In 9.1, how would you solve numbers 43 and 44, where the y coordinate point is neither in radians nor degrees, but just a plain decimal number. I tried solving it using the formula, but didn't get the answer.
Also, on numbers 65-71, how would you KNOW which graph the equation is supposed to make? I tried attempting one, but got stuck. Please help if you can.
Yeah i had the same questions as Chandan, I tried the last ones and got them wrong. I also needed help on number 37. for some reason i did something wrong but i can't figure out what.
For number 37, your potential error could have been CALCULATING the cos and sin of the angle incorrectly, which would have given you the wrong answer.
So your final coordinate point is:
Note: read the root3/2's as (root3)/2. Sorry about having no parentheses.
On number 49 from 9.1...could the y coordinate, which is according to the answers -pi/4...so can I write 7pi/4 for this as well? If you know or have an idea about this, please reply. Thanks.
My bad. The book says you can't do it that way. Just ignore that.
How on Earth do you solve 67??? Looking at the answer in the back of the book made it more confusing.
I was doin tonight's homework and i really needed help on problems 53 and 59. could someone help me please?
For problem 67, you more or less use the same method you used to solve all of the other ones. So you multiply both sides by r to get r^3 = r cos theta. You can replace the r cos theta with x, so you have r^3 = x. So we know r^2 = (x^2 + y^2), but we need r^3. So if bring each side of the r^2 equation up to the 3/2 power we get (r^2)^3/2 = (x^2 + y^2)^3/2. For the left side of the equation, the powers multiply, giving us r^3 = (x^2 + y^2)^3/2. Now, if we substitute this in for r^3 in the first equation, we get (x^2 + y^2)^3/2 = x. Then you just subtract the x over to get (x^2 + y^2)^3/2 – x = 0.
Let’s see so for problem 53 you have to use a calculator to figure it out. To find r, you use the formula r^2 = x^2 + y^2. So 1.3^2 + -2.1^2 = 6.1. Take the square root of 6.1 and you find that r = 2.47. Then, to find theta, you say tan theta = -2.1/1.3. When you use the calculator, you get -58.24, but since that isn’t in the range (I think, I don’t remember exactly what was in the book!), you add 360, to find that theta = 301.75. Therefore, the polar coordinates are (2.47, 301.76 degrees). For 59, you simply substitute x = r cos theta and y = r sin theta into the equation. You get (r cos theta)^2 = 4 (r sin theta). After squaring and distributing, you get r^2 cos^2 theta = 4r sin theta. Then, you subtract 4r sin theta from each side to set the equation equal to 0, resulting in r^2 cos^2 theta – 4r sin theta = 0.
I neeeded help doing problems 43 and 45 on tonights homework. For 43, i got the graph wrong. and for 45, i just had no idea how to figure out the coordinate points or anything.
Thanks Kelly! :)
In that example at the end of class the other day the graph showed it was symmetrical about the polar axis but when we tested its symmetry it said it wasnt, does anyone know why that is?
In order to do polars in the
calculator, does it always have to
be in radians
Julianne, the symmetry tests are kind of inconsistent. The symmetry tests only prove that a graph is symmetrical, not that it isn’t symmetrical. If the test works, then you know the graph is symmetrical somehow. But if it doesn’t work, it doesn’t necessarily mean it isn’t symmetrical, but you just have to plot more points to check. Hope this helps.
And Robert, no, it doesn’t always have to be in radians. However, when you graph it in degrees, the graph is really stretched out because it is graphed over the course of 360 degrees, which, numerically, is much larger than the 2 pi radians if you graph in radians. Basically, if you want to graph in degrees, you just have to modify the window.
For 43, you can determine that it is a rose with 4 petals by looking at the equation and comparing it to the chart on page 582 in the book. I’m horrible at describing stuff, but it should look like a rose with four petals, centered at the origin, with each petal 3 units from the pole. Using the symmetry tests, you know it is symmetrical with respect to the x-axis so I only found points from 0 to pi. For 45, you should get a rose with 3 petals, with each petal reaching to 4 units from the pole. Using the symmetry test, you can determine it is symmetrical about the y-axis, so you only have to find points ranging from 3 pi/2 to pi/2. Then, you simply follow the same process of finding points that you have been using. For example, if we say theta is pi/6, then r = 4 sin (3 (pi/6)). Next: r = 4 sin (3pi/6), r = 4 sin (pi/2), r = 4 sin (1), r = 4. So the coordinate we just found is (4, pi/6). Then you would just plot this point and repeat the process until you can determine the graph.
on the first section of 9.3 homework does anybody know how to do number 10?
