230k views
0 votes
Find the length of the curve. r(t) = cos(2t) i + sin(2t) j + 2 ln(cos(t)) k, 0 ≤ t ≤ π/4

2 Answers

4 votes

Final answer:

To find the length of the curve, you need to use the arc length formula and find the derivatives of the given function. Integrate the resulting expression over the given interval to find the length.

Step-by-step explanation:

To find the length of the curve r(t) = cos(2t) i + sin(2t) j + 2 ln(cos(t)) k, 0 ≤ t ≤ π/4, we can use the arc length formula:

L = ∫√(dx/dt)² + (dy/dt)² + (dz/dt)² dt

where:

dx/dt, dy/dt, and dz/dt are the derivatives of the components of r(t) with respect to t

Let's differentiate r(t) to find dx/dt, dy/dt, and dz/dt:

dx/dt = -2sin(2t)

dy/dt = 2cos(2t)

dz/dt = -2sin(t)/cos(t)

Now, we can substitute dx/dt, dy/dt, and dz/dt into the arc length formula and evaluate the integral:

L = ∫√(-2sin(2t))² + (2cos(2t))² + (-2sin(t)/cos(t))² dt

L = ∫√12sin²(2t) + 4cos²(2t) + 4sin²(t)/cos²(t) dt

L = ∫√4(sin²(2t) + cos²(2t) + sin²(t)/cos²(t)) dt

L = ∫√4 dt

L = 2t

Evaluating the integral from 0 to π/4, we get:

L = 2(π/4) - 2(0)

L = π/2

Therefore, the length of the curve is π/2.

User Oisdk
by
8.2k points
4 votes

dx/dt = -2sin2t: dy/dt = 2cos2t: dz/dt = -2sint/cost
\int\limits^ ( \pi )/(4) _0 { √((dx/dt)^2+(dy/dt)^2+(dz/dt)^2) } \, dt
integral (0 to \pi /4){\sqrt{4sin^22t+4cos^22t+4[tex]int ( √(4 + 4tan^22t))dt
\int\ {sect} \, dt ::from::t= 0 to \pi/4
ln(sect+tant) from t = 0 to pi/4
ln(sec \pi +tan \pi )
Length = -1sin^2t/cos^2t}} dt [/tex]
User TechRemarker
by
7.0k points