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# p4 polars watch

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1. arg
from the heinmann book, exercise 7a q6

6. The polar equation of a curve C is r^2 = 2sin2\$
\$ = theta

Show that C could be represented parametrically by the equations

x= 2t/(1+t^4)
y=2t^3/(1+t^4)

where t is a parameter

the cartesian equation is (x^2 + y^2)^2 = 4xy btw

danke!
2. damn... i would try to help ya... but i got phy6 today... if its not done by the time i get back from phy6... ill give it a shot for ya
3. (Original post by kikzen)
arg
from the heinmann book, exercise 7a q6

6. The polar equation of a curve C is r^2 = 2sin2\$
\$ = theta

Show that C could be represented parametrically by the equations

x = 2t/(1+t^4)
y =2t^3/(1+t^4)

where t is a parameter

the cartesian equation is (x^2 + y^2)^2 = 4xy btw

danke!
x = 2t/(1+t^4)
y = 2t³/(1+t^4)

also,

x = rsinø
y = rcosø

dividing y by x gives

y/x = t²
t = √(y/x)

now,

y - x = (2t³ - 2t)/(1+t^4)
y - x = 2t(t²-1)/(1+t^4)

substituting for t and t²,

y - x = 2√(y/x)(y/x - 1)/(1 + (y/x)²)

multiplying top and bottom of rhs by x²,

y - x = 2√(yx)(y - x)/(x² + y²)
(x² + y²) = 2√(yx)

now substitute for x = rsinø and y = rcosø,

r²(sin²ø + cos²ø) = 2r√(cosø.sinø)
r = 2√(½sin(2ø))
r² = 2sin(ø)
========

bitte!
4. (Original post by Fermat)
...
danke(this is the only german i know)

BUT...

i get how you need to find t in terms of y and x

but then why do you do y - x = ...
5. (Original post by kikzen)
danke(this is the only german i know)

BUT...

i get how you need to find t in terms of y and x

but then why do you do y - x = ...
I want to find a (mathematical) relationship between r and ø.
I started off getting an expression for t and t² in terms of x and y.
Then I simply noticed that subtracting x and y would also give me an expression in t and t² (actually t^4).
So I just did that and substituted, for t and t², giving an eqn in x and y only. Then it was just a matter of using the x = rsinø and y = rcosø eqns
6. (Original post by Fermat)
I want to find a (mathematical) relationship between r and ø.
I started off getting an expression for t and t² in terms of x and y.
Then I simply noticed that subtracting x and y would also give me an expression in t and t² (actually t^4).
So I just did that and substituted, for t and t², giving an eqn in x and y only. Then it was just a matter of using the x = rsinø and y = rcosø eqns
hey hey stay there ive got another

well, its more the integration i cant do

curve C has polar eq r=asintsin2t between 0 and pi/2
find the area of C

guh i used to be good at integration! see what happens when you dont do p3 !
7. I gotta go just now, BUT...
You have to do an integration involving

sin²t.sin²(2t)

which becomes,

sin(^4)t.cos(^2)t
sin(^4)t - sin(^6)t

convert the sin(^4)t using sin²t = ½(1-cos2t), expand the result and use cos²t = ½(1+cos2t)
Do similar for the sin(^6) term.
8. (Original post by Fermat)
I gotta go just now, BUT...
You have to do an integration involving

sin²t.sin²(2t)

which becomes,

sin(^4)t.cos(^2)t
sin(^4)t - sin(^6)t

convert the sin(^4)t using sin²t = ½(1-cos2t), expand the result and use cos²t = ½(1+cos2t)
Do similar for the sin(^6) term.
thanks ill try again
9. I've just been trying it again. I got an answer of the area of,

A = (aπ)/256

But I'm not at all happy with that. I can't find any fault in my working, but when I drew the graph in Graphmatica, I roughly measured the area and got (approx)

A = 0.28π (using a=1)

What did you get ?
10. (Original post by Fermat)
I've just been trying it again. I got an answer of the area of,

A = (aπ)/256

But I'm not at all happy with that. I can't find any fault in my working, but when I drew the graph in Graphmatica, I roughly measured the area and got (approx)

A = 0.28π (using a=1)

What did you get ?
ive been doing some other things but im having a go now (just a few lines in i can tell this will take a bit of time )
but the book got (pi*a^2) / 16
11. (Original post by kikzen)
ive been doing some other things but im having a go now (just a few lines in i can tell this will take a bit of time :p)
but the book got (pi*a^2) / 16
But I found my mistake :)
I had divided by 4 instead of multiplying by 4. So I was out by a factor of 16. Which is right. That gives me,

A = (a²π)/256 * 16
A = πa²/16
=======

And, like you said, there is a bit of work involved!

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