H Pre Calculus Ch. 6

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H Pre Calculus Ch. 6

1/5/2011

Happy New Year! Get off to a great start by doing your homework.

Chammy

1/6/2011 10:32:23 am

have you guys done 89, 92, 95, or 98 in the homework? cause i don't quite understand how you are able to find out what theta is and how to find the 6 trig functions...its on page 394

maddie strick

1/6/2011 12:09:25 pm

for those problems i drew a coordinate plane with all the quadrants and then i plotted the points. then you have to find the hypotenuse with the pythagorean theorm. after that you are able to find the exact values of the 6 trig functions...hopefully that helps

maddie strick

1/6/2011 12:11:34 pm

101 and 104 anyone?

Eric Cheng

1/6/2011 12:57:44 pm

Actually, you don't even have to draw a coordinate plane. With the given points, you can get the hypotenuse and once you have that, you can start solving for the six trig functions. Remember that sine goes with the y part of the coordinate and cosine goes with the x part. So for example:
to solve for sin when given (x,y), it is the y value over the radius (the hypotenuse). to solve for cosine, it is x value over the radius.
Use the theorem box on page 391 to see what I am talking about.

Mihir Surati

1/9/2011 10:14:49 am

Does anyone know how to do problem number 102?

Max Jordan

1/9/2011 01:33:40 pm

For 102

First you find cosθ by using inverse cosine. You must be in radians to get the correct answer.

So with finding the inverse cosine you find the value of theta.
Then just plug it into the equation so it should look like this:

cos((cos^-1(.3))+ π)= -.3

Hopefully that helps

Kayla Simon

1/11/2011 08:13:56 am

Can anyone explain the difference between linear speed v. linear velocity problems? Is there are a difference in solving them? And also, what is the difference between angular speed and angular velocity. Which one does v=rw apply to?
Thanks!

Alyssa Pezan

1/11/2011 09:13:46 am

I'm having problems with setting up problem 89 in 6.1's second assignment. I know I should work backwards (starting with the linear speed equation and work backwards to the angular speed) to get the answer, but whenever I try to solve for theta I get a really weird fraction that makes no sense in context with the problem. Does anyone know how to do this problem?

Muhammed Alikhan

1/11/2011 09:31:52 am

Kayla: I'm not entirely sure, but I think linear speed and linear velocity are actually the same thing, just worded differently. In the same way, this also applies to angular speed and angular velocity. But I'm positive that v=rw is for linear speed and velocity. You use it only when you found out w, which is angular velocity, or (theta/time).

Max Jordan

1/11/2011 10:35:25 am

89. Ok, so we know that linear speed is 9.55 miles/hr and the diameter is 8.5 ft.

So we use the equation for linear speed v=r(omega)and we're trying to solve for omega since we know linear speed and radius.

First use dimensional analysis to get the linear speed into the same units as the radius and also to set it up for solving in rev/min.

So (9.55 mi/hr)x(5280 ft/mi)x(1 hr/60 min)= 840.4 ft/min

Then divide by the radius to find omega.

diameter = 8.5 ft ... radius = 4.25 ft

so 840.4 ft / 4.25 ft gives us 197.74 radians/min since we now have angular speed.

Finally just use dimensional analysis again to get to rev/min

(197.74 radians/min)x(1 rev/2pi radians) = 31.47 rev/min

Hopefully that makes sense

Shivani D

1/11/2011 10:37:24 am

Hi Alyssa!
For 89 these are the steps I followed:
1. Since there is 5280 feet per mile, multiply 9.55 by 5280, in order to get 50424 feet per hour.
2. Then you divide the 50424 feet by 60 in order to see how many feet are covered in 1 minute: 840.4ft/min
3. You then find the circumference of the wheel (8.5 times pi).
4. You then divide 840.4 ft by 8.5pift in order to see how many revolutions are necessary. 31.47 or 31.5 revolutions are necessary.

Vish Patel

1/11/2011 10:40:43 am

Hey I know the answers to number 88 on 6.1 are on the syllabus, but can someone explain to me how exactly to figure it out? I cant really figure it out. Thanks

Ayo Adewole

1/11/2011 10:49:05 am

Hey vish I can help with you the first part of #88:
what I did was find w. You do this by multiplying 480revs/min by 2pi radians/rev and you end up with 960piRadians/min.

Then you have to find the linear speed by using the formula r=rw:
13in x 960piRadians/min which =12480pi/min
Then you have to convert this to miles per hour:
12480pi/min x 60mins/hr x 1ft/12in x 1mile/5280ft and when you multiply everything including the pi you end up with the right answer.

Can someone explain the second part of #88?

