Yes, your solution is correct!

[Guest2]

Consider the segment of the curve $y = \cosh(ax)/a$ between $x = −l$ and $l$. Here $a$ and $l$ are positive constants. Find an explicit expression for the length of this curve segment in terms of $a$ and $l$, as well as its limit for $a \to 0$.

What I had done:

Using the formula $\displaystyle L = \int_a^b \sqrt{1+\bigg(\frac{dy}{dx}\bigg)^2}$

I get $\displaystyle L = \int_{-l}^l \sqrt{1+\left(\sinh{ax}\right)^2}\,{dx} = \int_{-l}^{l} \cosh(ax)\,{dx}=\sinh(al)-\sin(-al) = 2\sinh(al). $

$\displaystyle \lim_{a \to 0} 2\sinh(al) =2\sinh(0) = 0$

Could someone please confirm whether this is correct?

 

[Greg]

What do you get when you differentiate $\sinh(ax)$?

 

[Guest2]

What do you get when you differentiate $\sinh(ax)$?

I get $a\cosh(ax)$.

 

[Greg]

Right. So,

\(\displaystyle \int\cosh(ax)\,dx=\dfrac{\sinh(ax)}{a}+C\)

 

[Guest2]

Right. So,

\(\displaystyle \int\cosh(ax)\,dx=\dfrac{\sinh(ax)}{a}+C\)

Oh wow. I can't believe I got that wrong! (Giggle)

So I should have got $ \frac{2\sinh(al)}{a}$ and $2\lim_{a \to 0}\frac{\sinh(al)}{a} = 2l\lim_{a \to 0}{\cosh{a}} = 2l$

Is this correct?

 

[Greg]

Though your final answer is correct, there is an error. See if you can spot it.

 

[Guest2]

Though your final answer is correct, there is an error. See if you can spot it.

$\displaystyle 2\lim_{a \to 0}\frac{\sinh(al)}{a} = 2\lim_{a \to 0}\frac{\frac{d}{da}\sinh(al)}{\frac{d}{da} a} =2l\lim_{a \to 0}{\cosh{a}} = 2l$?