# A cable hangs between two poles of equal height and 35 feet apart. At a point on the ground...

## Question:

A cable hangs between two poles of equal height and {eq}35 {/eq} feet apart. At a point on the ground directly under the cable and {eq}x {/eq} feet from the point on the ground halfway between the poles the height of the cable in feet is {eq}h(x)=10+(0.5)x^{1.5} {/eq}. The cable weighs {eq}18.3 {/eq} pounds per linear foot. Find the weight of the cable.

## Arc Length:

To find the length of the curve {eq}f(x) {/eq} over the interval {eq}[a,b] {/eq}, we can use the following arc length integral formula:

{eq}\displaystyle \int_{a}^{b} \sqrt{1 +(f'(x))^2)} dx {/eq}

Here {eq}f'(x) {/eq} is the first derivative of {eq}f(x) {/eq}. Once we have our definite integral set up, we can evaluate it using standard integration techniques, such as u-substitution.

Given Data:

• The distance between the two poles is: {eq}d = 35\;{\rm{ft}} {/eq}
• The height of the cable is: {eq}h\left( x \right) = 10 + 0.5\left( {{x^{1.5}}} \right) {/eq}
• The cable weight per unit feet is: {eq}{w_o} = 18.3\;{\rm{lb/ft}} {/eq}

Differentiate the height of cable equation with respect to x:

{eq}\begin{align*} \dfrac{{d\left( {h\left( x \right)} \right)}}{dx} &= \dfrac{{d\left( {10 + 0.5\left( {{x^{1.5}}} \right)} \right)}}{{dx}}\\ &= 0 + 0.5\left( {1.5} \right)\left( {{x^{0.5}}} \right)\\ &= 0.75\left( {{x^{0.5}}} \right) \end{align*} {/eq}

The expression for length of the cable is

{eq}l =2\times \int_0^{\frac{d}{2}} {\sqrt {1 + {{\left( {\dfrac{{d\left( {h\left( x \right)} \right)}}{{dx}}} \right)}^2}} dx} {/eq}

This is because the problem defines x as the distance from a point halfway between the cables,

so our upper limit of integration is the full distance between the cables divided by two,

but we multiply our entire arc length integral by two to account for the full cable length.

Substituting our values into the arc length integral, we have

{eq}\begin{align*} l &=2\times \int_0^{\dfrac{{35}}{2}} {\sqrt {1 + {{\left( {0.75\left( {{x^{0.5}}} \right)} \right)}^2}} dx} \\ &= 2\times\int_0^{17.5} {\sqrt {1 + 0.5625\left( x \right)} dx} \cdots\cdots\rm{(I)} \end{align*} {/eq}

Let {eq}1 + 0.5625x = u \cdots\cdots\rm{(II)} {/eq}

Differentiate the above expression

{eq}\begin{align*} \dfrac{{d\left( {1 + 0.5625x} \right)}}{{dx}} &= \dfrac{{du}}{{dx}}\\ 0.5625 &= \dfrac{{du}}{{dx}}\\ dx &= \dfrac{{du}}{{0.5625}} \end{align*} {/eq}

Substitute the value in expression (II) at {eq}x = 0 {/eq}

{eq}\begin{align*} 1 + 0.5625\left( 0 \right) &= u\\ u &= 1 \end{align*} {/eq}

Substitute the value in expression (II) at {eq}x = 17.5 {/eq}

{eq}\begin{align*} 1 + 0.5625\left( {17.5} \right) &= u\\ u &= 10.84375 \end{align*} {/eq}

The limits of the new integration are

{eq}1 \le u \le 10.84375 {/eq}

Substitute the value in expression (I)

{eq}\begin{align*} l &= 2\times\int_1^{10.84375} {\sqrt u \dfrac{{du}}{{0.5625}}} \\ &= 2\times\dfrac{1}{{0.5625}}\left( {\dfrac{2}{3}} \right)\left[ {{{\left( u \right)}^{\dfrac{3}{2}}}} \right]_1^{10.84375}\\ &= 2\times\dfrac{2}{{1.6875}}\left[ {{{\left( {10.84375} \right)}^{\dfrac{3}{2}}} - 1} \right]\\ &= 82.27\;{\rm{ft}} \end{align*} {/eq}

The expression for weight of the cable is

{eq}{W} = {w_o}l {/eq}

Applying our values, we find that

{eq}\begin{align*} {W_o} &= 18.3\;{\rm{lb/ft}} \times 82.27\;{\rm{ft}}\\ &= {\rm{1505}}{\rm{.541}}\;{\rm{lb}} \end{align*} {/eq}

Thus, the weight of the cable is {eq}{\rm{1505}}{\rm{.541}}\;{\rm{lb}} {/eq} 