Find the length of the curve. r(t) = cos (7t) i + sin (7t)j + 7 ln cos t k, 0 ≤ t ≤ π/4.

Solution:

Given the curve r(t) = cos (7t) i + sin (7t)j + 7 ln cost k

Length of curve or the arc length =\(\int\limits_0^{\pi/4}\) √{(dx/dt)² +(dy/dt)² +(dz/dt)²}dt

=\(\int\limits_0^{\pi/4}\) √{49sin²7t + 49cos²7t + 49tan²7t}dt

= 7 \(\int\limits_0^{\pi/4}\) √(1+ tan²t)dt [since cos²t + sin²t=1]

= 7\(\int\limits_0^{\pi/4}\) sect dt

= 7 log | sect + tant t |\(^{\pi/4}_0\) [since ∫ sec x dx = ln(sec x + tan x) + c]

= 7(log(sec(π/4) +tan(π/4))  - log((sec(0) +tan(0)))

= 7(log (√2 +1) - log(1 +0)) [Using the trigonometric ratios]

= 7(log (√2 +1))

Find the length of the curve. r(t) = cos (7t) i + sin (7t)j + 7 ln cos t k, 0 ≤ t ≤ π/4.

Summary:

The length of the curve. r(t) = cos (7t) i + sin (7t)j + 7 ln cos t k, 0 ≤ t ≤ π/4 is 7(log (√2 +1)).