Chandan Yashraj

1/11/2011 11:15:18 am

Hi,

Ok I get the angular and linear velocity and speed problems once I know the appriach, but I am having an extremely difficult time figuring out when to use angural, and when to use linear, and HOW to approach it. Yes it says when to use angular or linear in some of the problems, but I find myself completely stuck. Please help!

Chandan Yashraj

1/11/2011 11:19:47 am

sorry, in the third line that's supposed to say "approach."

Beatrice Koka !

1/12/2011 07:48:54 am

ayee! so i was wondering if anyone know hknows how to convert radians to feet? for problem #86 in section 6.1 im not sure how to convert radians to feet because i already found the linear speed which is "pie(radians)/35secs" so after that you would multiply that times 30ft (because the equation w= vr.
How would you convert the radians to feet?

Madeline Zehnal

1/12/2011 09:03:16 am

Hey guys! What did you get as the answer to number 30?

The Mattinator

1/12/2011 11:02:42 am

for angular how do you change radians to the correct unit? like problem 6.1 # 76? <(") penguin

the mattinator is matt latham

1/12/2011 11:05:51 am

Shivani D

1/12/2011 11:06:26 am

Hi!!
Can somebody explain problem 109 on 6.2 to me?

Shivani D

1/13/2011 01:51:38 am

I can help with number 76:
Since the problem asks for the angular speed, first you have to find the radians per second.
The problem states that in 20 seconds, 5 meters are traveled. You use the formula s=r(theta) and substitute 5=2(theta)
Divide both sides by 2.
you get 2.5 = theta
theta equals 2.5 radians, you don't have to change it.

Chandan Yashraj

1/19/2011 12:08:58 pm

Hi guys,

So as you all know, for this chapter one of the main things that we are focusing on is graphing of the different trig functions, and I used the internet to help me with 6.4 and 6.5 homework. These videos come from a website called www.schooltube.com and it is an extremely helpful place to get lessons from, if you look in the right places. It's literally like Youtube, but for school purposes only (weird, i know).

The Unit Circle and Graphing Part 1: http://www.schooltube.com/video/90b318c585c5ab0b7eff/21-The-Unit-Circle-Graphing-Part-1

The Unit Circle and Graphing Part 2:
http://www.schooltube.com/video/fdb9170cd09217b5a278/21-The-Unit-Circle-and-Graphing-Part-3

The General Sine Curve Part 1:
http://www.schooltube.com/video/732a38fb418d98bbe2cc/22-The-General-Sine-Curve-Part-1

The General Sine Curve Part 2:
http://www.schooltube.com/video/501cbc974354295981fc/22-The-General-Sine-Curve-Part-2

The Graphs of Secant and Cosecant part 1:
http://www.schooltube.com/video/255822879e6d652f51d0/23-The-Graphs-of-Secant-Cosecant-Part-1

The Graphs of Secant and Cosecant part 2:
http://www.schooltube.com/video/1bd8ab2f5204c5bb03e9/23-The-Graphs-of-Secant-Cosecant-Part-2

Basic Sine, Cosine, and Tangent Equations:
http://www.schooltube.com/video/e403d8bd35f553b98895/Basic-Sine-Cosine-and-Tangent-Equations-Part-2

I know that these are helping me alot with the homework and other practice problems, and I hope they do the same to you as well.

Chandan Yashraj

1/19/2011 12:20:05 pm

**Yes I know that the teacher teaching in the videos is a hardcore perfectionist, but bear with her, it really helps.

Sejzelle Erastus-Obilo

1/19/2011 01:30:09 pm

Can anyone help me with the limit problem on the worksheet? I feel like I'm over-thinking it. It says what is the limit as x approaches cos(theta) from the right and left. Thanks!

Sejzelle Erastus-Obilo

1/19/2011 01:31:58 pm

Forget that, I meant "what is the limit of cot(theta) as x approaches zero from the left and right". My mistake.

Jamez Hunter

1/19/2011 01:58:59 pm

It really helps to look at the graph of cot on a calculator, then you can clearing see what it is talking about. But if you cannot, try drawing the graph out by making a t-chart and you can see that as the graph approaches 0 from the left it is going to negative infinity and from the right it is going to infinity

Lexy Neville

1/20/2011 05:47:54 am

Can someone tell me how to graph cot, csc, and sec on the caclulator to check my work? I forgot how to do it

Mihir Surati

1/20/2011 09:27:45 am

I don't think that there is an actual button for cot, csc, and sec functions on the calculator. But for cot, you could type cos(x)/sin(x) into the calculator. For csc, you could type in 1/sin(x). And for sec, you could type in 1/cos(x).

