1. Find the length of f(x) = mx from x = 0 to x = 1. Does this agree with basic geometry?
Answer
1. Length = √(1 + m2)
Solution
1. First, f'(x) = m, so √(1 + (f'(x))2) = √(1 + m2). The arclength is ∫01√(1 + m2) dx = √(1 + m2)x|01= √(1 + m2). This agrees with basic geometry, because the path is the straight line from (0, 0) to (1, m), the hypotenuse of a right triangle with base 1 and height m.
2. Find the length of r(t) = < cos(t), sin(t) > from t = 0 to t = π/2.
Answer
1. The length is π/2.
Solution
1. First, note r'(t) = < -sin(t), cos(t) >, so |r'(t)| = √(sin2(t) + cos2(t)) = 1. The arclength is: ∫0π/2 1 dt = t|0π/2 = π/2
3. Find the length of r(t) = < cos(t2), sin(t2) > from t = 0 to t = π/2.
Answer
1. Length = π2/4.
Solution
1. First, note r'(t) = < -sin(t2)2t, cos(t2)2t >, so |r'(t)| = √(sin2(t2)4t2 + cos2(t2)4t2) = 2t. The arclength is ∫0π/2 2t dt = t2|0π/2= π2/4
4. Find the length of f(x) = x2/2 from x = 0 to x = 1.
Answer
1. The arclength is (√2)/2 + ln(1 + √2)/2.
Solution
1. First, f'(x) = x, so √(1 + (f'(x))2) = √(1 + x2). Applying formula 16 from the integral table, we find the arclength is (√2)/2 + ln(1 + √2)/2.
5. Find the length of f(x) = (2/3)x3/2 from x = 0 to x = 1
Answer
1. The arclength is (4√2 – 2)/3.
Solution
1. First, f'(x) = x1/2, so √(1 + (f'(x))2) = √(1 + x) and the arclength is: ∫0π/2√(1 + x) dx = (2/3)(1 + x)3/2|01= (4√2 – 2)/3.