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

Solution:

Given r(t) = cos (4t) i + sin (4t) j + 4 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 (4t) , y = sin (4t) and z = 4ln cos(t)

Arc length is given by s = \(\int_{a}^{b} \sqrt{(dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2} dt \)

Length of the curve

s = \(\int_{0}^{π/4} \sqrt{(cos(4t))^2 + ( sin(4t))^2 + (4 ln(cos(t)})^2 dt \) 

Consider

x = cos(4t) ⇒ dx / dt = - 4sin(4t) [derivative of cos(ax) is - asin(ax)]

y = sin(4t) ⇒ dy / dt = 4cos(4t) [derivative of sin(ax) is acos(ax)]

z = 4 ln(cos(t)) ⇒ dz / dt = (4 / cos(t)) × [-sin(t)) [by Chain rule]

dz / dt = - 4 (tant) [sint/cost = tant]

(dx / dt )² + (dy / dt )² + (dz / dt )² = ( - 4sin (4t) )² + (4cos (4t) )² + ( - 4 (tant) )²

 = 16sin² (4t) + 16cos² (4t) + 16tan² (t)

= 16 (sin² (4t) + cos² (4t) + tan² (t))

 = 16(1+ tan² (t)) {since sin² (4t) + cos² (4t) = 1}

= 16sec² (t)

s = \(\int_{0}^{π/4} \sqrt{(dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2} dt \)

 =S = \(\int _ {0} ^ { \pi / 4} \sqrt{16sec² (t)} dt \)

∴ S = \(\int _ {0} ^ { \pi / 4} 4 sec (t) dt \)

S = 4ln(sec(t)+ tan(t))]\(_{0}^{π/4}\)

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

S = 4 [ln (√2+ 1) - 4 [ ln (1 + 0)] (Using the trigonometric ratios)

S = 4 [ln (√2+1)] - 0

∴ S = 4 [ln (√2+1)]

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

Summary:

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