word problem hurting my brain

[strugglingwithAlg]

A model of a bridge is built. In one section of the suspension bridge, cable runs from the top of one tower down to the roadway, touching the roadway, and up again to the top of a second tower. Both towers are 12.25 inches tall and stand 70 inches apart. At some point along the road from the lowest point of the cable, the cable is 1.96 inches above the roadway. Find the distance between that point and the base of the nearest tower.


I don't even know where to start with this one. Thanks.

 

[stapel]

Are you perhaps studying conics? Are you perhaps supposed to assume that the cable is hanging in the shape of a parabola?

If not, then I don't see how you're supposed to do this.

Eliz.

 

[galactus]

This problem isn't as bad as it may appear. You can use the standard parabola

equation: \(\displaystyle y=a(x-h)^{2}+k\).

Where h and k are the vertex coordinates. Use x=35 and y=12.25 in your equation.

Let the vertex lie at the origin. Solve for a. You will then have your equation.

It will be a rather short one. Set it equal to 1.96 and solve for x. Remember, that

will be the distance form the origin though.

 

[strugglingwithAlg]

stapel - I'm assuming that the cable is supposed to be in the shape of a parabola. The question is part of a beginning of semester review sheet assignment we got, I guess to see how much we know. I've never seen anything like this before.

galactus - where does the x=35 come from?

 

[soroban]

Hello, strugglingwithAlg!

There's something seriously wrong with the wording . . ,
\(\displaystyle \;\;\)Are you sure that's the way it's given?

A model of a bridge is built.
In one section of the suspension bridge,
the cable runs from the top of one tower down to the roadway,
touching the roadway, and up again to the top of a second tower.
Both towers are 12.25 inches tall and stand 70 inches apart.

At some point along the road from the lowest point of the cable,
the cable is 1.96 inches above the roadway.\(\displaystyle \;\) . . . We can't have it both ways!

Find the distance between that point and the base of the nearest tower. \(\displaystyle \;\;\) . . . which point?

If "that point" is the "point along the road", it is 35 inches from each tower
\(\displaystyle \;\;\)assuming the cable is hanging symmetrically.

If "that point" is "the lowest point of the cable", we have a right triangle.
The base is 35, the height is 1.96 . . . now find the hypotenuse.

 

[strugglingwithAlg]

oops, I did write some of it wrong.


cable runs from the top of one tower down to the roadway, just touching it there, and up again to the top of a second tower.

 

[galactus]

stapel - I'm assuming that the cable is supposed to be in the shape of a parabola. The question is part of a beginning of semester review sheet assignment we got, I guess to see how much we know. I've never seen anything like this before.

galactus - where does the x=35 come from?

The 35 comes from half of 70. The problem says, "at some point along the road from the lowest point of the cable". The lowest point will be the vertex at (0,0).

You have: \(\displaystyle 12.25=a35^{2}\); \(\displaystyle a=.01\)

The equation is \(\displaystyle y=\frac{1}{100}x^{2}\)

At some point it is 1.96 from the road:

\(\displaystyle \frac{1}{100}x^{2}=1.96\)

\(\displaystyle x^{2}=196\)

\(\displaystyle x=sqrt{196}=14\)

Therefore the point is 35-14=21 feet from the nearest tower.

Since we have symmetry, it will be the same on the other side.

 

[stapel]

There's something seriously wrong with the wording...Are you sure that's the way it's given?

I took the exercise to mean that one is to find the distance from the nearest tower to the point, between that tower and the center where the cable touches, at which the cable is 1.96 inches above the roadway.

Eliz.