Two poles of heights m and n are separated by a horizontal dis...

Question

Two poles of heights m and n are separated by a horizontal distance d. A rope is stretched from the top of one pole to the ground and then to the top of the other pole. Show that the configuration that requires the least amount of rope occurs when Θ₁ = Θ₂ (see figure).

Accepted Answer

Below there today we're going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. In a park, a zip line is to be installed from the top of a tower to the ground and then up to the top of a hill for an adventure activity. The tower is 60 m high. The hill is 40 m high, and the horizontal distance between them is 100 m. To minimize the length of the cable used, what should be the angle between the cable and the horizontal at the point where the cable touches the ground? Report your final answer in units of degrees rounded to two decimal places. Fantastic. So ultimately our final answer that we're trying to solve for our angle is we're trying to figure out what should be the angle between the cable and the horizontal at the point at which this cable touches the ground in order to minimize the length of cord. That is used to create this zip line. So now that we know that we're trying to figure out what the angle between the cable and the horizontal at the point of where the cable touches the ground is, let us read off our multiple choice answers noting once again that they're all in the same units of degrees. So A is 38.66, B is 48.66, C is 40.21. And D is 45.86. Fantastic. So first off, let us recall and note that in order to minimize the amount of cable used for the zip line, we will need to figure out at what angle at which the cable should touch the ground. And this problem, as we should recognize, will involve the optimizing of the total cable length, which will consist of two parts to figure this out. So, firstly, we need to note that the cable from the top of the tower to the point at which it touches the ground, and secondly, the cable from the point where it touches the ground to the top of the hill. Let us also note that we are told that the tower is 60 m high. So we're told that the tower is 60 m high, and we're also told that the hill is 40 m high. And let us note that the horizontal distance between the tower and the hill is also 100 m. Fantastic. So, with that said, our goal now is we want to maximize or rather, our goal is we want to minimize the length of the zip line, which involves determining the optimal angle theta between the cable and the horizontal at the point where the cable touches the ground. So once again, that is our goal. So in order to accomplish that, we must first define the geometry that we will need to use in order to help us solve this particular problem. Let us state that X is to be the horizontal distance from the tower to a point where the cable touches the ground, so the horizontal distance from the point where the cable touches the ground to the top of the hill will be, and let's write this down, 100 minus X. And with that said, we now need to note that the vertical distance from the top of the tower to the ground will be 60 m, and the vertical distance from the point where the cable touches the ground to the top of the hill, this will be 60 minus 40, which will equal 20 m. Fantastic. So now that we have all these key pieces of information, let's create a diagram to help us better visualize this problem. So when you go ahead and create a diagram, you will notice very quickly that we are dealing with triangles. So as you can see, we have our tower, which is taller than our hill, since our tower is 160 m in height, whereas our hill is in 40 m of 40 m in height, and the hypotenuse of our tower triangle is going to be L1, whereas the hypotenuse of our hill triangle is gonna be L2. And the distance between our tower and hill is 100 m, and it's a distance X to where these two meet the two triangles touch in the center. So from our tower to where the edge of our hypotenuse on the hill triangle is, will be Value X, whereas focusing on our hill going to where the tower triangle touches the hill triangle is going to be 100 minus X. And then, of course, we have some angle theta that looks up towards the height, the top of our tower's height. So with that said, our next step now is we need to determine the length of the cable so we can express the lengths of the two segments of the cable. So firstly, we can say that the cable from the top of the tower to the point where the cable touches the ground, we could write this expression as L1. Is going to be equal to the square root of X2 + 60 squad. Fantastic. And secondly, the cable from the point where the cable touches the ground to the top of the hill, we can call this L2 and L2 is equal to the square root of 100 minus X to the power of 2 plus 20 of 2. Fantastic. So, with that said, our next step now is we need to note that therefore, the total length, which we'll call L of X, the total length L of X of the cable is going to be equal to L1 plus L2, which will mean that L of X is going to be equal to the square root of X2 + 602 plus the square root of 100 minus X2 + 20 of 2. So now we need to minimize the length of the cable. That is our next step. So now that we're minimizing the length of the cable, we need to recall a note that in order to so to minimize, so in order to minimize this total cable length, we need to differentiate L of X with respect to X and find the value of X such that it minimizes this particular length. So when we take the derivative of L of X, we will find that D of, so we need to take D of, so D L X divided by DX. So DL of X by DX is equal to X divided by the square root of X2 + 60 to the power of 2. Plus 100 minus x divided by the square root of 100 minus X2 + 20 squad fantastic. So now we need to set the derivative equal to 0 in order to find the critical points. So we need to take. X divided by the square root of X2 + 60 square. is equal to 100 minus X divided by the square root. Of 100 minus X2 + 202. And when we rearrange this to isolate and solve for X all by itself, we will find that X is equal to 75. So now we need to verify the minimum. So in order to verify the minimum, we need to confirm that this critical point is a minimum. In order to do that, we need to check the second derivative of D squad of L of X by D of X2. So this will be equal to 3600 divided by X2. Plus 3600. Multiplied by the square root of X2. Plus 3600. Fantastic. And this will be subtracted from Drum roll. 400 divided by the square, so so it's gonna be, so 400 divided by X2 minus 200. X + 10,400, so it's going to be 10,400. So, it's gonna be X2 minus 200 X plus 10,400, multiplied by and I'm running out of room, so I'll draw a little arrow, and this is going to be multiplied by the square root of X2. This will be multiplied by X2 minus 200 X plus 10,400. Fantastic. So, With that said, we can now safely say, without a doubt that since D2d of L of X by DX2 is greater than 0, and this is 4. X is greater than 0. So we could say that the function is going to be concave up as a result, which I'll denote this as when I draw a little arrow, concave up, which I'll call CU and I'll draw a little arrow so we know that it's pointing up, so it's concave up. So, this confirms that X equals 75 is a minimum, so I'll give it a red check mark since it is correct. It it confirms that X equals 75 is indeed. A minimum value. So with that said, our next step now is we need to find the angle theta. So we need to recall and note that at the point where the cable touches the ground, the vertical distance will be 60 m, and 60 m once again is the height of the tower. And the horizontal distance is X, and the angle theta is given by the tangent of the angle, which will be the ratio of the vertical distance to the horizontal distance, meaning we can go ahead and write that tangent. Of theta is equal to 60 divided by X. So therefore, when we rearrange this expression to isolate and sulfur theta all by itself, we will find that theta is equal to inverse tangent of 60 divided by X, which will mean that theta is equal to inverse tangent of 60 divided by 75, since we know that X is equal to 75, so when we take Inverse tangent of 60 divided by 75, and we round to approximately two decimal places like we're told by the problem itself, we will find when making sure our calculator this is very important. You need to make sure your calculator is in degrees mode because if it's not in degrees mode, you will not get this answer. So you will get, when you plug this expression into your calculator, making sure it's in degrees mode, you will get that it's approximately equal to 38.66 degrees, and that is our final answer. Hooray, we did it. So looking at our multiple choice answers, the correct answer without a doubt has to be multiple choice answer letter A, which once again is 38.66 degrees. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Two poles of heights m and n are separated by a horizontal distance d. A rope is stretched from the top of one pole to the ground and then to the top of the other pole. Show that the configuration that requires the least amount of rope occurs when Θ₁ = Θ₂ (see figure). <IMAGE>

Key Concepts

Optimization

Trigonometric Functions

Symmetry in Geometry

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