# More Math Help (Calc/Trig related)

Hello all. I’ve just started my first semester in Calculus and Analytic Geometry, and it’s been going rather rough. While I enjoy doing math, I just can’t seem to grasp all the rules and tricks to solve for some of these problems. So, a few days ago, our teacher gave us a story problem, to guage how well we can handle more involved, lengthy steps. He’s allowed anyone to stop by for help with their assignment, which I’ve done, but I still can’t figure out how to solve this problem in any way. I was hoping some smart math dopers could at least point me in the right direction.

Problem: We have an 8 foot high fence which seperates a clear area and a building. The building’s height is unimportant. The distance from the wall and the building is 4 feet. The goal is to place a ladder on the opposite side of the wall and lean it up against the building, obviously in such a way that it clears the 8 foot high fence. The shortest length of ladder to clear the wall and touch the building is the answer.

In drawing a picture, I get two triangles. One that connects the base of the ladder to the wall (I’m assuming the shortest possible ladder would just touch the wall) and one that goes from the base of the wall to the building. The one at the building would have a width of 4, and would be 8 feet high. I draw an inverse triangle that went down to the base of the building, instead of going up, thus giving me a triangle with two known sides (8 feet high, 4 feet wide). I figure distances would be the same for the hypotenuse of the triangle in either direction (may be wrong on that). Pictures would help, here.

I solved and had 8.944 ft. as the length of the hypotenuse of the second triangle. My triangle that connects from the ground to the wall is still unknown, and I don’t know what to do with it. Everything is unknown (as far as I know) except it’s height (8 feet). How do I figure the first triangles hypotenuse length or the angle and manage (I.E. get lucky) to find the length of the shortest possible ladder that can touch the building?

It’s the same, answer is 2 x root(80) = 2 x 8.944

To elaborate, the angles that make up the top of the triangles in your second drawing must be the same, because they were straight angles in the first drawing. Since they have two angles (that one and the right angle) and the side between them the same, the triangles must be congruent. So the hypotonuses would be the same, no?

Maybe I’m misunderstanding you, but this doesn’t sound right. There should be one ladder segment that extends from the ground to the top of the ladder, making a triangle with a height of 8 feet; and there should be another ladder segment extending from the top of the wall to the side of the building, making a triangle with a base of 4 feet.

The two segments are part of the same ladder so they make the same angle (call it theta) with respect to the horizontal. You can express the lengths of each of the two segments as a function of theta and add them together to get a function which gives the length of the ladder vs. theta. Now find the value of theta which minimizes the length of the ladder.

Assuming I read the problem correctly, the ladder is 16.65 feet long.

Well, it extends from the ground to the top of the wall (I think). Other than that, exactly.

Yes, yes! This sounds exactly right. However, I’m not quite sure on how I would make a function which gives the length of the ladder vs. theta. And I’m not sure how I would minimize the length of the ladder. But from your answer, it sounds like you’re spot on. I’m just not sure how you arrived at it.

Asking for homework help on the SDMB is generally considered bad form, however since you seem so nice and earnest:

The length of the ladder squared is (4+x)^2+h^2 where h is the distance up the building and x is the distance fromt the fence to the ladder base. Also from similar triangles 8/x=h/(4+x). Take the derivative of the length squared and set it equal to 0 to find the extremes… One real root, plug this in and get the answer…

(I got 16.6478 ft)

It’s pretty late, if this is wrong, my apologies.

Ladder length = 4/cos[symbol]q[/symbol] + 8/sin[symbol]q[/symbol]

Doing it that way, you get [symbol]q[/symbol] = 51.56[sup]o[/sup]

I get the same angle (arctan(h/(x+4)))

Yep. It’s atan(8[sup]1/3[/sup]/4[sup]1/3[/sup])

May as well fill in the rest:

L = 4 / cos[symbol]q[/symbol] + 8 / sin[symbol]q[/symbol]

dL/d[symbol]q[/symbol] = 4 sin[symbol]q[/symbol] / (cos[symbol]q[/symbol])[sup]2[/sup] - 8 cos[symbol]q[/symbol] / (sin[symbol]q[/symbol])[sup]2[/sup] = 0

4 sin[symbol]q[/symbol] / (cos[symbol]q[/symbol])[sup]2[/sup] = 8 cos[symbol]q[/symbol] / (sin[symbol]q[/symbol])[sup]2[/sup]

4 (sin[symbol]q[/symbol])[sup]3[/sup] = 8 (cos[symbol]q[/symbol])[sup]3[/sup]

4[sup]1/3[/sup] sin[symbol]q[/symbol] = 8[sup]1/3[/sup] cos[symbol]q[/symbol]

tan[symbol]q[/symbol] = 8[sup]1/3[/sup] / 4[sup]1/3[/sup]

[symbol]q[/symbol] = atan(8[sup]1/3[/sup] / 4[sup]1/3[/sup])