Asked by COFFEE
i'm still getting this question wrong. please check for my errors:
Use Simpson's Rule with n = 10 to estimate the arc length of the curve.
y = tan x, 0 <or= x <or= pi/4
.. this is what i did:
y' = sec(x)^2
(y')^2 = [sec(x)^2]^2
[f'(x)]^2 = sec(x)^4
Integral of
sqrt( 1 + sec(x)^4 ) dx
from x=0 to x=pi/4
deltaX = (pi/4 - 0) / 10
deltaX = (pi/4) / 10
deltaX = pi/40
= (pi/40)/3 [f(0) + 4f(pi/40) + 2f(2pi/40) + 4f(3pi/40) + 2f(4pi/40) + 4f(5pi/40) + ... + 4f(9pi/40) + (pi/4)]
= (pi/120) [sqrt(1+sec(0)^4) + 4sqrt(1+sec(pi/40)^4) + ..etc..etc..
and after calculating.. i got these:
= (pi/120) [1.414 + 5.674 + 2.864 + 5.822 + 2.981 + 6.161 + 2.837 + 6.802 + 3.652 + 7.991 + 0.785]
= (pi/120) [46.983]
= 1.230012064
= 1.23
........ and this answer was wrong. can someone point out how i calculated this wrong or if i missed something? i thought i did everything by the book & notes. thanks :)
never mind, I figured it out...1.277995
Use Simpson's Rule with n = 10 to estimate the arc length of the curve.
y = tan x, 0 <or= x <or= pi/4
.. this is what i did:
y' = sec(x)^2
(y')^2 = [sec(x)^2]^2
[f'(x)]^2 = sec(x)^4
Integral of
sqrt( 1 + sec(x)^4 ) dx
from x=0 to x=pi/4
deltaX = (pi/4 - 0) / 10
deltaX = (pi/4) / 10
deltaX = pi/40
= (pi/40)/3 [f(0) + 4f(pi/40) + 2f(2pi/40) + 4f(3pi/40) + 2f(4pi/40) + 4f(5pi/40) + ... + 4f(9pi/40) + (pi/4)]
= (pi/120) [sqrt(1+sec(0)^4) + 4sqrt(1+sec(pi/40)^4) + ..etc..etc..
and after calculating.. i got these:
= (pi/120) [1.414 + 5.674 + 2.864 + 5.822 + 2.981 + 6.161 + 2.837 + 6.802 + 3.652 + 7.991 + 0.785]
= (pi/120) [46.983]
= 1.230012064
= 1.23
........ and this answer was wrong. can someone point out how i calculated this wrong or if i missed something? i thought i did everything by the book & notes. thanks :)
never mind, I figured it out...1.277995
Answers
Answered by
dude
dude me too i cant lol
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