we can start the question.

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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Now that it’s easy to understand what we should start doing first… We can proceed with answering the question. So 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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Okay! Moving along. Now that we know the original coordinates for each point on the graph, A, B, C and D, we can take those coordinates and apply what we’re given. Let’s look at the formula and the original question again:

Af(B(x-C))+DLooking back, we remember that A stretches the y-

coordinate and B stretches the x-coordinate. C and D we’ll look at later because it’s Stretches before Translations!

Taking the first point, A(-6,-3), we can apply these to them. A((-6)(3)),(-3)(-2))

Just in case you’re wondering where in the world did the 3 come from… remember that we have to use the reciprocal of B and the reciprocal of is 3. (:

You’re then left with a new coordinate: A(-18,6).BUT you’re not finished yet.

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You’ve stretched the coordinate to A(-18,6) but you haven’t yet shifted it (Translations) on the graph.

Af(B(x-C))+DWith the new coordinates we can now shift them up, down

or side to side knowing that C shifts the x-coordinate and D shifts the y-coordinate. Examining the question, we see that C is (-4) and D is (5).

We know that 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 question, 4 must be a negative number.

(x+4) (x-(-4))Looking above, we know that 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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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.

The remaining points are:B(-3,-3)C(1,1)D(4,1)

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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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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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Problem Two – answers (b)

For this part of the question, we’re asked to find the INVERSE of the function. It might sound a little intimidating if you’re not exactly sure how to approach this question, but it’s actually really, really easy!

Since we already have the coordinates for each points we don’t have to find those anymore. To solve this problem, it’s as easy as switching the x-coordinate with the y-coordinate! Sounds easy right? Here’s what you do…

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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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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Problem Two – answers (c)

This part of the question is a little more tricky than the previous one. It’s asking for the reciprocal of the original function. First thing to do is to look for the “invariant points.” Those are at (-1,-1) and (1,1). We look for these points because we know that these points are going to be on both the original function and the reciprocal function. This is because the reciprocal of 1 is 1. Once we’ve found the invariant points, you can examine the straight horizontal lines. One of them is y=(-3) and the other is y=(1). To find out where the reciprocal part of the graph is, you use the reciprocal of these. So, these would now be y=(1/3) and y=(1).

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Problem Two – answers (c)

From here we can use the “Biggering, smallering” game. If a number increases, its reciprocal decreases and vice versa. So when the numbers increase on the graph, the new graph will have the numbers decreasing and vice versa again. Once it’s finished, it should look like this:

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Don’t forget to add an asymptote where the graph has zero(s) or where x = 0.

Problem Three

Prove:

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Problem Three - answers

This is a unit that you REALLY need to practice, otherwise you’ll never completely grasp the repetitions that occur in problems. First thing to always do is to draw the “Great Wall of China.”

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We basically expanded 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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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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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 = 1If we rearrange the order, we can see thatcos2Θ = 1-sin2Θ

If you notice, 1-sin2Θ is a difference of squares. SO, it is easily factorable.


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The 1-sinΘ’s reduce.

And you’re left with 1+sinΘ!

Q.E.D*When You’re finished solving, you must always put Q.E.D to indicate that you are finished.

Problem FourAt 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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Problem Four – answers(a)

For part “a)” of this question 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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1135 = 960(Model)2

Divide both sides by 960 to leave (Model)2 by itself and reduce:

There’s two ways to solve this from here. You can use “ln” or you there’s an alternate way. I’ll first explain it in ln. Take the ln of both sides:

Multiply by ½ on both sides to remove the 2 from the right side.

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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1135 = 960(Model)2

Another way to solve this problem, which is a lot quicker, is to do it like this:

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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These reduce.

Problem Four – answers(b)

We have the Model now, and 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:

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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