Chandan Yashraj

1/20/2011 10:33:30 am

To Sejzelle: Okay so this is basically the same thing as James said, but to make things really easy, when it says what is the limit of cot(theta) as x with a negative sign, then you know that it will be from the left, and if it has a plus sign then you know that it will be from the right. Knowing this, use your finger and from the far LEFT OR RIGHT, trace it till you get to that spot and see what the value is at that specific point. In the instance of cot, if it is coming from the right (positive) then it will always be positive infinity, and from the left side (negative), it will be negative infinity.

To Lexy: The first thing you need to make sure is that your calculator is set to RADIANS, and NOT degrees!!!! This is the most important factor in order for you to be able to see your graph. To chance/check this, just so hit second MODE, and then go down and click enter on radians. THEN go back to Y= and enter in the function, which will be, as mihir said, when you want sec, csc, or cot, you have to put it in as 1/cos(x) for sec(x), 1/sin(x) for csc(x), and cos(x)/sin(x) for cot(x).

As Mrs. Johnson said, AVOID writing cot(x) as 1/tan(x)!! just make your life easier by just doing cos(x)/sin(x). you may find that it will help you more on a trick question on the quiz/test.

I hope this helped (and made sense)!

1/20/2011 11:57:02 am

I agree with Chandan

Kayla Simon

1/22/2011 06:12:27 am

On 6.3 #73-78, how do you solve the problems without using a calculator. For example, 73 is sin^2(40)+cos^2(40) and 40 degrees is not on the unit circle...

Chandan Yashraj

1/23/2011 03:41:31 am

To Kayla: This problem at first may look confusing and not make sense, but you have to remember the pythagorean identities, which in this case you should utilize sin^2(theta)+cos^2(theta)=1. THIS IS CRUCIAL to remember in order for you to do this problem. Comparing the appearance of the identity and #73, wich is sin^2(40)+cos^2(40), you see the similar form. *you do not need the "theta" to make it able to be solved*. So in this case, the answer is just 1. It's that simply.

If you look at #74 in the same section, then you can do the same thing with the tan/sec identity, which is on page 402 of the book. The answer for 74 is also 1...

I hope this helped.

Chandan Yashraj

1/23/2011 03:50:24 am

I should have asked this before, but I need help on #s 75-78 in 6.3. How would you solve those; it says not to use the calculator and 80 degrees is not defined on the unit circle...

Chandan Yashraj

1/23/2011 03:54:15 am

NEVER MIND THAT QUESTION. I just figured out that one of the terms is just a reciprocal so the answer for all 5 is wither 1 or 0. Sorry about that.

Chandan Yashraj

1/23/2011 04:43:00 am

The last line is supposed to say "either"...NOT wither. My bad.

Sejzelle Erastus-Obilo

1/23/2011 02:05:11 pm

Thanks, Chandan

Chandan Yashraj

1/24/2011 10:33:52 am

Ok so Mrs. Johnson was not here today and she gave us a review packet to do in-class. There was one problem which I copied down and I need help understanding it:

If cosθ=lambda/T, and tanθ=K/lamda with 3pi/2<x<2pi, find sin(θ+pi).

Ok so you know that this is located in the 4th quadrant given the information, but how would you approach this problem. I was getting the start to this problem in class but the bell rang.

Also...what does it mean by EVEN AND ODD MULTIPLES OF PI or PI/2. PLEASE EXPLAIN! Is even basically 2pi, 4pi, 6pi, etc. and odd is 3pi, 5pi, 7pi, etc.? Even if I got that right, can someone please elaborate more?

Chandan Yashraj

1/24/2011 10:36:13 am

Also, it is true that in order to find the horizontal shift of a trig function, you have to solve for the x if the there is a number in front of the "x"?

For example: would the phase shift of the function tan(2x-(pi/3)) be to the right *PI/6 units*??

Kayla Simon

1/24/2011 01:30:53 pm

Chandan,
For the phase shift, it would be pi/6. You factor out the 2 from the equation leaving 2(x-pi/6).

Anika Gupta

1/24/2011 01:39:15 pm

Could someone please help me with the mathematical explanation on the worksheet that we got for homework tonight? I'm really confused.

Chandan Yashraj

1/24/2011 01:43:44 pm

Thanks Kayla!

Yea, I'm with Anika on this one...the worksheet is just confusing me even more the longer I read it. I'm just extremely stuck.

Sohaila Mali

1/24/2011 02:14:27 pm

hey guys im just doing some review problems from sections 6.4-6.5. the problem says y=cot(x+ pi/8). I understand how to graph the problem and i know u wud shift it over to the left pi/8 and the period would still be 2 pi, but im just confused on how i would label the x axis on the graph.

