Find the length of the curve over the given interval:
x=t+1
y=ln cos(t)
for t=0 ---> t=pi/4
To find the length of the curve defined by the parametric equations x = t + 1 and y = ln(cos(t)) over the interval t = 0 to t = π/4, we need to use the arc length formula:
L = ∫√[(dx/dt)² + (dy/dt)²] dt
Step 1: Compute dx/dt and dy/dt
Given x = t + 1, take the derivative with respect to t:
dx/dt = d/dt(t + 1) = 1
Given y = ln(cos(t)), take the derivative with respect to t using the chain rule:
dy/dt = d/dt(ln(cos(t))) = -sin(t)/cos(t) = -tan(t)
Step 2: Calculate √[(dx/dt)² + (dy/dt)²]
Now, substitute the values of dx/dt and dy/dt into the equation:
√[(dx/dt)² + (dy/dt)²] = √[1² + (-tan(t))²] = √[1 + tan²(t)] = √(sec²(t))
Step 3: Evaluate the integral
Now we can rewrite the length integral as:
L = ∫√(sec²(t)) dt
= ∫sec(t) dt
This is now a standard integral. Integrating secant (sec(t)) produces the natural logarithm of the absolute value of the secant of t plus the tangent of t:
L = ln|sec(t) + tan(t)| + C
Step 4: Evaluate the integral within the given interval
Substitute the limits of integration (t = 0 to t = π/4) into the formula:
L = ln|sec(π/4) + tan(π/4)| - ln|sec(0) + tan(0)|
Secant and tangent values for π/4 and 0 can be found from trigonometric identities:
sec(π/4) = √2 and tan(π/4) = 1
sec(0) = 1 and tan(0) = 0
Substituting these values:
L = ln|√2 + 1| - ln|1 + 0|
= ln(√2 + 1) - ln(1)
= ln(√2 + 1)
Therefore, the length of the curve over the interval t = 0 to t = π/4 is ln(√2 + 1).