Error in Catenary Derivation in Text Book?

Hi All!

I'm going over the derivation of the catenary presented in Hugh Neill's Teach Yourself Calculus, Ex. 21.3.2, p. 287 (See scanned pages below).

Could anyone confirm that the below line is correct, and if so how it works?

(P. 288, second image below)

It seems to be implying that T/δs = T, but shouldn't T/δs be closer to ∞?

---start---

(T + δT)sin(Ψ+δΨ)-TsinΨ=wδs

Expanding the left hand side and dividing by δs,

(T + δT/δs)sinΨcosδΨ+(T+δT)cosΨ(sinδΨ/δs)-TsinΨ=w

---end---

Thanks!

http://i1075.photobucket.com/albums/...psbe0bc869.jpg

http://i1075.photobucket.com/albums/...ps23b7a389.jpg

http://i1075.photobucket.com/albums/...ps3c1dca2a.jpg

Re: About Catenary Derivation in Teach Yourself Calculus, Hugh Neill, Ex. 21.3.2 p. 2

Note that there are frequent errors in this book.

If you agree that the above is strange please let me know!

Just not sure that it's an error because everything leading up to and following this seems to make sense...

Re: About Catenary Derivation in Teach Yourself Calculus, Hugh Neill, Ex. 21.3.2 p. 2

Yes, there is something wrong in the process:

$\displaystyle (T+\delta T)\sin{(\psi+\delta \psi)}-T\sin{\psi}=w\delta{s}$

and using the trig identity for $\displaystyle \sin{(a+b)}$:

$\displaystyle (T+\delta T)\sin{\psi}\cos{\delta \psi}+(T+\delta T)\cos{\psi}\sin{\delta \psi}-T\sin{\psi}=w\delta{s}$

which can be rearranged to:

$\displaystyle T(\sin{\psi}\cos{\delta \psi}-\sin{\psi})+\delta T\sin{\psi}\cos{\delta \psi}+(T+\delta T)\cos{\psi}\sin{\delta \psi}=w\delta{s}$

Dividing by $\displaystyle \delta{s}$:

$\displaystyle \frac{T}{\delta{s}}(\sin{\psi}\cos{\delta \psi}-\sin{\psi})+\frac{\delta T}{\delta{s}}\sin{\psi}\cos{\delta \psi}+(T+\delta T)\cos{\psi}\frac{\sin{\delta \psi}}{\delta{s}}=w$

Now let $\displaystyle \delta{s}$ approach 0. The first term is zero since $\displaystyle \cos{\delta \psi}$ approaches 1, and also $\displaystyle \cos{\delta \psi}$ drops out of the second term. In the third term $\displaystyle T+\delta T$ approaches T, and as the book says, $\displaystyle \frac{\sin{\delta \psi}}{\delta{s}}$ approaches $\displaystyle \frac{d\psi}{ds}$. The result is:

$\displaystyle \frac{dT}{ds}\sin{\psi}}+T\cos{\psi}\frac{d\psi}{d s}=w$

which is the equation at the top of page 289.

- Hollywood

Re: About Catenary Derivation in Teach Yourself Calculus, Hugh Neill, Ex. 21.3.2 p. 2

Ah, that's much better!

Thanks Hollywood!!