Help with some trigonometry problems please

[mathzeroh]

help with some trigonometry problems please!

hello everybody! how's everyone doing? i hope you guys are well and your homework is going great!
i posted another help thread but it was in the wrong section! sorry! but i got great help there!

ok here's my new problem. i tried, but i just don't know how i could do it! it's a bit long but please bear with me.

Write each rectangular equation in polar form.
26. y=-5
27. x=10
28. x^2+y^2=7 (latex version: [tex] x^{2}+y^2=7[/tex])
29. 2x^2+2y^2=5y (latex version:[tex]2x^{2}+2y^{2}=5y[/tex])



This is what i did:
26: y=-5 ===> I used the pythagorean theorem: r=√(x^2+y^2)
r=√(x^2+(-5)^2) ---> I plugged in the value for y into this formula
r=√(x^2+25) ---> -5 squared is 25
r=5+√(x^2) ---> 25 square rooted is 5
r=5+x ---> the square root of x squared is x...isn't it?[?]

and also, i said that θ=Undefined because they did not give me an "x" value, so i just said that x=0; because, the formula

for converting a rectangular coordinate to polar coordinate is:
[tex]\theta=tan^{-1}(y/x)[/tex], when x>0

and x cannot be zero in the denominator spot right?

and that's what my answer is. but i have this feeling that it is not right! because i did the next problem which is very similar

to this one. (and sadly the answers are not in the back of the book for even number problems)

27: basically i did the same exact thing as number 26, and got this:
r=10+y --->again, here, θ=zero because of the same reason up there (except this time i didn't have a "y" value, so i said

y=0 and got zero because arctan 0=0).

so i checked in the back to see if this one was right, but it wasn't.[b(] [b(] this is what was there (the answer for #27)
r=10secθ

HELP!

then, i got to number 28, and i didn't have a clue as to how to even begin this problem! same with 29! can someone please

help me with these? any help is appreciated! thanks a lot in advance!

 

[cookiemonster]

Whew! I think you're working way too hard!

Okay, let's start from the beginning. When we have some coordinate pair (x,y), we draw a right triangle by dropping a line straight down (or up, if y is negative) to the x-axis. Then we have a nice little triangle.

Now let's figure out what we know from that triangle. Let's say we were given r and theta and we wanted to find x and y. How would we do it? Well, we know that sin = opposite/hypotenuse and cos = adjacent/hypotenuse, so let's see where that takes us.

[tex]\sin{\theta} = \frac{\textrm{opposite}}{\textrm{hypotenuse}} = \frac{y}{r}[/tex]

which we can rearrange to get

[tex]y = r\sin{\theta}[/tex]

You can do the same with the cosine to get an expression for x.

So how does this help you? Well, let's say that y = -5. So what happens when we plug that into the equation we just got?

[tex]y = r\sin{\theta}[/tex]
[tex]-5 = r\sin{\theta}[/tex]
[tex]r = \frac{-5}{\sin{\theta}}[/tex]

That's what we want, right? A function for r in terms of theta. All done!

Now try it for 27, except this time use the equation you derived for x using cosine.

As for 28 and 29, I think you should try the method you tried to use for 26 and 27 in your post. I think I see some x^2 + y^2's in there that will turn into some nice r^2.

So the moral of the post: if you see a lonely y or a lonely x, use r*sin(theta) or r*cos(theta) (respectively), but if you see x^2 + y^2's, turn them into r^2's!

Why don't you give them all a try again and see what you get.

cookiemonster

 

[mathzeroh]

thank you for that awesome reply! ok, I've taken all of the advice. here's what I've got:

27: x=10
[tex]x= rcos\theta[/tex]
[tex]10=rcos\theta[/tex]
then, i divided both sides by [tex]cos\theta[/tex] and got:
[tex]\frac{10}{cos\theta}[/tex]
...and since 1/cos theta is the same as sec theta, the answer would be:
r=10sec theta

28: i don't want to make it too long, but my answers that i got were:
r=7 and r=-7

and 29: r=5/2 sin theta

just like in the back of the book! thanks a lot for your help man!

 

[cookiemonster]

Uh oh, you were a little careless on #28.

cookiemonster

 

[mathzeroh]

what? how?[b(] [b(]

ok here's my work:

x^2+y^2=7
(rcosθ)^2+(rsinθ)^2=7
r^2(cosθ^2+sinθ^2)=7
r^2=7
I SEE IT!
r=THE SQUARE ROOT OF PLUS OR MINUS SEVEN!

thanks a lot man! you're so smart!

can i ask you for help on another problem? please?

 

[cookiemonster]

Sure, sure.

cookiemonster

 

[mathzeroh]

cool thanks a lot

ok. here they are. :

Write each polar equation in rectangular form.
30: r=12
x=rcosθ
x=12cos&theta; <---that's my answer for the "x-value"

here's the "y-value":
y=12sin&theta;

is that right?

can you just give me some more advice on these problems, like the ones you did up there? thanks a lot btw for that.

