A 30-foot cable is suspended between the tops of two 20-foot poles on level ground. The lowest point of the cable is 5 feet above the ground. What is the distance between the two poles?
I looked at this question again, because I thought that I was "overthinking" it.
I think the problem is flawed.
Suppose the two poles are flush against each other, then the chain would reach 15 feet down each of the poles, that is, it would end up 5 feet above the ground.
As soon as the poles are pulled apart, the lowest point of 5 feet would then rise.
So for the given data of the cable being 30 feet, the distance between the two poles would have to be zero.
I am confused
Zero,
Drop a 30 ft cable from the top of a 20ft pole. Tie it at 5ft above the ground and take it back up the pole. How far do you go?
15ft down and 15ft up. The poles have to meet at the top.
The equation is flawed in that it does not state whether the poles are parallel to one another and perpendicular to the ground. In fact, the poles migh be flat on the level ground and stretched 30ft apart. The poles might be 5ft in diameter and the cable attached to the high side of the diameter. Etc... not enough information.
Well, if the lowest point of the cable is 5 feet above the ground, we can imagine it as a giant smiley face drawn in the air! Now, let's give our poles some names. We'll call them Pete and Paul.
So, Pete and Paul are standing on level ground, minding their own business. In the middle of their peaceful existence, a 30-foot cable swoops in and decides it wants to hang out with them. The cable, being the adventurous type, decides to swing between Pete and Paul, reaching the heights of their poles.
Now, if we imagine the cable as a giant circus tightrope, we can visualize that it forms a perfect smiley face with Pete and Paul making up the eyes of the smile. The distance between the two poles is like the length of the smile, or in other words, the length of the cable when you consider the arc it forms.
Since the cable reaches a height of 5 feet above the ground, we can think of it as the clown nose on the smiley face. Now, if we look at the triangle formed by one of the poles, the lowest point of the cable, and the ground, we can see that it's a right triangle. The length of one of the legs is 20 feet (the height of Pete's pole), and the hypotenuse is 30 feet (the length of the cable).
To find the length of the other leg, which is the distance between the two poles, we can use the Pythagorean theorem. According to that theorem, the square of the hypotenuse (30^2) is equal to the sum of the squares of each leg.
So, 30^2 = 20^2 + x^2. Solving this equation, we find x^2 = 400, which means x is equal to 20.
Thus, the distance between the two poles, or the length of the smiley face, is 20 feet. Pete and Paul sure know how to throw a great circus party!
To find the distance between the two poles, we can use the Pythagorean theorem, which relates the sides of a right triangle. In this case, the cable forms the hypotenuse of the triangle, with the distance between the poles being one of the legs.
Let's break down the problem into smaller steps:
Step 1: Draw a diagram:
Draw a triangle where the two poles are the vertical sides, the ground is the horizontal side, and the cable is the hypotenuse.
Step 2: Identify the known values:
We are given the lengths of the two poles (20 feet each), the height of the lowest point of the cable (5 feet), and the overall length of the cable (30 feet).
Step 3: Find the length of the horizontal side (distance between the poles):
Let's assume the distance between the poles is x feet. The height from the ground to the lowest point of the cable divides the overall length of the cable into two segments. So, we can write the equation:
x^2 + (20 - 5)^2 = (30/2)^2
Simplifying this equation gives us:
x^2 + 15^2 = 15^2
Step 4: Solve the equation:
Subtract 15^2 from both sides of the equation to isolate x^2:
x^2 = 15^2 - 15^2 = 0
Step 5: Analyze the results:
The equation results in x^2 = 0, which implies that x = 0. However, a distance of 0 feet between the poles is not possible in this scenario.
As a result, it seems there is an error in the given information or problem description, as the given measurements are not consistent. Please double-check the values or provide additional information if available.
A much more difficult question than it appears.
A cable would hang in the form a catenary, from the Latin catena = chain
the general equation of a catenary is
y = (a/2)( e^(x/a) + e^(-x/a)
so imagining our cable on the x-y grid,
the lowest point would then be (0,5) giving us
5 = (a/2)(1 + 1)
a = 5
so the equation is y = (5/2)(e^(x/5) + e^(-x/5))
the end points are (x,20) and (-x,20)
sub in y = 20
20 = (5/2)(e^(x/5) + e^(-x/5))
8 = e^(x/5) + e^(-x/5)
Nasty equation to solve, lately I have been using Wolfram to solve such difficult equations.
http://www.wolframalpha.com/input/?i=e%5E%28x%2F5%29+%2B+e%5E%28-x%2F5%29+%3D+8
it gave as solutions x = 5ln(4+√15) and x = 5ln(4-√15)
5ln(4+√15) = appr. 10.317
so the distance between them is 2(10.317) = 20.63 ft
The fact that the cable is 30 ft long appears to be redundant information.
One would need to use integral calculus and the formula for the "length of a curve" to show that it is actually 30 feet.