Find the length of the curve C

[chwala]

Homework Statement

see attached

Relevant Equations

Integration

This question is from a Further Maths paper;

Part (a) is pretty straight forward...No issue here...one has to use chain rule...

Let $U=\dfrac{e^x+1}{e^x-1}$ to realize $\dfrac{du}{dx}=\dfrac{-2e^x}{(e^x-1)^2}$

and let
$y=\ln u$ on taking derivatives, we shall have $\dfrac{dy}{du}=\dfrac{e^x-1}{e^x+1}$

therefore,
$\dfrac{dy}{dx}=\dfrac{e^x-1}{e^x+1}×\dfrac{-2e^x}{(e^x-1)^2}=-\dfrac{2e^x}{(e^{2x}-1)^2}$Now my question (reason for posting this problem is on part b). Why do we have ± on the highlighted...i thought we are taking absolute value of $\dfrac{dy}{dx}$ which would be a positive.

i.e Arc length = $$\int_a^b \sqrt{1+\left[f^{'}(x)\right]^2} dx$$

Thanks.

 

[chwala]

Homework Statement:: see attached
Relevant Equations:: Integration

This question is from a Further Maths paper;

View attachment 315769

Part (a) is pretty straight forward...No issue here...one has to use chain rule...

Let $U=\dfrac{e^x+1}{e^x-1}$ to realize $\dfrac{du}{dx}=\dfrac{-2e^x}{(e^x-1)^2}$

and let
$y=\ln u$ on taking derivatives, we shall have $\dfrac{dy}{du}=\dfrac{e^x-1}{e^x+1}$

therefore,
$\dfrac{dy}{dx}=\dfrac{e^x-1}{e^x+1}×\dfrac{-2e^x}{(e^x-1)^2}=\dfrac{-2e^x}{(e^{2x}-1)^2}$Now my question (reason for posting this problem is on part b). Why do we have ± on the highlighted...i thought we are taking absolute value of $\dfrac{dy}{dx}$ which would be a positive.

i.e Arc length = $$\int_a^b \sqrt{1+\left[f^{'}(x)\right]^2} dx$$

View attachment 315771Thanks.

Aaaaargh, I've seen the reason... ...the sign on the derivative may either be positive or negative ...in our problem the derivative is having a negative sign. i.e $\dfrac{dy}{dx}=-\dfrac{2e^x}{(e^{2x}-1)^2}$