# 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.

$$\frac{d(|\vec{T}|\cdot \cos(\theta))}{dx} = \lim_{\Delta x \to 0}\frac{|\vec{T} + \vec{\Delta T}|\cdot \cos(\theta + \Delta \theta)}{\Delta x}$$ Any pointers as to where my error is? Appreciate any advice offered. Thanks.

• First of all use don't use vectors but scalars, as your book suggests. Then $T\cos\theta$ must be eliminated before diving by $\Delta x$.
– N74
Apr 22 '17 at 9:07
• Are you able to elaborate on this any further? I am taking the norm of the vector, which is a scalar. I don't see how I can eliminate $T\cos(\theta)$ before dividing by $\Delta x$? Apologies if I am missing something obvious. I tried to apply the sum angle identity but was unable to make anything cancel in the way you suggest. Apr 23 '17 at 5:24

First of all is better to use $T$ as a scalar and write: $$( {T} + {\Delta T})\cdot \cos(\theta + \Delta \theta) - {T}\cdot \cos(\theta) = 0$$ next expand the cosine: $$( {T} + {\Delta T})\cdot (\cos(\theta)\cos( \Delta \theta)- \sin(\theta)\sin( \Delta \theta) )- {T}\cdot \cos(\theta) = 0$$ and regroup: $${T} \cos(\theta)(\cos( \Delta \theta) -1) + {\Delta T}\cdot \cos(\theta)\cos( \Delta \theta)- (T+ \Delta T) (\sin(\theta)\sin( \Delta \theta) ) = 0$$ now we can divide and take the limit: $$\lim_{\Delta x \rightarrow 0} {T} \cos(\theta){(\cos( \Delta \theta) -1) \over \Delta \theta} {\Delta \theta \over \Delta x} + \lim_{\Delta x \rightarrow 0} {\Delta T \over \Delta x}\cdot \cos(\theta)\cos( \Delta \theta)- \lim_{\Delta x \rightarrow 0} (T+ \Delta T) \sin(\theta){\sin( \Delta \theta)\over \Delta \theta} {\Delta \theta \over \Delta x}= 0$$ I also introduced a couple of ${\Delta \theta \over \Delta \theta}$ where needed.
Now, remember that also $\Delta \theta$ and $\Delta T$ go to $0$, so the first limit is $0$, in the second $\cos \Delta \theta$ goes to $1$ and in the third we can neglect $\Delta T$, we remain with: $${d T \over d x}\cos \theta-T \sin \theta {d \theta \over d x}=0$$ that is the expression we were looking for.