Maths Aid!

Competition - post a photo you've taken of nature >>

Scroll to see replies

mikesgt2

If the sector is in a circle of radius r and has an angle t (in radians), then:

Area = ( tr^2 )/2
Perimeter = 2r + rt = r(t+2)

So a square with the same perimeter has sides of length r(t+2)/4; therefore, the area of this square is ( r(t+2)/4 )^2.

We require the area of the sector to be equal to the area of the square, so:

( tr^2 )/2 = ( r(t+2)/4 )^2
( tr^2 )/2 = ( r^2(t+2)^2 ) /16
8tr^2 = r^2(t+2)^2

The r^2 terms cancel, so:

8t = (t+2)^2
8t = t^2 + 4t + 4
t^2 -4t +4 = 0
(t-2)^2 = 0
t = 2

That's really interesting, I'd never thought of doing it that way. Only thing is, how does it show that this is the maximum area a sector could take?

Reply 21

sazzles

Jonatan

They clean up lots of equations. If you want to use degrees you very often end up with a 360/2pi factor.

Also, when you need to raise a number to a complex exponent you probably want to use radians when expressing the complex numbers in polar form. This gives rise the following beautiful equation:

e^(i * pi) + 1 = 0

Also, do you take this task (Its a physics one, but still)

At what height do you find geostationary satelites?

A geostationary satellite goes around the earth with an angular velocity equal to that of earth.
Consider the satellite to have mass m, height h and velocity v and the earth to have mass M, radius r and time period T
force due to circular motion=force due to gravitational attraction
(mv^2)/(r+h)=GMm/(r+h)^2
v^2=GM/(r+h)
v=(GM/(r+h))^0.5
The time period can be expressed as
T=2π(r+h)/v
v=2π(r+h)/T
Put both equations equal to eachother and rearrange.
(GM/(r+h))^0.5=2π(r+h)/T
GM/(r+h)=4π^2(r+h)^2/T^2
(r+h)^3=GMT^2/4π^2
r=(GMT^2/4π^2)^(1/3)-r
Substitute in T = 24*60*60= 86400s, G = 6.67*10-11 N m2/kg2, M = 5.97*10^24kg and R = 6.37*10^6m, and then h=3.59*10^7m

Oh, and e^(i*pi)+1=0, derived from Taylor expansions - again differentiating trig functions. I will admit they tidy things up, though.

Reply 22

It'sPhil...

sazzles

My ridiculously complicated method didn't because I wasn't integrating trig functions, that's the only time I can think of when radians are useful.

Radians are essential as, unlike degrees, they represent angles and lengths therefore equations such as y = xcosx, y = x - tanx etc can only be studied using radians. You cannot have a length of 360 degrees but you can have a length of 2pi. So obviously in calculus where you say as x->0, sinx -> x you can only use radians because you have changed x from an angle to an approximate length

Reply 23

sazzles

It'sPhil...

I see you point about the trig function, I was wrong;I don't really like radians (I always try to work things out in radians when my calculator is in degrees mode and vice versa). What you're saying about the approximations is the same my previous post-I only use radians for calculus.

Reply 24

theone

I think that this problem I did in class today is quite neat, it's an extension of the sine rule:

Say that ABC is a triangle and P is an point inside ABC. Let a = <BPC - <BAC, b = <CPA - <CBA, c = <APB - <ACB. ( < means angle here).

Show that (PA.sin<BAC)/sin(a) = (PB.sin<CBA)/sin(b) = (PC.sin<ACB)/sin(c).

Reply 25

LimpNoodle

yeah nice,

see told u it was radians sazzles.

i got a good problem here take a look at this puzzle (it was easier to draw than type out) and you use common denominator to work out the second line, if u didnt understand

maths.gif5.3KB

Reply 26

sazzles

theone

That's really interesting.
On the subject of triangles and formulae, you just reminded me of Heron's formula for the area of a triangle, A with sides of length a, b and c.
A=(s*(s-a)*(s-b)*(s-c))^(1/2)
where s=(a+b+c)/2
How random!?

Reply 27

greenie787

please please please help me on this question, the exam is 2moro!

Find the value's of x between 0 and 180 satisfying the equation

tan3x=1

i know you tan-1 both sides to give 45 and divide by 3 because of the 3 before the x, but how do you find the other values? thanks!

Reply 28

sazzles

greenie787

You done the first step, work out the acute angle (45)
Look at what other quadriles tan is positive (1st and 3th). This means that;
3x=45, 180+45, 360+45, 360+180+45, 2*360+45, 3*360+180+45 etc
3x=45, 225, 405, 585, 1035 etc
x=15, 75, 135, 195, 255, 345
Hope this is useful.

Reply 29

sazzles

LimpNoodle

lol.
Graph the function f(x)=x/(x-1) + (x-l)/x -2
f(x)=2 is an asymptote.
I liked that one it was funny.

Reply 30

greenie787

sazzles

You done the first step, work out the acute angle (45)
Look at what other quadriles tan is positive (1st and 4th). This means that;
3x=45, 360-45, 360+45, 2*360-45, 2*360+45, 3*360-45 etc
3x=45, 315, 405, 675, 1035 etc
x=15, 105, 135, 225, 255, 345
Hope this is useful.

thank you. with the sin and cos angles i draw the graphs, but with tan its a lot harder

Reply 31

sazzles

greenie787

thank you. with the sin and cos angles i draw the graphs, but with tan its a lot harder

No problem. I agree with you about tan graphs, horrible things.

Reply 32

mikesgt2

sazzles

That's really interesting, I'd never thought of doing it that way. Only thing is, how does it show that this is the maximum area a sector could take?

Well, when you equate the area of the sector with the area of the square you find that there is only one angle for which the areas are equal. Since there is only one, it must be maximum.

Reply 33

sazzles

mikesgt2

Well, when you equate the area of the sector with the area of the square you find that there is only one angle for which the areas are equal. Since there is only one, it must be maximum.

Can you explain it again? I'm getting confused :confused:

It seems as though you are using a circular argument.

Reply 34

Nylex

sazzles

No problem. I agree with you about tan graphs, horrible things.

I never drew graphs for those types of questions. We were taught to draw a quadrant diagram and used that.

Reply 35

mikesgt2

sazzles

Can you explain it again? I'm getting confused :confused:

It seems as though you are using a circular argument.

Why is it circular?

All I am saying is that you do not end up with an equation which you have to maximise using calculus. You simply do the maths and come up with the answer of 2 radians. The way I did the question seemed to show that there was only one angle for which the area of the sector was equal to the area of the square, so there is not an issue of maximum.

Reply 36

sazzles

mikesgt2

I'm probably being really stupid here, but with your arguement isn't there the possibility of having a sector with a different angle, that has a larger area than the square. I thought I had phrased the question so it was prove that with a set perimeter, the maximum area of a sector is equal to the area of a square. I still can't see how your logic proves this (try using one syllable words - I'm kind of dumb).