Not your typical curve problem
This is problem 38 in Section 4.9 of James Stewart’s Calculus, 5th edition, but you don’t need any calculus to solve it. Good trig knowledge, some algebra skills, and a capable graphing calculator or some other technological means will get you there. Note points A and B and what they would be called in a horizontal curve diagram and think about tangents and how they could be derived. Solving it gets you two plus marks in the Great Problem-Solving Book in the Sky.
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Here is one that has bothered me for years. It looks so simple but it’s not.
Two ladders, one 30 feet long and one 20 feet long are propped in the corners of an alley. From where the ladders cross to the ground is 10 feet. What is the width of the alley?
Well, there goes any constructive work this morning …
.Here is one that has bothered me for years. It looks so simple but it’s not.
An algebraic approach runs into a lot of complication, and I’ll have to work some more on it. Using a computer yields 12.3119 ft.
.some other technological means will get you there.
I can’t see how tangents help. I quickly derived the equation sin(theta/2) = 0.4 theta and calculated 2.2622 radians or ~ 130 degrees.
That comes from r*theta =5, and divide the triangle with a vertical line so you have r*sin(theta/2) = 2
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This is problem 38 in Section 4.9 of James Stewart’s Calculus, 5th edition, but you don’t need any calculus to solve it. Good trig knowledge, some algebra skills, and a capable graphing calculator or some other technological means will get you there. Note points A and B and what they would be called in a horizontal curve diagram and think about tangents and how they could be derived. Solving it gets you two plus marks in the Great Problem-Solving Book in the Sky.
Solving by Newton’s method I get a central angle of 129?ø36’53” (2.2622 radians) and a radius of 2.21.
Jacob Wall
2.2622 radians
129?ø36’53”
The alley problem is indeterminate, since the ladders are at a corner of an alley and no mention is made that they cross from the bottom of the left building with the other at the base of the right building. One must state that both ladders are across the alley, also assume that the building walls are perpendicular or that they even exist and that the alley is level. Solve my assumed problem.
Paul in PA
The alley problem is indeterminate
His diagram shows the perpendicularity and placements.
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Right. The tangents are irrelevant to the problem. I was just extending it to a more common format.
I saw at least 1 problem like that ladder question on each of the exams– LSIT, Fundamentals and Principles. They weren’t terribly hard but they’re deceptively time consuming, which was probably how they ended up on the exams in the first place.
The width is 12.311857 feet and the intersection is 7.8114542 feet from the left wall.
Those ladders must be of a very interesting design. Either they are infinitely thin or the top and bottom have been perfectly trimmed (at an angle) to match the building wall. Otherwise the top of the ladder side would hit the wall at ground level and the bottom of the ladder side would hit the wall on the opposite side of the alley. Thus, introducing reasoning to knock off a whole bunch of those numbers to the right of the decimal point.
A bit of overkill, and pretty useless, but I have a simple excuse. That’s what the calculator got! :silly:
If you make the distance from the wall equal to 7.81 feet and the width of the alley 12.3 feet, both numbers have 3 significant figures, and the height of the crossing point is 10.013653, which is 10.0 feet to 3 significant figures.
Actually, if you can justify accurate measurements to hundredths of a foot, then the 12.3 can be changed to 12.31 and the height of the intersection becomes 10.000567, or 10.00 to hundredths of a foot.
Either way, all those digits aren’t needed.
That doesn’t solve the shape problem, but it does add some reasonableness to the answer.
I assume that all those that have solved the problem used CAD to derive their answers. What I was REALLY looking for was the formulae by which the answer was derived.
First off you must assume that the ladders are 20.00′ and 30.00′ to have any precision. Throwing in a rail width of 0.33′ or more complicates the solution. Finally I have never seen a 20 or 30 foot ladder that did not have an extentsion, so the actually ladder is not even a straight line.
Snow on the ground here, what is everyone else doing inside?
Paul in PA
I assume that all those that have solved the problem used CAD to derive their answers. What I was REALLY looking for was the formulae by which the answer was derived.
I’ve worked a little with algebra and seem to get into 4th order polynomials. Haven’t gotten around to solving tor roots, but am still looking for a simplification. I was hoping to get a quadratic equation.
.The wording of the problem was to get the general idea across. The actual problem was the diagram. Therefore there is no need to invent practical difficulties that weren’t in the diagram.
.Here’s how I did it. (Remember, my students always accused me of doing things in the hardest way, so there are probably easier and more elegant solutions in the heads of others.) I added the little proportion at the top afterwards as a check. Note the use of the Pythagorean Theorem, the distances 0 to x and 0 to w, where x is the distance from the left wall to the intersection point and w is the width of the alley. Also, the 10-foot altitude splits each of the right triangles formed by the ladders and walls into two similar right triangles, which is what gives the information needed to form the two equations.
As @Bill93 pointed out, the equations are messy. I graphed them in a TI89 and let it find the intersection point, the value of x in the diagram. Then I solved for w and used the proportion at the top of the page for a check. I also did an iterative thing in a spreadsheet that kinda worked, but some technological help is valuable.
That’s too much problem for an exam, but if either the width of the alley or the distance to the intersection point were also given, it would be fair, in my opinion. Too much math and too many different skills required as it’s written.
By the way, if you replace x and w in the proportion at the top of the page with rounded versions of the final answers, and replace 10^2 with y^2, you can solve for the height of the intersection produced by reasonable measurements.
The real world differs from the theoretical one; that’s why I like Einstein’s little quote so much.
The functions that we use to represent real-world situations are perfect mathematically while the situations are not. So, as you pointed out, assumptions always have to be made. Not knowing what the assumptions are and how significant they are can lead to huge real-world problems.
No snow here, but we may have a round of the dreaded freezing rain later in the week.
I dug these out of my archives. They aren’t the same problem, but very much like it: given chord and arc find radius.
They may be available on Profsurv.com under the Problem Corner archives there, too. (Problems 47 and 71)
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