Find the length of the curve. r(t) = cos(8t) i + sin(8t) j + 8 ln(cos(t)) k, 0, 0 £ t £

Solution:

Given r(t) = cos (8t) i + sin (8t) j + 8 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 (8t) , y = sin (8t) and z = 8ln cos(t)

The 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 =\(\int_{0}^{π/4} \sqrt{(cos 8(t))^2 + (sin 8(t))^2 + (8 ln cos (t))^2} dt\)

Consider

x = cos(8(t)) ⇒ dx / dt = - 8sin(8(t)) [derivative of cos(ax) is - asin(ax)]

y = sin(8(t)) ⇒ dy / dt = 8cos(8t) [derivative of sin(ax) is acos(ax)]

z =8ln(cos(t)) ⇒ dz / dt = (8 / cos(t)) × (-sin(t)) [log(x) derivative is 1/ x]

dz / dt = - 3 (tan(t)) [sin(t)/ cos(t) = tan(t)]

(dx / dt )² + (dy / dt )² + (dz / dt )² = ( - 8sin (8(t)) )² + (8cos (8(t)) )² + ( -8(tan(t)) )²

=64 sin² (8(t)) + 64cos² (8(t)) +64tan² (t)

= 64 (sin² (8(t)) + cos² (8(t)) + tan² (t))

=64(1+ tan² (t)) {since sin² (8(t)) + cos² (8(t)) = 1}

= 64sec² (t)

⇒\(\sqrt{(dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2} dt \)= 8sec (t)

∴ S = 8\(\int _ {0} ^ { \pi / 4} {sec\;t}\; dt \)

S = 8ln(sec(t) + tan(t))]\(_{0}^{π/4}\)

S = 8[ln(sec(0) + tan(0) -ln(sec(0) + tan(0))]

S =8 [ln (√2+ 1) -[ln (1 + 0)] (Using the trigonometric ratios)

S =8[ln (√2+1)] - 0

∴ S =8 [ln (+1)]

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Find the length of the curve. r(t) = cos(8t) i + sin(8t) j + 8 ln(cos(t)) k, 0, 0 £ t £

Summary:

The length of the curve. r (t) = cos (8t) i + sin (8t) j +8 ln (cos (t)) k, 0 ≤ t ≤ π / 4 is 8 [ln (√2+1)]