Find the length of the curve x = t^2, y = t from t = 0 to t = 1.

• A. ∫_0^1 √(4t^2+1) dt = (1/2)√5 + (1/4) ln(2 + √5) ≈ 1.479
• B. ∫_0^1 √(4t^2+1) dt = √5
• C. ∫_0^1 √(4t^2+1) dt = (1/4)√5 + (1/2) ln(2 + √5)
• D. ∫_0^1 √(4t^2+1) dt = (1/2)√5

Answer: (B)

When a curve is given parametrically by x(t) and y(t), its length from t = a to t = b is the integral of the speed: ∫ from a to b of sqrt[(dx/dt)^2 + (dy/dt)^2] dt.

Here, dx/dt = 2t and dy/dt = 1, so the integrand becomes sqrt(4t^2 + 1). The arc length is thus

L = ∫_0^1 sqrt(4t^2 + 1) dt.

A convenient antiderivative is

∫ sqrt(4t^2 + 1) dt = (t/2) sqrt{4t^2 + 1} + (1/4) ln(2t + sqrt{4t^2 + 1}) + C.

Evaluating from t = 0 to t = 1:

L = [ (1/2) sqrt{5} + (1/4) ln(2 + sqrt{5}) ] − [ 0 + (1/4) ln(1) ] = (1/2) sqrt{5} + (1/4) ln(2 + sqrt{5}).

Numerically, that's about 1.479.

So the exact length is (1/2)√5 + (1/4)ln(2 + √5), which is not simply √5. The discrepancy comes from the logarithmic term that appears in the antiderivative.