Thread: Parabolic Arch Bridge

1. Parabolic Arch Bridge

A bridge is to be built in the shape of a parabolic arch and is to have a span of 100 feet. The height of the arch, a distance of 40 feet from the center, is to be 10 feet. Find the height of the arch at its center.

2. Originally Posted by magentarita
A bridge is to be built in the shape of a parabolic arch and is to have a span of 100 feet. The height of the arch, a distance of 40 feet from the center, is to be 10 feet. Find the height of the arch at its center.
Draw the parabola with the y-axis as axis of symmetry and the bed of the road will be the x-axis. See diagram.

The points (-50, 0) and (50, 0) lie on the parabola on the x-axis since the span is 100.

The points (-40, 10) and (40, 10) also lie on the parabola.

We use $\displaystyle (x-h)^2=4p(y-k)$ for our equation of a parabola with vertex (h, k) since the axis if symmetry is vertical. We know our vertex is at (0, k). We need to find k.

Substituting point (50, 0) into this equation, we get:

$\displaystyle (50-0)^2=4p(0-k)$

$\displaystyle \boxed{2500=0p-4pk}$

Substituting point (40, 10) into this equation, we get:

$\displaystyle (40-0)^2=4p(10-k)$

$\displaystyle \boxed{1600=40p-4pk}$

Use the two boxed equations to solve for p.

$\displaystyle 2500= \ \ 0p-4pk$
$\displaystyle 1500=40p-4pk$

Subtract the two equations to get:

$\displaystyle 900=-40p$

$\displaystyle p=-\frac{45}{2}$

Now, to find k, we substitute p back into one of our boxed equations.

$\displaystyle 2500=0\left(\frac{45}{2}\right)-4\left(-\frac{45}{2}\right)k$

$\displaystyle 2500=90k$

$\displaystyle k=\frac{350}{9} \approx 27.8$ feet.

3. ok....

Originally Posted by masters
Draw the parabola with the y-axis as axis of symmetry and the bed of the road will be the x-axis. See diagram.

The points (-50, 0) and (50, 0) line on the parabola on the x-axis since the span is 100.

The points (-40, 10) and (40, 10) also lie on the parabola.

We use $\displaystyle (x-h)^2=4p(y-k)$ for our equation of a parabola with vertex (h, k) since the axis if symmetry is vertical. We know our vertex is at (0, k). We need to find k.

Substituting point (50, 0) into this equation, we get:

$\displaystyle (50-0)^2=4p(0-k)$

$\displaystyle \boxed{2500=0p-4pk}$

Substituting point (40, 10) into this equation, we get:

$\displaystyle (40-0)^2=4p(10-k)$

$\displaystyle \boxed{1600=40p-4pk}$

Use the two boxed equations to solve for p.

$\displaystyle 2500= \ \ 0p-4pk$
$\displaystyle 1500=40p-4pk$

Subtract the two equations to get:

$\displaystyle 900=-40p$

$\displaystyle p=-\frac{45}{2}$

Now, to find k, we substitute p back into one of our boxed equations.

$\displaystyle 2500=0\left(\frac{45}{2}\right)-4\left(-\frac{45}{2}\right)k$

$\displaystyle 2500=90k$

$\displaystyle k=\frac{350}{9} \approx 27.8$ feet.
What a great reply.

4. Originally Posted by magentarita
What a great reply.
You are toooooo kind! Blush Blush

5. ok.....

Originally Posted by masters
Draw the parabola with the y-axis as axis of symmetry and the bed of the road will be the x-axis. See diagram.

The points (-50, 0) and (50, 0) lie on the parabola on the x-axis since the span is 100.

The points (-40, 10) and (40, 10) also lie on the parabola.

We use $\displaystyle (x-h)^2=4p(y-k)$ for our equation of a parabola with vertex (h, k) since the axis if symmetry is vertical. We know our vertex is at (0, k). We need to find k.

Substituting point (50, 0) into this equation, we get:

$\displaystyle (50-0)^2=4p(0-k)$

$\displaystyle \boxed{2500=0p-4pk}$

Substituting point (40, 10) into this equation, we get:

$\displaystyle (40-0)^2=4p(10-k)$

$\displaystyle \boxed{1600=40p-4pk}$

Use the two boxed equations to solve for p.

$\displaystyle 2500= \ \ 0p-4pk$
$\displaystyle 1500=40p-4pk$

Subtract the two equations to get:

$\displaystyle 900=-40p$

$\displaystyle p=-\frac{45}{2}$

Now, to find k, we substitute p back into one of our boxed equations.

$\displaystyle 2500=0\left(\frac{45}{2}\right)-4\left(-\frac{45}{2}\right)k$

$\displaystyle 2500=90k$

$\displaystyle k=\frac{350}{9} \approx 27.8$ feet.

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