Problem

A woman at a point $ A $ on the shore of a circul…

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Problem 49 Hard Difficulty

Solve the problem in Example 4 if the river is $ 5km $ wide and point $ B $ is only $ 5 km $ downstream from $ A $.

Name: solve the problem in example 4 if the river is 5km wide and point b is only 5 km downstream from a
Uploaded: 2021-04-02
Duration: 6 min 19 s
Channel: Numerade
Description: Solve the problem in Example 4 if the river is $ 5km $ wide and point $ B $ is only $ 5 km $ downstream from $ A $.

Answer

$$T(0) \approx 1.46$$
$$T(5) \approx 1.18$$

Grace H.

Catherine R.

Missouri State University

Heather Z.

Oregon State University

Joseph L.

Boston College

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Video Transcript

whereas to solve the problem. An example. Four of this section if the river is five kilometers wide and the point B is only five kilometers downstream from the point A. So again drawing a diagram like an example. Four. We have a triangle with three points a B and C. Great. It's a right triangle and then we have another point. D on the side opposite of a and now we have the distance from A to C is five kilometers and the distance from a to B is five kilometers. I'm sorry. Sorry. B is only five kilometers downstream from a. This means the distance from B to C is five kilometers. Not for me to be now let's call this mystery distance, but from D to see X mm. Why so X is the distance downstream from the point directly across the river? Clearly now the running distance. This is the length of DB, which is five minus X and the rowing distance once they're on the river. This is the distance from a T. D. Using Pythagorean theorem. This is the square root of 25 plus X squared. Also, our rowing speed is six kilometers per hour and running speed is eight kilometers per hour. So it follows the total time running and rowing, which will call T F X. Well, this is our rowing distance divided by our rowing rates. Six. So we have the square root of X squared plus 25/6, plus the rowing running distance five minus X over the running speed eight. And it's clear from our picture that the domain includes all X between zero and five. Now we're trying to minimize this time so we'll take the derivative of t Want to solve t prime of X equals zero. So we get t prime of X equals. This is one half times X squared plus 25 the negative one half time is two x over six minus 1/8 mhm which is equal to yes x over or X, basically over six times the square root of X squared plus 25 minus 1/8. Again, we want to set this equal zero So this is X over six times the square root of X squared plus 25 equals 1/8. So we have eight x equals six times the square root of X squared plus 25 solving this radical equation, you can square both sides Eventually you should get seven. X squared equals 225 and so X is equal to the square root of 225 over seven, which is approximately 567 and recalled. The units are in kilometers now. The critical number that we have this is outside of our domain from 0 to 5 time the absolute minimum max in a closed interval. It therefore follows that our answer must be at the end points. So let's find t of zero and t of five t of zero. If you plug this in is approximately 1.46 and the unit is in ours and t of five. This is approximately 1.18 and again the unit isn't ours. Looking at these two times, we see the time is shorter when X equals five and so police T has an absolute men at X equals five. Therefore, what's the man's strategy? To minimize the time the man should row directly from point A to point B

Chris T.

Ohio State University

Related Courses

Calculus 1 / AB

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 4

Applications of Differentiation

Section 7

Optimization Problems