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A Tibetan monk leaves the monastery at 7:00 AM and takes his usual path to the top of the mountain, arriving at 7:00 PM. The following morning, he starts at 7:00 AM at the top and takes the same path back, arriving at the monastery at 7:00 PM. Use the Intermediate Value Theorem to show that there is a point on the path that the monk will cross at exactly the same time of the day on both days.

Define $u(t)$ to be the monk's distance from the monastery, as a function of time $t$ (in hours), on the first day, and define $d(t)$

to be his distance from the monastery, as a function of time, on the second day. Let $D$ be the distance from the monastery to

the top of the mountain. From the given information we know that $u(0)=0, u(12)=D, d(0)=D$ and $d(12)=0 .$ Now

consider the function $u-d,$ which is clearly continuous. We calculate that $(u-d)(0)=-D$ and $(u-d)(12)=D$

So by the Intermediate Value Theorem, there must be some time $t_{0}$ between 0 and 12 such that $(u-d)\left(t_{0}\right)=0 \Leftrightarrow$ $u\left(t_{0}\right)=d\left(t_{0}\right) .$ So at time $t_{0}$ after $7: 00 \mathrm{AM}$, the monk will be at the same place on both days.

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Harvey Mudd College

University of Michigan - Ann Arbor

Boston College

This is problem number seventy three of this Stewart Calculus eighth edition, Section two point five A Tibetan monk leaves the monastery at seven AM and takes his usual path to the top of the mountain, arriving at seven p. M. The following and to model this problem using some equations were going to call the a distance of the monk on the first day. A za function of you, a function of time teeing in ours and this is going up the mountain and then going down the mountain will be a separate function tisa d of t, which is the function of team in ours from the first function. What we know is that initially, at zero hours, the monk is exactly where the monetary is. So we say that he had travels. He has has yet to travel any distance, so he's at zero. But we say at twelve hours when he reaches the top of the mountain ah, he has traveled a certain distance. D, for example, this second function for the second day, initially at seven a. M. The second day, we say that his distance from his initial starting point of the monastery is Capital D the distance that he had travelled the first day after two hours and then as he's coming back down the mountain after twelve hours, he has reached his initial point. Ah, again, he has arrived of the monastery such that his final position is zero, So he has returned back to zero. So what we want to do now is we want to call her name, any function, a new function, you minus D. And this is the difference between the position that he was on the first day on the position that he is on the second day and what we should confirm is that first of all, the first day function, this position function is continuous. As we assumed that the monk continuously caught walks up the mountain. There are no jump discontinuities due to teleportation or flying for any reason. Ah, and down the mountain is also a continuous function since they're both continues function. The difference of the two continues functions is also a continuous function. And our goal is to see if there is a possibility that this difference of their position equals zero. Because in that case that would mean that there was a position where it was equal you nd abdomen and down the mountain at the same time of day, he would have been in the same position. So what we want to do is we want evaluate dysfunction. Ah, you might miss teen. Let's evaluated at two different times and t equals zero for zero hours. It will be the same as you minus t at zero hours which is zero minus deal or just negative capital D. And then we take the difference function evaluated at home hours and we see that the difference in their positions is D minus zero. So capital D minus zero We see that for the first difference function at the beginning of this interval of twelve hours, the total distance traveled is considered a negative or the difference in the positions and considered negative. And in the second case, after twelve hours, a different function, it turns out to be a positive number. So I should write that as he is greater than zero and they had ideas, lessons here also, we see that in this interval from seventy m to seven p. M. Or zero hours traveled and then told our shoveled this difference function Perry's from different values from Negative de Capital D, which is the distance from the monastery to the top of the moment and capital D positive. So it goes from negative to positive lesson zero to greater than zero. So according to the interim, your Valley intermediate values here, Um since this function is a is definitely a continuous function, that it must take all the values between negative the end capital D for this interval from zero to twelve hours. Therefore, there must have been a point in between zero in twelve hours where thie difference function of the position up the mountain down the mountain must have been equal to zero. Therefore, it is proven using them to me the value to that the monk or cross Ah, a certain point in the path at exactly the same time of the day on both days