# P5 Intergration (Sector Area)Watch

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Thread starter 14 years ago
#1
Ok im stuck again. I can do all questions except these 4:

Thanx
0
14 years ago
#2
For 1 and writing t for theta:

s = Int[0->pi] rt((dx/dt)^2 + (dy/dt)^2))

= a Int[0->pi] rt((1+cost)^2 + sin^2t)

= a Int[0->pi] rt(2 + 2cost)

= a Int[0->pi] rt(2 + 4 cos^2(t/2) -2)

= 2 a Int[0->pi] cos(t/2)

= 2 a [2sin(t/2)] [0->pi]

= 4a
0
Thread starter 14 years ago
#3
aww dam... no wonder i couldnt do that 1. LoL didnt read the question properly was finding the area
0
14 years ago
#4
(Original post by Syncman)
Ok im stuck again. I can do all questions except these 4:

Thanx
x=a(m+sinm) y=a(1-cosm)

dx/dm=a(1+cosm) dy/dm=asin
(dx/dm)^2+(dy/dm)^2=a^2(1+2cosm+cos^2m+sin^2m) =2a^2(1+cosm)
{(dx/dm)^2+(dy/dm)^2}^1/2=rt(2)a(1+cosm)^1/2
so
S= a rt(2) int {0 to pi} (1+cosm)^1/2 dm
but cos m=cos(m/2+m/2)=2(cosm/2)^2-1
S=2a int 0 to pi cosm/2
S=2a{2sinm/2} 0 to pi
S=4a
3)
x=a(ln(sect+tant)-asint y=acost

dx/dt=a{(secttant+sec^2t)/sect+tant-cost}
=a{sect(tant+sect)/(sect+tant)-cost
=a{1/cost-cost}
=a{1-cos^2t}/cost
=asin^2t/cost
dy/dt=-asint
so {(dx/dt)^2+(dy/dt)^2}^1/2=a{sin^2t+sin^4t/cos^t}^1/2
=asint({cost+sin^2t}/cost)^1/2
A=2pia^2 int {0,pi/2}costsint ({cost+sin^2t}/cost)^1/2
=2pia^2int{0.pi/2}sint..........by moving cost into ({cost+sin^2t}/cost)^1/2
=2pia^2{-cos(pi/2)+cos(0)}
=2pia^2
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Thread starter 14 years ago
#5
(Original post by evariste)
=a{sect(tant+sect)/(sect+tant)-cost
=a{1/cost-cost}
=a{1-cos^2t}/cost
Just working through it now. i dont understand what uve done here. could someone please explain. Thanx!
0
14 years ago
#6
(Original post by Syncman)
Just working through it now. i dont understand what uve done here. could someone please explain. Thanx!
Just sect(tant+sect)/(sect+tant) = sect = 1/cost. That's all
0
14 years ago
#7
(Original post by Syncman)
Just working through it now. i dont understand what uve done here. could someone please explain. Thanx!
=a{sect(tant+sect)/(sect+tant)-cost
=a{(1/cost)-cost}...........cancelling tant+sect and using sect=1/cost
=a{1-cos^2t}/cost.......... putting over common denominator
0
Thread starter 14 years ago
#8
ah that bracket makes more sense
=a{(1/cost)-cost}
Thanx alot!
0
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