1. Find the length of f(x)= x3/6 + 1/(2x) from x = 1 to x = 2.
Answer
1. The arclength is 17/12.
Solutions
1. First, f'(x)= x2/2 – 1/(2x2), so √(1 + (f'(x)2) = √(x4/4 + 1/2 + 1/(4x4)) = √((x2/2 + 1/(2x2))2). Then the arclength is: ∫12x2/2 + 1/(2x2) dx = ((x3/6) – 1/(2x))|12 = 17/12.
2. Find the length of f(x) = x2/2 – ln(x)/4 from x = 1 to x = 2.
Answer
1. The arclength is 3/2 + ln(2)/4.
Solution
1. First, f'(x)= x – 1/(4x), so √(1 + (f'(x))2) = √(x2/4 + 1/2 + 1/(16x2) = √((x + 1/(4x)(2). Then the arclength is ∫12(x + 1/(4x)) dx = ((x2/2) – ln(x)/4)|12 = 3/2 + ln(2)/4.
3. Find the length of f(x)= ln(cos(x)) from x = 0 to x = π/4.
Answer
1. The arclength is ln(√2 + 1).
Solution
1. First, f'(x)= -sin(x)/cos(x) = -tan(x), so √(1 + (f'(x)2) = √(1 + tan2(x)) = √(sec2(x)) = sec(x), because sec(x) is positive in this range of x values. Then the arclength is: ∫0π/4sec(x) dx = ln|(sec(x) + tan(x)||0π/4(by 42 from the integral table) = ln(√2 + 1)
4. Find the length of r(t) = < et, (1/2)e2t> from t= 0 to t= 1
Answer
1. The arclength is ((e/2)√(1 + e2) + (1/2)ln(1 + √(1 + e2)) – ((√2)/2 + (1/2)ln(1 + √2)).
Solutions
1. First, r'(t)= < et, e2t > and |r'(t)| simplifies to et√(1 + e2t) Then the arclength is: ∫01et√(1 + e2t) dt= (e/2)√(1 + e2) + (1/2)ln(1 + √(1 + e2)) – ((√2)/2 + (1/2)ln(1 + √2)), where we substituted x = et and applied 16 from the integral table.
5. Find the length of f(x) = ln(x) from x = 1 to x = 2.
Answer
1. The arclength is (√5 – ln((1 + √5)/2)) – (√2 – ln(1 + √2)).
Solution
1. First, f'(x)= 1/x, so √(1 + (f'(x)2) = √(1 + 1/x2) = (√(x2 + 1))/x. Applying formula 18 from the integral table, we find the arclength is: (√5 – ln((1 + √5)/2)) – (√2 – ln(1 + √2)).