Question

You spend most of Wednesday hiking up a hill. Your ascent begins at 6 AM, and you reach the summit at 6 PM. Of course you stop to eat lunch and take rests along the way. The next day, Thursday, you climb back down the hill. Again, you start down at 6 AM and reach the bottom at 6 PM, though you don't necessarily take the same path. Show that there is a time of day, 1:09:43 for example, and an elevation, 1,453 feet for example, such that you were precisely that high, at that time, on both Wednesday and Thursday. You're not going to know exactly when, or how high, but you can show there is

Answer 1

a time that produces the same elevation for both days.

Answer 2

Intermediate value theorem

Answer 3

explain

Answer 4

We will call the functions representing the height with respect to time for the two trips \(f_1\) and \(f_2\), we will assume that they are continuous.

We will call the starting time \(a\) and the ending time \(b\).

At \(a\) for the first trip we are at lowest height \(f_1(a)\) and on the second trip we are at peak height \(f_2(a)\)

So therefore \(f_1(a) - f_2(a) < 0\)

And at \(b\) for the first trip we are at the peak height \(f_1(b)\) and on the second trip we are at the lowest height \(f_2(b)\).

So therefore \(f_1(b) - f_2(b) > 0\)

So we have \(f_1(a) - f_2(a) < 0 < f_1(b) - f_2(b)\) and by the intermediate value theorem and by the continuity of \(f_1\) and \(f_2\) there is some time \(t\) such that \(f_1(t) - f_2(t) = 0\) and so \(f_1(t) = f_2(t)\)

Answer 5

Clear enough?

Answer 6

yes thanks

Answer 7

I should have rewritten \(f_1(x)-f_2(x)\) as a new function before applying IVT, but its clear.