from a handpicked tutor in LIVE 1-to-1 classes

# Find the length of the curve. r(t) = cos(5t) i + sin(5t) j + 5 ln(cos(t)) k, 0 ≤ t ≤ π/4.

**Solution:**

Given r(t) = cos(5t) i + sin(5t) j + 5 ln(cos(t)) k

r(t) is a parametric equation of t .

Also if r(t): R→R^{3} 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(5t) , y = sin(5t) and z = 5ln 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(5t) ⇒ dx/dt = -5sin(5t)

y = sin(5t) ⇒ dy/dt = 5cos(5t)

z = 5 ln(cos(t)) ⇒ dz/dt = (4/cos(t)) × (-sin(t))

dz/dt = -5(tant)

(dx/dt)^{2} + (dy/dt)^{2} + (dz/dt)^{2} = (-5sin(5t))^{2} + (5cos(5t))^{2} + (-5(tant))^{2}

= 25sin^{2}(5t) + 25cos^{2}(5t) + 25tan^{2}(t)

= 25[sin^{2}(5t) + cos^{2}(5t)] + tan^{2}(t))

= 25(1+ tan^{2}(t))[Since sin^{2}t + cos^{2}t= 1]

⇒ (dx/dt)^{2} + (dy/dt)^{2} + (dz/dt)^{2} = 25sec^{2}(t)

⇒ \(\sqrt{(dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2}dt\) = √25sec^{2}(t)

= √(5sec(t))^{2}

= 5sec(t)

S = \(\int_{0}^{π/4} 5sec(t)\,dt\)

∫sec(t)= ln[sect + tant]

S = 5ln(sect + tant)]\(_{0}^{π/4}\)

= 5[ln(sec(π/4) + tan(π/4))] - 5 [ln(sec(0) + tan(0))]

= 5[ln(√2 + 1) - ln(1 + 0)]

= 5[ln (√2 +1 )] - 0

S = 5[ln(√2 + 1)]

## Find the length of the curve. r(t) = cos(5t) i + sin(5t) j + 5 ln(cos(t)) k, 0 ≤ t ≤ π/4.

**Summary:**

The length of the curve. r(t) = cos(5t) i + sin(5t) j + 5 ln(cos (t)) k, 0 ≤ t ≤ π / 4 is 5[ln(√2 + 1)]

visual curriculum