Find the Length of the Curve: r(t) = cos(4t)i + sin(4t)j + 4ln(cos(t))k, where 0 ≤ t ≤ π/4

To find the length of the given curve, we first need to determine the parameterization of the curve and then use the arc length formula.

The arc length L of a curve defined by a vector function r(t) from a to b is given by the formula:

L = ∫_a^b ||r'(t)|| dt

where r'(t) is the derivative of r(t), and ||r'(t)|| is its magnitude.

Given the curve:

r(t) = cos(4t)i + sin(4t)j + 4ln(cos(t))k

We first compute the derivative:

r'(t) = d/dt [cos(4t)]i + d/dt [sin(4t)]j + d/dt [4ln(cos(t))]k

Using the chain rule:

r'(t) = -4sin(4t)i + 4cos(4t)j + 4[-sin(t)/cos(t)](1/cos(t))k

Which simplifies to:

r'(t) = -4sin(4t)i + 4cos(4t)j - 4tan(t)k

Next, we find the magnitude:

||r'(t)|| = sqrt((-4sin(4t))^2 + (4cos(4t))^2 + (-4tan(t))^2)

Calculating the squares:

||r'(t)|| = sqrt(16sin^2(4t) + 16cos^2(4t) + 16tan^2(t))

Using the identity sin^2(x) + cos^2(x) = 1 simplifies this to:

||r'(t)|| = 4sqrt(1 + tan^2(t))

Knowing that 1 + tan^2(t) = sec^2(t), we get:

||r'(t)|| = 4sec(t)

The arc length is now computed by integrating from 0 to π/4:

L = ∫_0^(π/4) 4sec(t) dt

The integral of sec(t) is ln|sec(t) + tan(t)| + C. Therefore:

L = 4[ln|sec(t) + tan(t)|]_0^(π/4)

Evaluating the limits:

At t = π/4, sec(π/4) = √2 and tan(π/4) = 1. Thus:

ln|√2 + 1|

At t = 0, sec(0) = 1 and tan(0) = 0. Thus:

ln|1| = 0

Therefore, substituting in these values gives:

L = 4[ln(√2 + 1) - 0] = 4ln(√2 + 1)

So the length of the curve from t = 0 to t = π/4 is:

L = 4ln(√2 + 1)