For #10, section 9.3- it is done identically to the example I did in class:
1. Plot the point (2, root3) in quadrant one and form a triangle.
2. Use r^2= x^2 + y^2 to find that r= root7
3. Polar form: z=r(cos theta + i sin theta) yields:
z= root7(cos 40.9deg + i sin 40.9 deg)
Note: To find theta, complete the triangle in Quadrant one and use pythagorean's theorem to find the missing side. Then use inverse tangent to find theta - which is 40.9 deg.
How do you distinguish the difference between the graph with the loop in the middle and the graph without it? And how do you know how many petals there are?
Hey guys. This is mainly for lesson 9.3, but I found this great video (or sets of videos) that explain the complex plane and how to write complex numbers in rectangular form and vice versa. Enjoy! :
Thanks that really helped a lot. But can anyone re-explain to me how to go from polar to rectangular...
Are you asking about like the Lamacene(not spelled right) with out and the Lamacene with loop?
for polar to rectangular, you have to use section 9.1 to help you. this is when you know that the x value is rcosx, and y is rsinx.
Knowing this, you have to find x and y respectively by computing what the rcosx and rsinx are, and then you have your x and y coordinates.
After that, you just put those together to make your coordinate point.
Madeline, if you are told (6, pi/6) in the form of p= (r, theta) then all you have to do is remember the equations to transform each.
So in this case, 6 is equal to r so we can subsitute that and we also know that theta is pi/6. So all we have to do is plug it in!
x= 6cos(pi/6) so we find on the unit circle cos is at pi/6 which is root 3/2 and multiply it by 6 to get 3root3!
and do the same thing for y!
hope that helps!
Sivani, thank you so much for the link...I'm not only using it for math now, but for chem as well ;).
How do you do number 22 on 9.3? I don't understand how to solve when you have to find the cosine and sine of pi/10
On the worksheet it says to transform the equation r=2cosx so that it indicates a rotation of pi/2 counterclockwise and a rotation of pi/3 clockwise. How do you do that?
I'm confused on problem 25 in 9.3. The answers say that it should be 12(cos40+isin40) but I keep getting 400. Is there a missing zero or am I missing something?
I had the same problem and I'm not exactly sure if this is right, but I figured since 400 is 40 degrees over 360, you should just take the difference. If you do problem 28 theres a similar situation but you end up with negatives and you have to make it positive.
Rotating the graph for a polar graph is very similar to a horizontal shift for a rectangular graph. To rotate the graph r = 2 cos theta counterclockwise pi/2, then it just looks like this: r = 2 cos (theta – pi/2). So to rotate the graph clockwise pi/3, it would look like r = 2 cos (theta + pi/3). So for counterclockwise, you subtract the angle from theta, and for clockwise, you add the angle to theta. Hope that helps!
For 22, there is no way to find cos and sin of pi/10 using a unit circle. So you just have to use a calculator! You just put the calculator in radian mode and put in pi/10 for theta and solve.
I don't get how you would do the transformation for part 1b on the polar worksheet. I get the pi/2 one, but how would you know which coordinates will work for the pi/3 one...?
I'm a little confused on #25 in section 9.3, i understand how to do:
z(times)w= 12[(cos 130+270)+(isin130+270)]
but then in the back of the book it says it equals:
12 (cos40 + isin40)
why would you need to subtract 360?
(Kayla the reason is because they subtracted 400-360 but im not understanding why....)
its not really subtracting 360 because 400 is the same angle as 40 they are just using it in terms that we can use on the unit circle
and for chandan i think you would just put r = 2cos(theta - PI/3)
I was a little confused about the "More Polar Graphing" Worksheet
on problem 8 and 11
8. 4 = r^2 + 9 - 6r * cos(theta - (pi/12))
I wasn't sure how to graph it
I would appreaciate if someone could help me!
Vish (way up there),
To tell how many petals there are on a rose, look at the coefficiant on the theta. asin(theta) is 4 petals, and asin(theta) is 3 petals.
