Jane is 2 mi offshore in a boat and wishes to reach a...

Name: Problem 44, Chapter 4: University Calculus: Early Transcendentals
Uploaded: 2020-05-18T18:51:51-08:00
Duration: 8 min 5 s
Channel: Regina Hays
Description: Jane is 2 mi offshore in a boat and wishes to reach a coastal village 6 mi down a straight shoreline from the point nearest the boat. She can row 2 mph and can walk 5 mph. Where should she land her boat to reach the village in the least amount of time?

Jane is 2 mi offshore in a boat and wishes to reach a coastal village 6 mi down a straight shoreline from the point nearest the boat. She can row 2 mph and can walk 5 mph. Where should she land her boat to reach the village in the least amount of time?

Key Concepts

Optimization

This problem is an example of an optimization task, where the goal is to minimize the total traveling time. In such problems, a function (in this case total time as a function of landing point) is defined based on the different segments of the journey, and then techniques are used to find where that function reaches its minimum value.

Calculus-based Minimization

Calculus is used to solve this problem by setting up the total time as a function of a variable (for instance, the landing point along the shore) and then differentiating this function with respect to that variable. Setting the derivative equal to zero helps locate the critical points, one of which will be the point where the total travel time is minimized.

Geometric Distance Computation

The problem involves calculating distances using geometric principles such as the Pythagorean theorem. The rowing distance is determined as the straight-line distance from the boat's starting point to a point along the shore, which is a function of the horizontal and vertical separations.

Speed and Time Relationship

A key concept is the relationship between distance, speed, and time, where time is the ratio of distance to speed. Since the speeds for rowing and walking are different, the total time is a sum of the two segments’ times, and finding the optimal landing point involves balancing these two based on their respective speeds.

Transcript

00:01 Okay, what we're going to do is we have a boat out here in the water, and the boat is two miles offshore.

00:17 And then we have a village that is six miles down the shore.

00:28 And what we want to determine is where the boat should land to minimize the time to get to the village.

00:55 Okay.

00:56 And so what we do know is that by boat you can row two miles per hour or you can walk five miles per hour.

01:13 Okay, and so what we want to do is if i arbitrarily pick some point x away from going straight from the boat to the shoreline, then this distance is 6 minus x.

01:33 And so if you're going to row your boat at an angle, that distance will be 4 plus x squared.

01:42 Okay.

01:43 And what we do know is that we have all of our distances in terms of x.

01:50 And so x can be anywhere from 0 to 6.

01:59 And so now what we need to do is we also, just by using that distance is equal to rate times time, or time is equal to distance over the rate, we are going to write our equation for the total time.

02:15 And so the total time would be the distance that we're going to row by boat divided by the rate at which we row the boat, plus the distance we're going to travel by foot divided by how fast we can walk.

02:36 And so this will be the total time that it would take us to get from, to go from, the boat in the ocean to the village.

02:47 Okay, so now what we want to do is to take the derivative because we want to minimize.

02:56 So the derivative of the time with respect to x is the one -half times the one -half times four plus x squared to the negative one -half times 2x plus a negative one -fifth.

03:15 So there is that derivative...

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