Find the length of the curve. r(t) = cos(3t) i + sin(3t) j + 3 ln(cos(t)) k, 0 ≤ t ≤ π/4.

Solution:

Given r(t) = cos(3t) i + sin(2t) j + 3 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(3t) , y = sin(3t) and z = 3ln 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(3t) ⇒ dx/dt = -3sin(3t) [derivative of cos(ax) is - asin(ax)]

y = sin(3t) ⇒ dy/dt = 3cos(3t) [derivative of sin(ax) is acos(ax)]

z = 3 ln(cos(t)) ⇒ dz/dt = (3/cos(t)) × (-sin(t)) [by Chain rule]

dz/dt = -3(tant)

(dx/dt)² + (dy/dt)² + (dz/dt)² = (-3sin(3t))² + (3cos(3t))² + (-3(tant))²

= 9sin²(3t) + 9cos²(3t) + 9tan²(t)

= 9[sin²(3t) + cos²(3t)] + tan²(t))

= 9(1+ tan²(t))

= 9 sec²(t)

⇒(dx/dt)² + (dy/dt)² + (dz/dt)² = 9 sec²(t)

\(\sqrt{(dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2} dt\)= 3 sect 

S = \(\int_{0}^{\pi/4} 3 sec (t) dt\)

But sect integral is ln[sect + tant]

S = 3 ln(sect + tant)]\(_{0}^{π/4}\)

S = 3[ln(sec(π/4) + tan(π/4))] - 3[ln(sec(0) + tan(0))]

S = 3[ln(√2 + 1) - ln(1 + 0)] (Using the trigonometric ratios)

S = 3[ln (√2 +1 )] - 0

S = 3[ln(√2 + 1)]

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Find the length of the curve. r(t) = cos(3t) i + sin(3t) j + 3 ln(cos(t)) k, 0 ≤ t ≤ π/4.

Summary:

The length of the curve. r(t) = cos(3t) i + sin(3t) j + 3 ln(cos (t)) k, 0 ≤ t ≤ π / 4 is 3[ln(√2 + 1)]