And to tell whether a limacon has a loop or not:
if an equation is in the form
* if a<b then it has an inner loop
* if a>b then it does not have an inner loop
For the PISA essay, does anyone know if we need to make an outline? If you look at the sheet under the College Test Prep page, it says to on the sheet, but I thought that was only pertaining to the class that has the assignment...(<--?)
No outline, just the essay.
are we supposed to watch the video to write the paper?
i'm really confused about this writing assignment. Can someone explain it?
I didn't watch the videos that were provided by the power point PDF file. Basically, the prompt is asking you if you agree with the PISA method of testing or what we have today, which is the SAT and ACT. This is basically all you have to do; take a side and use supporting details to support your point.
Hopefully that made sense.
Im a little confused on how to do problem #40 in 9.3...if anyone could explain it, thatd be great!
oh and Beatrice, for #25 they put the answer as 12 [cos 40 + isin 40]. They do subtract 360, but when you think about it, the sin and cos of 40 degrees is actually the same as 400 degrees, because they are just going around the unit circle again.
oh. nvm. alex already answered that..
Does anyone know how to do number 43 from 9.3? Thanks!
Hey Sejzelle, this is what I did for that one...
The complex cube roots of 1+i
> 1(cos 45 + i sin 45)
> square root of (2)^1/3 (cos (45+360k/3) + isin(45+360k/3))
> 6th root (2) (cos 135k + isin 135k)
> 6th root(2) [cos 15 + isin 15]
> then take 135 and plug it in for 15 and do the same with 255
I hope that made sense. :]
So that means you have three answers for 43, by the way.
For the writing assignment, do we have to cite sources within the essay? and how long does it have to be?
Chinar, 360/3 = 120.
Shouldnt it be:
2^1/6 * (cos45 + isin45)
2^1/6 * (cos165 + isin165)
2^1/6 * (cos285 + isin285)
I did the same thing, but for k dont we use 0, 1, and 2 because there are 3 answers? Yours dont seem to match mine at all. I'm confused on where the 15 and 135 came from is all and want to make sure i havent been doing them all wrong.
Shivani,i think if you used sources then you will need to cite it, she did say MLA format. Although it was supposed to be like a act writing prompt so im not sure if you really needed to use outside information... i dont know i could be wrong. it probably should be four to five paragraphs as well. i have a question though, do we turn it in tomorrow or e-mail it to her tonight????
I know she said it has to be 5 paragraphs, and yes i do believe that we have to email it to her tonight because she is going to check it into turnitin.com for legibility reasons.
Yes, that is correct. She said you have to turn in your soft copy into her via email, and she'll take care of the rest with the turnitin stuff.
but just to be safe, i would have a hard copy with you in class tomorrow.
I think what you did wrong is that it is (45 +360k)/3 not 45 + (360k/3). You're taking the quantity of [(45+360k)/3].
It's just a simple algebra mistake. If you fix that, the rest of your answer should be fine (also according to the back of the book).
On number 9 in the review section, why can't the coordinate point (2,-pi/2), which is the answer in the back of the book for one of the answers, be written as:
can someone please explain the logic that is needed to differentiate between on and the other. thanks.
How would you solve number 11 from the review section?
Chandan, you are correct in saying that there are several ways to write each polar coordinate. However, reading the directions will yield the following:
"Find two pairs of polar coordinates (r,Θ) for each point, one with r > 0 and the other with r < 0."
In other words, the book wants you to write the polar coordinates with one positive distance from the origin, and one negative distance from the origin.
But Kyle, isn't r the x coordinate...the y coordinate is what I am concerned about. I got the negative and positive x coordinate, but is 3pi/2 or -pi/2 right for the y coordinate?
yea, i would think so, since they are the same point. i guess there would be four answers.
i think the first set of answers are just simpler to comprehend/less movement and thats why the book chose them.
feel free to correct me if im wrong, but im gonna go ahead and agree with Chandan.
For problem 41 of the review, this is vectors correct? I dont think we need to know this for the test friday since the test is over 9.1-9.3 but can anyone just confirm or deny this for me? Thanks
Hi Jacob! We do need to know that, since it is in section 9.3. It is part of Demoviare's theorem:)
Hey! for number 11 on the review, are we supposed to use a calculator to figure it out? isn't that the only way, since the unit circle does not tell us what angle has a tangent of 3/4?
hey shivani, for problems like
those were you cant use the unit
circle i just use my calculator
I don't understand how to do number 15 on the review. I keep trying to simplify it but i can't get it to a point where i can make it polar coordinates. How do i get it into an equation where i can substitute r for (x^2)+(y^2) or
rsin(theta) for y and rcos(theta) for x.
