Question

Find the length of the curve y = ln(5cosx) from 0 <= x <= pi/4 . I don't know where to start on this problem. I can't seem to find any formulas or strategies to use. Thanks!

Answer 1

\[y=\ln (5cosx)\] from \[0 \le x \le \frac{ \pi }{ 4 }\]

Answer 2

Do I need to make it an integral? \[\int\limits_{0}^{\frac{ \pi }{ 4 }} \ln(5cosx)\]

Answer 3

The length of a curve is given by:\[\bf L=\int\limits_{a}^{b}\sqrt{1+\left( \frac{ dy }{ dx } \right)^2} \ dx\]So find the derivative of the given function, square it and plug it in. Then plug in the upper/lower bounds for 'a' and 'b' and evaluate the integral. You might need to use a graphing calculator. @bigmatt

Answer 4

I will give it a try.

Answer 5

The derivative is \[-\frac{ 5sinx }{ 5cosx } \rightarrow -\frac{ sinx }{ cosx }\] which is: \[-tanx\]
then square it:
\[\tan^2x\]

Answer 6

Now I will integrate the equation.

Answer 7

\[\int\limits_{0}^{\pi/4} 1 + tanx dx\]
which is:
\[|x+\ln|secx|+C|_{0}^{\pi/4}\]

Answer 8

Yup. Now the stuff under the square root is:\[\bf 1+\tan^2(x)\]But note that:\[\bf 1+\tan^2(x)=\sec^2(x)\]Hence the square-root can be simplified as:\[\bf \int\limits_{0}^{\pi/4}\sec(x) \ dx\]

Answer 9

uh oh I did it wrong. :)

Answer 10

ahh the trig identity.

Answer 11

I think it's \[\ln (1+\sqrt{2})\]

Answer 12

@bigmatt Correct.

Answer 13

You are amazing! Thank you!

Answer 14

You can derive this formula yourself so it's important to know where it comes from. If you know where it comes from and are aware of the proof, than anytime you forget the formula, you can always remember it back. That's why I always expose myself to proofs and explanations for formulae and don't simply memorise or try to remember them. @bigmatt