# A cable hangs between two poles of equal height and 32 feet apart. At a point on the ground...

## Question:

A cable hangs between two poles of equal height and 32 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 + (0.3)(x^{1.5}).{/eq}

The cable weighs 10.9 pounds per linear foot.

Find the weight of the cable.

## Integration

Integration is used to determine the length of the curve, volume, area etc. depending on the condition. The integration along any axis or curve is evaluated to know the desired entity. The integration method is the summation of the finite or infinite small elemental entities.

## Answer and Explanation:

**Given Data**

- The length of the cable is {eq}L = 32\;{\rm{ft}} {/eq}

- The height of the cable is {eq}h\left( x \right) = 10 + \left( {0.3} \right){x^{1.5}} {/eq}

- The intensity of weight of the cable is {eq}w = 10.9\;{\rm{lb/ft}} {/eq}

Length of the curve is calculated by using the integration method.

The length of the cable is,

{eq}L = 2\int\limits_0^{\dfrac{L}{2}} {\sqrt {1 + {{\left[ {h'\left( x \right)} \right]}^2}} dx}\cdots\cdots\rm{(I)} {/eq}

Differentiate the height function to calculate the slope of the function as,

{eq}\begin{align*} h'\left( x \right) &= 0 + 1.5\left( {0.3} \right){x^{1.5 - 1}}\\ h'\left( x \right) &= 0.45{x^{0.5}} \end{align*} {/eq}

Substitute the known values in equation (I).

{eq}\begin{align*} L &= 2\int\limits_0^{16} {\sqrt {1 + {{\left[ {0.45{x^{0.5}}} \right]}^2}} dx} \\ &= 2\int\limits_0^{16} {{{\left( {1 + 0.2025x} \right)}^{\dfrac{1}{2}}}dx} \end{align*} {/eq}

Let us assume that {eq}1 + 0.2025x = t {/eq}.

Differentiate the above equation.

{eq}\begin{align*} 0.2025dx &= dt\\ dx &= \dfrac{{dt}}{{0.2025}} \end{align*} {/eq}

From the length equation,

{eq}\begin{align*} L &= 2\int\limits_0^{4.24} {{{\left( t \right)}^{\dfrac{1}{2}}}\dfrac{{dt}}{{0.2025}}} \\ &= \dfrac{2}{{0.2025}}\left( {\dfrac{2}{3}} \right)\left( {{{\left( t \right)}^{\dfrac{3}{2}}}} \right)_0^{4.24}\\ &= \dfrac{2}{{0.2025}}\left( {\dfrac{2}{3}} \right)\left( {{{\left( {4.24} \right)}^{\dfrac{3}{2}}}} \right)\\ &= 57.486\;{\rm{ft}} \end{align*} {/eq}

The weight of the cable is,

{eq}W = wL {/eq}

Substitute the values,

{eq}\begin{align*} W &= \left( {10.9\;{\rm{lb/ft}}} \right)\left( {57.486\;{\rm{ft}}} \right)\\ &= 626.597\;{\rm{lb}} \end{align*} {/eq}

Thus, the weight of the cable is {eq}626.597\;{\rm{lb}} {/eq}.

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