Find the length of the curve. r(t) = cos(2t) i + sin(2t) j + 2 ln(cos(t)) k, 0 ≤ t ≤ π/4.

Solution:

Given r(t) = cos(2t) i + sin(2t) j + 2 ln(cos(t)) k

r(t) is a parametric equation of t .

Also if r(t): R→R³ is a vector valued function of a real variable with independent scalar output variables x, y & z

r(t) = {x, y, z}

Where x = cos(2t) , y = sin(2t) and z = 2ln cos(t)

Length of the curve, S = \(\int_{0}^{π/4} \sqrt{(dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2}dt\)

Consider

x = cos(2t) ⇒ dx/dt = -2sin(2t) 

y = sin(2t) ⇒ dy/dt = 2cos(2t) 

z = 2 ln(cos(t)) ⇒ dz/dt = (2/cos(t)) × (-sin(t)) 

dz/dt = -2(tant)

(dx/dt)² + (dy/dt)² + (dz/dt)² = (-2sin(2t))² + (2cos(2t))² + (-2(tant))²

= 4sin²(2t) + 4cos²(2t) + 4tan²(t)

= 4[sin²(2t) + cos²(2t)] + tan²(t))

= 4(1+ tan²(t))[Since sin²t + cos²t= 1]

= 4sec²(t)

\(\sqrt{(dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2}dt\) = √4sec²(t)

= √(2sec(t))²

= 2sec(t)

S = \(\int_{0}^{π/4} 2sec(t) dt\)

∫sec(t)= ln[sect + tant]

\(S = 2ln(sect + tant)]_{0}^{π/4}\)

S = 2[ln(sec(π/4) + tan(π/4))] - 2[ln(sec(0) + tan(0))]

S = 2[ln(√2 + 1) - ln(1 + 0)]

S = 2[ln (√2 +1 )] - 0

S = 2[ln(√2 + 1)]

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Find the length of the curve. r(t) = cos(2t) i + sin(2t) j + 2 ln(cos(t)) k, 0 ≤ t ≤ π/4.

Summary:

The length of the curve. r(t) = cos(2t) i + sin(2t) j + 2 ln(cos (t)) k, 0 ≤ t ≤ π / 4 is 2[ln(√2 + 1)]