Find the length of the curve. r(t) = cos(5t) i + sin(5t) j + 5 ln(cos(t)) k, 0 ≤ t ≤ π/4.

Solution:

Given r(t) = cos(5t) i + sin(5t) j + 5 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(5t) , y = sin(5t) and z = 5ln 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(5t) ⇒ dx/dt = -5sin(5t) 

y = sin(5t) ⇒ dy/dt = 5cos(5t) 

z = 5 ln(cos(t)) ⇒ dz/dt = (4/cos(t)) × (-sin(t)) 

dz/dt = -5(tant)

(dx/dt)² + (dy/dt)² + (dz/dt)² = (-5sin(5t))² + (5cos(5t))² + (-5(tant))²

= 25sin²(5t) + 25cos²(5t) + 25tan²(t)

= 25[sin²(5t) + cos²(5t)] + tan²(t))

= 25(1+ tan²(t))[Since sin²t + cos²t= 1]

⇒ (dx/dt)² + (dy/dt)² + (dz/dt)² = 25sec²(t)

⇒ \(\sqrt{(dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2}dt\) = √25sec²(t)

= √(5sec(t))²

= 5sec(t)

S = \(\int_{0}^{π/4} 5sec(t)\,dt\)

∫sec(t)= ln[sect + tant]

S = 5ln(sect + tant)]\(_{0}^{π/4}\)

 = 5[ln(sec(π/4) + tan(π/4))] - 5 [ln(sec(0) + tan(0))]

= 5[ln(√2 + 1) - ln(1 + 0)]

= 5[ln (√2 +1 )] - 0

S = 5[ln(√2 + 1)]

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Find the length of the curve. r(t) = cos(5t) i + sin(5t) j + 5 ln(cos(t)) k, 0 ≤ t ≤ π/4.

Summary:

The length of the curve. r(t) = cos(5t) i + sin(5t) j + 5 ln(cos (t)) k, 0 ≤ t ≤ π / 4 is 5[ln(√2 + 1)]