seems too easy to be true

[allegansveritatem]

Here is the problem:


Here is exercise 68 that is alluded to above--in the book it is divided by a page break so I am going to type the first part (in order not to stuff too many images into one post) and then I will upload the second part that contains the diagram):

Length of a Tightrope: The figure illustrates the apparatus for a tightrope walker.Two poles are set 50ft apart,



Here is my solution to this:



This is the only way I could see to work this out. It surprised me that it was so easy....which puts me on my guard. Is the the way to go with this problem?

 

[Dr.Peterson]

It's mostly that easy. The only detail you missed is that the h you got is 2 feet less than his actual height above ground. You would have seen that if you had actually drawn a horizontal line 2 feet above the ground in order to make similar triangles.

And, of course, you didn't finish by expressing h as a function of t (replacing d with 2t).

 

[allegansveritatem]

It's mostly that easy. The only detail you missed is that the h you got is 2 feet less than his actual height above ground. You would have seen that if you had actually drawn a horizontal line 2 feet above the ground in order to make similar triangles.

And, of course, you didn't finish by expressing h as a function of t (replacing d with 2t).

Right...of course. They are asking for height above the ground. I think I was being influenced by fact that before I worked this problem out I went back and redid #68 where the the length of the shorter pole on the right bears on the accuracy of the result. And yes, I jumped the gun a little with the naming of the function. Thanks for checking my solution.

 

[allegansveritatem]

It's mostly that easy. The only detail you missed is that the h you got is 2 feet less than his actual height above ground. You would have seen that if you had actually drawn a horizontal line 2 feet above the ground in order to make similar triangles.

And, of course, you didn't finish by expressing h as a function of t (replacing d with 2t).

I was thinking again about this problem while listening to the birds chirping outside my early morning window. It suddenly occurred to me that I somehow must account for that 24 inch deficit on the right side. Here is how it could be done and done with ease: h(2t)= 28/57.3 (2t) +2. No?

 

[Dr.Peterson]

I think that's just what I said, and you agreed: The h you are looking for is 2 feet more than the h you had calculated.

Now, they asked for height above ground as a function of t, so the final answer will be either just h = 0.9773t + 2, or h(t) = 0.9773t + 2 if you want to explicitly indicate a function.

 

[allegansveritatem]

AS an addendum to my last post: If I kept the first factor such: 30/

I think that's just what I said, and you agreed: The h you are looking for is 2 feet more than the h you had calculated.

Now, they asked for height above ground as a function of t, so the final answer will be either just h = 0.9773t + 2, or h(t) = 0.9773t + 2 if you want to explicitly indicate a function.

Yes, I know that you pointed out the problem with my solution. But I starting thinking about that 2 feet on the right...I mean...if I use 30 and and 50 as the squared factors to find the hypotenuse squared won't I come up with a slightly off figure for the denominator of the ratio that I include in my function? That was one of my thoughts..

 

[Dr.Peterson]

Yes, that would be off. But you used 28 and 50, which is correct. There is no right triangle with legs 30 and 50 here.

I wonder if you have yet drawn the picture I suggested you needed in order to see everything clearly. Draw a horizontal line through the right attachment point 2 feet off the ground. This forms a right triangle with legs 28 and 50, and when you draw a vertical line through the person, you get a similar triangle with leg h-2 and hypotenuse d.

 

[allegansveritatem]

Right. What I was thinking re hypotenuse would't make sense. We have to shave two feet off bottom of the left pole to achieve a right triangle. And really, I knew that because I worked out (again!)the problem referenced in this problem just before I tangled with this one. So...just not thinking (again!). Thanks.