Muhammed Alikhan

1/24/2011 03:00:34 pm

I am stuck on the worksheet too, but I think that the mathematical explanation might have something to do with the fact that the Range equation has sin(2theta) and when you plug in 45, you get sin(90)=1, which is the largest sin value in the first quadrant. But I'm not sure how to explain it for the Height equation.

Cody Essling

1/24/2011 03:51:38 pm

For the worksheet, did anyone else get that the 45 degree angle maximized the range.

Vish Patel

1/25/2011 09:10:24 am

If you plug in the equation that the worksheet gave you on your calculator, you can check the range in the stat plot to check your answers

Ratuja Reddy

1/25/2011 10:09:48 am

Sohaila,
The period would be pi, not 2 pi because it's cotangent. For tangent and cotangent the period is pi and for everything else it's 2 pi. I'd just label the x axis by pi/4s and then put the asymptote in the middle of 3pi/4 and pi and in the middle of 7pi/4 and 2 pi and so on.

Chandan Yashraj

1/25/2011 01:12:50 pm

To Sohaila: Ok, remember that for the tangent and cotangent graphs' periods, YOU ALWAYS divide the term in front of the x by pi, not 2pi. What I do for labeling any of the graphs is I just put it starting it off with pi/2, then pi, then 3pi/2, 2pi, etc. Then I just visualize the original graph of the function, and knowing that I use the transformations and graph.

To Cody: Yes, that's right. At 45 degrees the equation results in the largest R value, and I got 90 for the maximum of the H value, because this value yields the largest degrees for that formula.

Madeline Zehnal

1/25/2011 01:13:30 pm

Hey guys when in the first problem y=5sin(4x) what does the 4 do. I know that the 5 changes the amplitude of the graph.

Chandan Yashraj

1/26/2011 08:48:18 am

To Maddie: Whenever there is a number in front of the x term, this defines the period of the function. For sin, csc, cos, and sec, you divide 2pi by the value, so here it is sin, so the period will by pi/2. BUT for tan and cot, you divide by pi instead of 2pi. You are right about the amplitude, but the 4 defines the period...hope I helped.

Chandan Yashraj

1/26/2011 02:16:07 pm

If anyone can answer the following questions I have related to the test on Friday, that would be great:

1) what happens if the period of a trig function graph is pi^2, which would happen if the value in front of the x will have a denominator of pi. or is this even possible?

2) can anyone explain in easiest terms the definition of an "inverse function" where it's something like sin^-1 of 1/2. The answer is 30 degrees for this...but am I right when I'm basically translating this problem to "the degree where sin(theta) is 1/2."

3) what is the easiest way that you use to picture the graph of any of the trig functions when the period is a term without pi...? i get it eventually, but i believe that my method of actually getting it takes too long and too complicated.

4) I think this is just a common sense question, but in 6.6 when it asks to you to find the equation for the graphs...are you solving it based off of sin or cos? I think it depends on the scenario, so if the x-intercept is at 0, then you'd use sin, but if it isn't then it would be cos. in essence sinx=cos(x-pi/2), so I'd think it wouldn't matter.

Thanks.

Anika Gupta

1/27/2011 08:01:28 am

Hi I was studying for tomorrow's test and one thing that I really need help on is when you know the linear speed and you need to find the angular speed. I can find the linear speed with the angular, but not opposite.

Chandan Yashraj

1/27/2011 08:08:18 am

Can someone please explain the projectile problems in 6.3 in simplest terms...? I am just completely stuck by those right now.

Shivani D

1/27/2011 08:36:43 am

Anika: Linear speed is when you find how much distance is traveled per unit of time (seconds). Angular speed is basically how many radians/degrees are covered per unit of time.
In order to find angular speed from linear speed, here's a practice practice problem:
Let's say, you have a wheel, with a diameter of 2, and it completes one revolution per every five seconds.
Then you realize that it covers 2pi per every 5 seconds, and you can simplify this further for how many radians are covered per second. I hope that helps, it is a little confusing. Did you have a specific problem that I can help you out on?

Chandan Yashraj

1/27/2011 09:22:51 am

This is for #71 for the review problems towards the end of the chapter. How is the answer pi/3...? Shouldn't this problem follow the rule for s=r(theta)? If you apply it here the answer turns out to be 60 feet. It is asking for the length of the arc.

Ashwin Chakilum

1/27/2011 10:09:18 am

@Chandan Y.

