Expressing Derivative of Cable Tension Component

I am looking at Hibbeler's section on a `Cable Subjected to a Distributed Load' in Engineering Mechanics - Statics, Ed13, as shown in the attached images.

I am getting stuck deriving equation (7-7). I derive the horizontal equilibrium equation just fine;

$$|\vec{T} + \vec{\Delta T}| \cdot \cos(\theta + \Delta \theta) - |\vec{T}|\cdot \cos(\theta) = 0$$

I rearrange to describe horizontal tension component as a function of $\Delta x$ (given $\Delta T$ and $\Delta \theta$ are ultimately functions of $\Delta x$);

$$|\vec{T}|\cdot \cos(\theta) = |\vec{T} + \vec{\Delta T}|\cdot \cos(\theta + \Delta \theta)$$

Then I express the differential;

$$\frac{\Delta(|\vec{T}|\cdot \cos(\theta))}{\Delta x} = \frac{|\vec{T} + \vec{\Delta T}|\cdot \cos(\theta + \Delta \theta)}{\Delta x}$$

Then when I go to express the derivative, I come up with a limit which does not exist.

$$\frac{d(|\vec{T}|\cdot \cos(\theta))}{dx} = \lim_{\Delta x \to 0}\frac{|\vec{T} + \vec{\Delta T}|\cdot \cos(\theta + \Delta \theta)}{\Delta x}$$

Further to this, it is not obvious to me how the resultant force from the distributed load acting over $\Delta x$ can be defined as $w(x)\Delta x$ (in line 8) as the author suggests. I feel it should be defined as;

$$\Delta F = \int_{x}^{x+\Delta x}w(x)dx$$

becuase the loading intensity function is not constant. Any pointers as to where my errors are? Appreciate any advice offered. Thanks.

• Your last doubt is a classic assumption in calculus. For an infinitesimal $\Delta x$ you can adopt that $w(x) = w(x+\Delta x)$ (that is, that the value of the load is constant along $\Delta x$). – Wasabi Apr 23 '17 at 6:06
• @Wasabi Ok, I didn't realise $\Delta x$ was intended as infinitesimal - which is why $dx$ is only formed after $\Delta x \to 0$. I thought $\Delta x$ was arbitrary until it was allowed to approach $0$. Am I incorrect? – James Izzard Apr 23 '17 at 6:14

$$-T\cos \theta + (T+\Delta T) \cos( \theta + \Delta \theta)=0.$$
It is important to note that $T$ and $\theta$ are both depending on $x$. Dividing the previous expression by $\Delta x$ and rewriting $T+\Delta T = T(x+\Delta x)$ and $\theta + \Delta \theta = \theta(x+\Delta x)$, will lead to:
$$\frac{T(x+\Delta x)\cos [\theta(x+\Delta x)]-T(x)\cos[\theta(x+\Delta x)]}{\Delta x}=0.$$
Now, take the limit $\Delta x \to 0$. As the above is the definition of the derivative of $T(x)\cos[\theta(x)]$ we obtain:
$$\dfrac{d}{d x}\left[T(x)\cos[\theta (x)]\right]=0 \implies T(x)\cos[\theta (x)] = \text{const.}$$