Find the length of the curve r(t)=cos(4t) i+ sin(4t) j + 4ln(cos(t)) k 0
I am trying to find the length of the curve defined by the parametric equation ( r(t) = \cos(4t) \, \mathbf{i} + \sin(4t) \, \mathbf{j} + 4 \ln(\cos(t)) \, \mathbf{k} ). Using the arc length formula, I arrived at the integral of ( 16 \sec(t) ) evaluated at the specified limits. However, I am struggling to obtain the correct answer. Can anyone help me identify where I might be going wrong?
1 Answers
Arc length = ∫√[(dx/dt)² + (dy/dt)² + (dz/dt)²] dt
= ∫√[16sin²(4t) + 16cos²(4t) + 16*tan²(t)] dt eval. from 0 to π/4
= ∫√(16 + 16*(sec²(t) - 1) dt eval. from 0 to π/4
= ∫4*sec(t) dt eval. from 0 to π/4
= 4*ln|sec(t) + tan(t)| eval. from 0 to π/4
= 4ln(√2 + 1) - 4ln(1)
= 4*ln(√2 + 1)
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