31.&theta;=-45 degrees (sorry i don't know how to make the degree sign)

this is what i did:
x=rcos-45 degrees
x=r(-0.53) <--my answer for the "X-Value"

"y-value":
y=rsin-45 degrees
y=r(-0.85)

then i checked the answer in the back, and they had:

y=-x

how??

32. rsin&theta;=4
33. r=-2sec&theta;



thanks agian!

 

[cookiemonster]

All right. 30 is technically correct, but it's not what they want. They want you to get rid of theta, too. So try doing this first, square r = 12 and see if you can't figure out what to do. If you're still stuck, think about #28 and how you got that answer, then work backwards.

For 31, what equation did you have relating theta to y and x? You used it somewhere in your first post, so why don't you go find it and see what you can do with it.

32: [itex]r\sin{\theta}[/itex]? That looks familiar...

33: Didn't your answer for #27 have a [itex]10\sec{\theta}[/itex]? Isn't [itex]-2\sec{\theta}[/itex] pretty similar to [itex]10\sec{\theta}[/itex]?

Sorry about the late reply.

cookiemonster

 

[mathzeroh]

it's alright man, i understand. it was pretty late anyway and i had some tough problems there. i noe i was pretty tired, and

you, well it might have been later even than it was here, so thanks a lot for sticking with me that much and that late

man, i appreciate it.

alright. I've taken that wonderful advice again (thank you for taking your time and giving it to me). here's what I've gotten:

30. well as you said, i worked it backwords. at first i was lost, but then i did wut you said and i looked back at

number 28 to see how i did that one, after all, they are pretty dang similar and I'm glad you pointed that out.

working it backwards, i got:

[tex]r=12[/tex]
[tex]\sqrt{r^2}=\sqrt{12^2}[/tex]
[tex]r^2(\cos\theta^2+\sin\theta^2)=144[/tex]
[tex](r\cos\theta)^2+(r\sin\theta)^2=144[/tex]
[tex]x^2+y^2=144[/tex]

and that's the answer, [tex]x^2+y^2=144[/tex] <--- because it's in the rectangular form...right?

31. (thanks a lot for reminding me of that formula, i was LOST man! [b(] )
[tex]\theta=-45\circ[/tex]

the formula:
[tex]\theta=\tan^{-1}\frac{y}{x}[/tex]

since i have already been given theta, i just plugged it in:
[tex]-45\circ=\tan^{-1}\frac{y}{x}[/tex]

take the tangent of both sides...
[tex]-1=\frac{y}{x}[/tex]

multiply both sides by x, to get the answer:
[tex]-x=y[/tex]

32. well this was pretty obvious too, but i just don't think sometimes
[tex]r\sin\theta=4[/tex]
rsin&theta; is the same thing as "y"!
[tex]r=4[/tex] DUH!

33.
[tex]r=-2\sec\theta[/tex]

sec&theta; is the same thing as 1/cos&theta; so i multiplied this to -2 then i multiplied both sides by "cos&theta;"
[tex](r=\frac{-2}{\cos\theta})*(\cos\theta)[/tex]
[tex]r\cos\theta=-2[/tex]
[tex]x=-2[/tex]

thank you very very much, i really appreciate your help! thanks a lot cookiemonster!

 

[mathzeroh]

hello! it's me again! ok I've been working on this problem and its about time that i give it up.

here it is:

Express each equation in polar form. Round "phi" to the nearest degree. ("phi" looks like θ (theta) but

except the line comes down from the top and it is more circular)

y=-1/3x+2

wut i did was bring over the y to that side and i got:
-1/3x-y+2=0

then i went like this:
A=-1/3
B=-1
C=2

i did that to put it into the normal form:
-sqrt((-1/3)^2+(-1)^2) <--negative sqrt because C is positive

i got sqrt(10/9) but i changed that to sqrt(10)/3 because you're not supposed to have a fraction in a sqrt...i think...

then i plugged this into the original formula ---> (-1/3x-y+2=0) to get:
((-1/3)/sqrt(10)/3)x+sqrt((10)/3)y+2sqrt(10)/3=0

i then worked that out to polish it up a bit. this is what i came to:
(3sqrt(10)/10)x+(sqrt10)/3)y+2sqrt(10)/3=0

then i went to solve "phi" which can be solved by this formula:
θ (pretend that is phi, and not theta)=arctan(B/A)
θ=arctan ((sqrt10)/3)/(3/sqrt(10)/10)
that (θ=arctan((sqrt10)/3)/(3/sqrt(10)/10)) would become this:
θ=arctan(10/9) after you clean it up a bit.
=48 degrees

so when i put this into the equation for the polar form, p=rcos(θ-phi)...
i got this:
2sqrt(10)/3=rcos(θ-48) <---theta here is the actual theta

but when after all this work, i cheked in the back, they had this answer:
(3sqrt(10)/5)=rcos(θ-72)

HOW?

Please help me, I AM LOST!

Thank you!

P.S., there might be more from where that came from

 

[mathzeroh]

comeon anybody?

 

[mathzeroh]

well ppl i just came on to say that i have solved the problem

thx for reading this thread though! i think it was ur guys' plan to make me work on it for hours until i get the answer! thanks very much for your help of not helping me. :tongue: