Find the length of the curve. r(t) = cos(6t) i + sin(6t) j + 6 ln(cos(t)) k, 0 ≤ t ≤ π/4.

Solution:

Given r(t) = cos (6t) i + sin (6t) j + 6 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)

Length of the curve

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

Consider

x = cos(6t) ⇒ dx / dt = - 6sin(6t) {derivative of cos(ax) is - asin(ax)}

y = sin(6t) ⇒ dy / dt =6cos(6t) {derivative of sin(ax) is acos(ax)}

z = 6 ln(cos(t)) ⇒ dz / dt = (6 / cos(t)) × (-sin(t)) {log(x) derivative is 1/ x}

dz / dt = - 6 (tant) { sint / cost = tant}

(dx / dt )² + (dy / dt )² + (dz / dt )² = ( - 6sin (4t) )² + (6cos (4t) )² + ( - 6(tant) )²

=36sin² (4t) + 36cos² (4t) + 36tan² (t)

= 36 (sin² (4t) + cos² (4t) + tan² (t))

 = 36(1+ tan² (t)) {since sin² (4t) + cos² (4t) = 1}

= 36sec² (t)

⇒\(\sqrt{(dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2} dt \) = 6sec(t))

∴ S = \(\int _ {0} ^ { \pi / 4} \sqrt{36sec² (t)} dt \)

S = \(\int _ {0} ^ { \pi / 4} 6 sec (t) dt \)

S = 6 ln(sect + tant)]\(_{0}^{π/4}\)

S= 6[ln(sec(π/4) + tan(π/4)]- 6 [ln(sec(0) + tan(0))]

S = 6 [ln (√2+ 1) -6 [ ln (1 + 0)] (Using the trigonometric ratios)

S = 6 [ln (√2+1)] - 0

∴ S = 6 [ln (√2+1)]

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Find the length of the curve. r(t) = cos(6t) i + sin(6t) j + 6 ln(cos(t)) k, 0 ≤ t ≤ π/4.

Summary:

The length of the curve. r (t) = cos (6t) i + sin (6t) j + 6 ln (cos (t)) k, 0 ≤ t ≤ π / 4 is 6 [ln(√2+1)]