Find the length of the curve over the given interval:
x=t+1
y=ln cos(t)
for t=0 ---> t=pi/4
Oh, you want to find the length of the curve! Well, it's a good thing I brought my measuring tape! Let's get started.
First, we need to find the derivative of x with respect to t. That will give us dx/dt, which in this case is just 1. Easy peasy!
Next, we need to find the derivative of y with respect to t. That'll give us dy/dt. Applying the chain rule, we get:
dy/dt = d/dt[ln(cos(t))] = -tan(t)
Great! Now we have both dx/dt and dy/dt. To find the length of the curve, we need to calculate the integral from t=0 to t=pi/4 of the square root of (dx/dt)^2 + (dy/dt)^2. I'll call that S for "super length".
S = ∫[0,pi/4]√[(dx/dt)^2 + (dy/dt)^2] dt
But since dx/dt = 1 and dy/dt = -tan(t), the integral simplifies to:
S = ∫[0,pi/4]√[1 + tan^2(t)] dt
Hmm, this looks like a job for a trigonometric identity. Recall that tan^2(t) + 1 = sec^2(t). So we can rewrite our integral as:
S = ∫[0,pi/4]√[sec^2(t)] dt
And since √[sec^2(t)] = sec(t), our integral becomes:
S = ∫[0,pi/4]sec(t) dt
Now, I must have left my trigonometric integration cape at home, so I'm going to have to look up the integral of sec(t). Oh, look at that! It's ln|sec(t) + tan(t)|. Fancy that!
Using this formula, we can find:
S = ln|sec(pi/4) + tan(pi/4)| - ln|sec(0) + tan(0)|
Now, sec(pi/4) = √2 and tan(pi/4) = 1. Also, sec(0) = 1 and tan(0) = 0. Plugging in these values, we get:
S = ln|√2 + 1| - ln(1 + 0)
But wait, ln(1) is just 0, so we can simplify further:
S = ln|√2 + 1|
And there you have it! The length of the curve is ln|√2 + 1|. Voilà!
(Note: Don't take my words too seriously! It's always best to check your work and consult your textbooks or teachers for the correct methods and solutions.)