Vish, you know that for this problem you have to do the OPPOSITE of what you mean...? Your goal is to go from terms of x and y to terms of an sin/cos/tan, etc. Are you referring to another problem?
In section 9.1 how would you do number 43 and 44...I am very confused on those.
Chandan, for those, you just do the same thing even though it is not in theta form.
example- X = r cos theta
=6.3 cos 3.8..........which equals -4.98
and do the same for Y (r sin theta).
can someone explain number 40 from 9.3 and number 49 from the review?
Ok, wow, sorry about the ch above.
Eric, ok I got it. But why is this one have to be set to RADIAN on the calc. instead of DEGREE...these are just numbers that have decimals, not in pi form.
40) Hint: convert root3-i to polar form, and then give the entire thing a power of 6. then solve from there using the theorem.
49) Use DeMoivre's Theorem. You maybe making this too hard; just plug everything in.
can you explain how to convert root3-i to polar form...thanks
Hey chandan, for problems that are in decmal form u put it as radians, because if u were to calculate out a fraction with pi you would get a decimal. You would only use degrees if the questions asked you to find the sin,cos,tan etc of a certain degree measure.
ok nevermind i get how to do 40...but i'm still stuck on 49
In 9.3, when you are trying to solve for Complex Roots...how do you know what k equals? In example 6, how does k go until 2 (including 0 and 1), and not 3...? We went over this in the last 2 minutes of class so I didn't catch this part.
the question was 2x^2-y^2 = y/x
I then made x = rcos(theta) and y = rsin(theta). So when i plugged those values in for x and y into the equation, i got (2r^2cos(theta)^2 - r^2sin(theta)^2 = rsin(theta) / rcos(theta). Then once you distribute the r from the right side of the equation you would get r^2[ (2cos(theta)^2 - sin(theta)^2)] = tan (theta). You get tan theta because rsin(theta) / rcos(theta) is equal to tan(theta). And finally, if you subtract the tan(theta) from the left side you would end up with r^2 [2cos(theta) ^2 - sin(theta)^2] - tan(theta) = 0
#49, Ch. 9 Review
(sqrt(2))^4 = 2^2 = 4
Now all you have to do is "multiply" each of the radians by the given exponent.
((5*PI) / 8) = 20*PI / 8 = 5*PI / 2
But the domain is between 0 and 2*PI so...
5*PI / 2 = PI / 2
If you know your unit circle, you can figure out each of the values from there and simplify further.
Thanks! what about 49 from 9.3?
"k" is based on how many roots you're supposed to obtain. The sequence goes 0, 1, 2, 3, 4, ..., n-1 depending on how many roots there are.
For example, if you need fifth roots, you need five roots. So your values for k are going to be 0, 1, 2, 3, and 4.
so if you're using the Example 6 in 9.3, k would basically be the thing you're trying to show as being the complex nth root of w. and when you're substituting it in the formula, you would only go up to however many roots the equation would have. In this case, this would be 3, including 0, 1, and 2 since you're trying to find the cube root.
I'm not sure about this one, since only one term is involved, but what I would do to set this up is to use a 0 in the beginning, and then you could pull out a 2 from both the 0 and the i. so it would just be 2(0+(1/2(i))). which would turn into 2(cos(90)+i(sin(30)). Then you take the 5th root of the 2 and everything else, and its pretty straigtforward from here.
When converting back to polars and you get like cos theta is -5/13 how do u find that angle without a calculator
Alex - I don't believe you can find it without a calculator because it comes out to some crazy angle not on the unit circle. I assume we wouldn't have noncalculator ones like that on the test :) hopefully
in #15 of the review, whis is it tan(theta) and not rtan(theta)
Thanks Ashwin and Muhammed. That cleared up a lot of stuff.
Alex: You can't...you need your calculator.
For that one, it is just tan(theta), because when you simplify rsintheta/rcostheta, the r's cancel out, leaving you with sintheta/costheta, which is tantheta.
Meghan, since you substitute rcostheta for x and rsintheta for y, the r's cancel out when you do y/x, and you subtract it to the other side.
yea on number 12 on the review i understand how they got 13 as r, but i am a little confused with how they got 1.96 as the theta, for one of the answers.