The conversion factor done here is:

(2 feet / radian)
*
(PI/6 radians / 1)

The "radians" unit cancels out in this dimensional analysis.
Just multiply PI/6 by 2 and you get
PI/3!

Sohaila Mali

1/27/2011 10:17:57 am

chandan,

im not sure if this is answering your question exactly but for the one where u asked if the value in front of the x had a denomiator of pi. so for example if the coeficiant was equal to 2/pi and it was a sin graph. Then you would do 2pi/ (2/pi) and you would get pi^2...and when i graphed it on my calculator the graph expanded. and also for ur question #2, you are right, you would find the angle like that. for #3, i would say the easiest way to picture your graph is by labeling the axis by whole # integers. So for example, if the period was 2 of a sin graph, then it would start at zero, intercept at 1 and would end at 2.

Ashwin Chakilum

1/27/2011 10:23:00 am

@Chandan Y.
Pertaining to the projectile problems in 6.2

The projectile RANGE and HEIGHT functions are given to you to make life much less darn painful.

The v0 variable represents the initial speed of the projectile.
v0^2 indicates the SQUARE of that initial speed.

The constant g is given to you as either 32.2 feet per second OR 9.8 meters per second. Choose units according to the initial speed's units depicted in the problem.

It's really all just pluggin'-n'-chuggin' if you think about it.

EX. #107

v0 = 100
v0^2 = 10000
THETA = 45 degrees
sin(2*THETA) = sin(90) = 1
sin^2(THETA) = sin(45)^2 = 0.5
g = 32.2 feet per second (NOT meters this time)

Plug (v0^2 * sin(2*THETA)) / 32.2 to obtain the RANGE.
Plug (v0^2 * sin^2(THETA)) / (2 * 32.2) to obtain HEIGHT.

RANGE = 310.56 ft.
HEIGHT = 77.64 ft.

Sohaila Mali

1/27/2011 10:31:24 am

And i just want to make sure. When you are trying to find out the phase shift of a graph and there is a coefficiant before the x value, so for example the problem says y=3cos(pix - 2)+1, you take the pix - 2 and set it equal to zero and solve for x.

Justin Temple

1/27/2011 10:41:30 am

Could someone help me with #26 from the review? the question is find the exact value of tan-20/tan200

Chandan Yashraj

1/27/2011 11:56:06 am

Ashwin, I could not have asked for a better explanation...you really cleared things up for me; you have absolutely no idea. One question for you though: what do you type into the calculator, or exactly what does this mean...for sin^2(THETA)?? is it basically sin of the value, and then you square it? I'm trying that but it's not working.

Sohaila: Yes, you are correct. You always do that and solve for x with whatever is inside the parenthesis.

Justin: Ok, I was confused at this at first, but you have to break the question into 2 parts...the top and the bottom. In the top, you see that its tan(-20)...which means you go CLOCKWISE 20 degrees starting at 0 degrees, and you end there. Then there is the other tan(200), which you go counter clockwise and if you compare the 2, they are 180 degrees apart. Knowing this, you know that this follows the rule where tan(theta) and tan(theta=pi) are both opposites. So the answer is -1...I hope that makes sense.

Sohaila Mali

1/27/2011 12:11:27 pm

Thanks chandan! Also, if when graphing an equation and you find that the period is in radians but the right or left shift is not, then how do you label the x-axis. would u convert one of them so that it is specifically either in radians or whole integers?

Shivani D

1/27/2011 12:16:42 pm

Hi Sohaila!
you would convert it to radians. like if the period was 2, you would convert it to 2pi/pi, like we did in class, i think.

Chandan Yashraj

1/27/2011 12:32:27 pm

Yea Sohaila, basically like Shivani just said. YOU ALWAYS LABEL THE X AXIS BASED ON THE PERIOD!!! if it is in terms of pi, then label in intervals of pi, but if just a normal integer, then label it using normal numbers.

Alex Tazic

1/27/2011 01:20:01 pm

For number 49 in the review, y=-2tan(3x) how would you label the x axis? i know the period is pie/3.

Sejzelle Erastus-Obilo

1/27/2011 01:46:34 pm

Alex: it would start at zero, and then go to pi/6, pi/3, etc, I believe.

Sejzelle ERast

1/27/2011 02:34:42 pm

Sejzelle Erastus-Obilo

1/27/2011 02:37:23 pm

If anyone is still awake, could you please help with number 63 in the recioew problems? It's not part of the homework. I just wanted to know how you would figure out which points to put on the x-axis?

Cody Essling

1/27/2011 03:46:06 pm

In the 6.1-6.2 quiz on the back page what is the answer to 4a. I know you can not distribute the cosine... so what are you supposed to do to find the solution.

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