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help's question at Yahoo! Answers regarding quadratic modeling

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MarkFL

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Feb 24, 2012
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Here is the question:

Pre-calculus, parabola problem?


A cable of a suspension bridge is attached to two pillars of height 55 m and has the shape of a parabola whose span is 40 m. the roadway is 5 m below the lowest point of the cable. If an extra support is to be placed where the cable is 30 m above the ground level, find the distance from the nearest pillar where the support is to be placed.
I have posted a link there to this topic so the OP can see my work.
 
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MarkFL

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Feb 24, 2012
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Hello help,

Let's orient our coordinate axes such that the origin is at the point 5 m below the vertex, or lowest point, of the cable. We know the axis of symmetry will be the $y$ axis, and we know the cable passes through the point $(0,5)$, and so we may write the function $c(x)$ that models the cable with:

\(\displaystyle c(x)=ax^2+5\)

Now, because the span between two pilalrs is 40 m., we therefore also know the cable passes through the point $(20,55)$, and so with this point, we may determine the parameter $a$:

\(\displaystyle c(20)=a(20)^2+5=400a+5=55\)

\(\displaystyle 400a=50\)

\(\displaystyle a=\frac{1}{8}\)

Hence:

\(\displaystyle c(x)=\frac{1}{8}x^2+5\)

Now, we want to find the $x$ value(s) such that $c(x)=30$:

\(\displaystyle 30=\frac{1}{8}x^2+5\)

\(\displaystyle x^2=200\)

We need only use the positive root by symmetry:

\(\displaystyle x=\sqrt{200}=10\sqrt{2}\)

Now, to find the distance from this value for $x$ and $20$, we simply subtract the $x$-value from 20 to get:

\(\displaystyle d=20-10\sqrt{2}=10\left(2-\sqrt{2} \right)\approx5.85786437626905\)

Here is a diagram of the parabolic function modeling the cable, and the added pillar along with the preexisting pillar:

help.jpg