Sohaila (kinda way up there) for number 43 I believe you have to use Demoivre's theorem but im still having difficulty with it. Can anyone help?
im having trouble graphing complex roots on the calculator, ive looked at the steps on page 592 but its still confusing, any advice?
For number 43
R= sqrt of 1+1 (1+i) is the equation
tan theta= 1/1=1 so
And k= 0,1, and 2
You just take those numbers and plug them into the complex roots equation on page 590
Hey! i'm really confused about how to solve 12 on the review....i keep getting -1.17 for theta, can someone help me out?
Hey BfRaNk for number 43 you do use Demoivre's theorem.
-First you want to graph 1+i, use the coefficients as x and y. x is 1 and y is also 1
-after plotting the triangle you solve for r which is sqrt((1)^2 + (1)^2)= sqrt(2)
- next step is to find tan theta which is opp/adj, 1/1= 1
-then take the 1 and find the inverse tan of it and you would get 45degrees. (depending on which quadrant the triangle was in)
-finally plug into Demoivre's theorem : cubed root (3) =n
theta nought= 45 degrees
-your final equation before plugging in values for k is:
(sixth root of 2)*(cos(15+120k)+isin(15+120k))
-now plug in values for k up to 360 which would be 0,1,2
hope this helps
Hey I had another question about 9.1, number 49. For my answer, I got root2, 7pi/4. The book says it's root2,-pi/4. Does it matter whether it's 7pi/4 or -pi/4?
No it wouldnt, both are correct. Although i'd say 7pi/4 would be the safer answer as most answers are contained to between 0 and 2pi
you wouldn't need to use a graphing calculator for that...when you graph complex numbers, you set the x axis as the real axis, and the y axis as the imaginary axis.
so for example, for something like 3-4i, you would go to the right 3, and down 4 units.
NOTE: the graphing calculator doesn't plot points, unless you know how to set it to that. a coordinate point is not a function, so this is why it's confusing you since nothing is coming up on the screen. I have no idea how to set the calculator to imaginary numbers.
oh nevermind jacob! since, the domain of tangent is from -pi/2 to pi/2.
For these problems, when the triangle is drawn in the 4th quadrant, you ALWAYS have to go CLOCKWISE from the x axis,'s 0 degree line, so the correct answer would be -pi/4 for that.
*Look at the top of page 566. I was confused on the same thing, but that cleared up the confusion.*
Setting mode to Polar lets you use polar equations but i dont think that'd help with imaginary numbers... Like Chandan said just graph with x as real and y as imaginary.
Jacob, I just checked setting to polar and graphing, but that doesn't change anything. So yea, just stick with setting the axes to respective settings and graph by hand.
For the graphing of the polar graphs, how do you know if the spiral continues forever or stops at a certain point...?
I believe it continues on forever, but if you were to graph it by hand you would probably stop at 2pi.
Chandan i think that all graphs on the calculator stop the spirals at a certain point. Because if you plug in the equation shown in example 13 on page 581 into your calculator, it stops at the polar axis but the picture shows it continuing. I'm guessing that they'res just a certain range that the calculator limits the equation to (which is weird because if you look at the table, it continues forever).
Hey!! I have one last question for 56 on the review. for my answers, i got 2, -2, 2i, -2i, but the actual answers say root 2, root 2i, etc...can someone explain this to me?
Hey Shivani, how did you get those answers?
Here are the equations I set up:
16 ^ (1/4) (cos2kpi/4 + isin2kpi/4) = complex roots
k=0 2(1 + 0) = 2
k=1 2(0 + 1) = 2i
k=2 2(-1 + 0) = -2
k=3 2(0 - 1) = -2i
Shivani, for 56, it would be asking for the fourth roots o -16, so these would be the roots of all your answers since you only went up to the third roots. (4i is the 2nd root, the root of that is the 3rd, and the root of that is the fourth).
Thanks, Shivani, and where did the 4 come from?
Sohaila...really high up
to get 13, you have to use the formula r=square root(x^2 + y^2), so it's square root(25+144) --> square root(169), and then the square root of 169 is 13. But I don't know how they got the 1.96 for theta, sorry.
Mrs. Johnson teaches math at Metea Valley High School in Aurora, IL.
Ap Ab Calculus
H Algebra 2 Trig
H Pre Calculus
Regular Pre Calculus