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 a curve given by parametric equations, we can use a technique called arc length integration.
The formula for arc length, L, over an interval [t1, t2] is given by the integral of the square root of the sum of the squares of the derivatives of x(t) and y(t) with respect to t, integrated with respect to t.
L = ∫[t1, t2] √(dx/dt)^2 + (dy/dt)^2 dt
In this case, we have the parametric equations:
x = t + 1
y = ln(cos(t))
First, we need to find the derivatives dx/dt and dy/dt.
Taking the derivative of x(t):
dx/dt = d/dt(t + 1) = 1
Taking the derivative of y(t):
dy/dt = d/dt(ln(cos(t)))
To find this derivative, we need to use the chain rule. The derivative of ln(u), where u is a function of t, is given by (1/u) * du/dt.
In this case, u = cos(t).
dy/dt = (1/cos(t)) * d(cos(t))/dt
dy/dt = (1/cos(t)) * (-sin(t))
Now, we can substitute these derivatives into the arc length formula and integrate over the given interval [0, π/4].
L = ∫[0, π/4] √(1)^2 + ((-sin(t))/(cos(t)))^2 dt
Simplifying the expression under the square root:
L = ∫[0, π/4] √(1 + sin^2(t)/cos^2(t)) dt
Recall the trigonometric identity: sin^2(t) + cos^2(t) = 1.
Using this identity, we can rewrite the expression under the square root:
L = ∫[0, π/4] √(cos^2(t)/cos^2(t) + sin^2(t)/cos^2(t)) dt
L = ∫[0, π/4] √(1/cos^2(t)) dt
Since the integral of (√(1/cos^2(t))) is equal to ∫sec(t) dt, we can rewrite L as:
L = ∫[0, π/4] sec(t) dt
Now, you can find the antiderivative of sec(t) and evaluate the integral to find the length of the curve over the given interval [0, π/4].