A suspension bridge has been constructed over a river. The suspension cable is parabolic in nature. The distance between the two towers holding the cable is 180 metres and the minimum height of the cable above the road is 15 metres. At a point 40 metres from the vertex of the cable the height above the road is 30 metres.

How high above the road are the cables attached up the tower?

Hello again nada! Here is a graphical interpretation of the problem:

The general equation for a parabola is y = ax^2 + bx + c. So with this general equation we need to find out what the equation for the suspension bridge is. To do that we need to find the values of the constants a,b, and c.

Note that with the given information we can say for sure what three points on the graph (the bridge) will be: (50,30), (130, 30), and (90, 15).

So plug this information into the general equation to get three simultaneous equation that you’ll want to solve:

\begin{cases} 2500a + 50b + c = 30 \\ 16900a + 130b + c = 30 \\ 8100a + 90b + c = 15 \end{cases}\tag*{}

I’ll leave it to you to solve this, but we get that a = \dfrac{3}{320}, b = -\dfrac{27}{16}, and c = \dfrac{1455}{16}.

So y = \dfrac{3}{320}x^2 - \dfrac{27}{16}x + \dfrac{1455}{16} is our equation.

So now we want to find the height of the towers. To do this we’ll just calculate the y-intercept of our equation. Set x=0 and solve!

y(0) = \dfrac{1455}{16} = 90.9 ~~\text{meters}\tag*{}.

I hope this helps! Also, in the future, it’d be more helpful if you try to solve the problem first, so I can identify what more specifically you need help with!

Thank you so much. I tried solving the question but I ended up with an answer in the thousands. Talk to you tomorrow!

Also when you did the 3 different equations, how did you solve them? I just tried and keep getting variables since there are 3.

That’s a good question! Since it’s definitely a messy system of equations, I have a question for you. Are you allowed to use a calculator for these problems, because I used a calculator to solve the system.

Yes I am, allowed to use a caluclator.

Great, using a calculator makes solving this system of equations easier.

\begin{cases} 2500a + 50b + c = 30 ~~~~~~~~(1) \\ 16900a + 130b + c = 30 ~~~~(2) \\ 8100a + 90b + c = 15 ~~~~~~~~(3) \end{cases}\tag*{}

Do **(2) - (1)** to get the equation 14400a + 80b = 0, and do **(2) - (3)** to get 8800a + 40b = 15. Now, multiply the second of these equations by 2 to get 17600a + 80b = 30, and then subtract one from the other to eliminate b and (finally) get 3200a = 30 \implies a = \dfrac{3}{320}. Now plug this in back into the first of the previous equations to get that 14400a + 80b = 0 \implies b = -\frac{14400(3/320)}{80} = -\frac{27}{16}. Now that we have both a and b, we can plug this information into either **(1)**, **(2)**, or **(3)**, to solve for c: c = 30 - 2500(\dfrac{3}{320}) - 50(-\dfrac{26}{17}) = \dfrac{1455}{16}.

Thank you so much for helping me out. This helped me understand the question and answer so much.

This topic was automatically closed 24 hours after the last reply. New replies are no longer allowed.