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Arc Length - Lecture Notes Section 6.
If a smooth curve is traced out EXACTLY ONCE by the parametric equations x f t and y g t for a ≤ t ≤ b then its length is
L
a
b dx dt
2
dy dt
2 dt
Find the arc length of the following curves:
1. x cos t , y sin t , 0 ≤ t ≤ 2 dx dt
− sin t
dy dt
cos t
L
0
2
− sin t ^2 cos t ^2 dt
0
2 sin 2 t cos 2 t dt
0
2
1 dt
0
2 dt 2 − 0 2
What if we looked at x r cos t , y r sin t , 0 ≤ t ≤ 2 ?
2. x e t^ e − t , y 5 − 2 t , 0 ≤ t ≤ 3 dx dt
e t^ − e − t^
dy dt
L
0
3 e t^ − e − t^ 2 − 2 2
dt
0
3 e^2 t^ − 2 e −^2 t^ 4 dt
0
3
e^2 t^ 2 e −^2 t^ dt
0
3 e t^ e − t^ ^2 dt
0
3 e t^ e − t^ dt e t^ − e − t^ | 03 e^3 − e −^3 − e^0 − e^0 e^3 − 1 e^3
If y f x , then we can regard x as the parameter so that the arc length for a ≤ x ≤ b becomes
L
a
b 1
dy dx
2 dx
If x f y , then we can regard y as the parameter so that the arc length for a ≤ y ≤ b becomes
L
a
b dx dy
2 1 dy
Find the arc length of the following curves:
1. y lncos x for 0 ≤ x ≤ /
dy dx
(^) cos^1 x − sin x − (^) cossin^ x x − tan x
L
0
/
1 − tan x ^2 dx
0
/ 1 tan 2 x dx
0
/
sec 2 x dx
0
/ sec x dx
14 ln|sec x tan x || (^0) /
ln 2 1 − ln|1 0| ln 2 1 ≈ 0. 881
2. x y − 1 3/2^ between 0, 1 and 8, 5. Here parameter is y. Sketch in parametric mode: X 1 T T − 1 ^3/2 and Y 1 T T. Use Window settings: tmin1, tmax5, tstep.05; − 1 ≤ x ≤ 9 and 0 ≤ y ≤ 6.
dx dy
y^ −^1 1/2 (^) dx dy
2 3 2
y^ −^1
y^ −^1 ^9 4
y − 9 4
L
1
5 9 4
y − 9 4
1 dy
1
5 9 4
y − 5 4
dy
Let u 94 y − 54 , then du 94 dy or 49 du dy. Also, change limits of integration: For
y 1 u 9 4
y 5 u 9 4
1
5 9 4
y − 5 4
dy
1
10 u^4 9
du 4 9
1
10 u 1/2^ du
u 3/ 1
10 8 27
10 3/2^ −^1 3/2^
10 3/2^ − 1 ≈ 9. 0734
Most integrals we get finding arc length are very difficult or impossible to integrate. Hence, we need to use a numerical approximation like Simpson’s Rule.
Use Simpson’s Rule with n 8 to approximate the length of y (^1) x from x 1 to x 3.
dy dx
x^2
L
1
3 1 − 1 x^2
2
dx
1
3 1 1 x^4
dx
Δ x 3 − 81 14 0. 25
