| [4.82s -> 19.36s] As always, please pause the video and try the question on your own before moving on. In this question, we are trying to minimize the amount of time it takes for a woman to go across the lake on the boat to point B and then travel by foot over to point C. |
| [19.36s -> 31.89s] In order to minimize her travel time, it's actually going to be useful to first look at the distances that she's traveling. From A to B, we can represent that distance as just D1 and from B to C, we can represent it as D2. |
| [31.89s -> 46.42s] And now at this point, a little high school geometry is going to kick in here. What we're going to do is draw a line from BC. And the reason that we do that is because we're going to end up forming a right triangle. We recall from high school geometry that in a circle, if you have... |
| [46.42s -> 59.18s] an inscribed angle, that's just an angle that touches the edge of the circle. If that inscribed angle is formed with a diameter of the circle, then the angle is 90 degrees. So we do indeed have a right triangle here. |
| [59.18s -> 72.02s] And since we have a right triangle, all the trigonometric ratios apply. One that's particularly useful here is the cosine. Now of course the cosine of an angle is equal to the adjacent over the hypotenuse. If we look at the right triangle that we just formed, |
| [72.02s -> 86.42s] In relation to theta, we can see that the adjacent side is marked as d1, and the hypotenuse would be 4 miles. So we can fill those into the cosine. We can easily solve this equation for d1 by multiplying both sides by 4. |
| [87.15s -> 101.74s] We'll hold on to this representation of D1 and come up with a representation for D2. Now we'll notice that D2 is a curved distance. It's an arc length from B to C. We know that arc length is equal to the radius times an angle. |
| [101.74s -> 115.90s] The only thing we have to watch out for is that angle. Of course, this right here is marked as theta. We recall from geometry that when you have an inscribed angle that's marked theta, then the arc that it intercepts is actually two theta. |
| [115.90s -> 129.87s] That's another idea from high school geometry. So, plugging into the arc length formula, we would have the radius of the circle, which was 2 miles, multiplied by the angle that the arc length is represented by, and that, again, is 2 theta, and therefore... |
| [129.87s -> 143.76s] we get 4 theta for the arc length from B to C. We've called that D2, so in essence, D2 is equal to 4 theta. Now, you might be asking, why do we need representations of the distances? We're trying to minimize the time. |
| [144.37s -> 158.27s] Well, we recall from perhaps a physics course that time is equal to a distance divided by a speed. So we can come up with an expression for t1, which was the time required to go across the length, by just plugging in that distance. |
| [158.27s -> 163.86s] D1 over the speed of the woman going across the lake. Now that speed was stated |
| [164.27s -> 174.22s] as being two miles per hour. So we can plug that in for the speed going across the lake. Remember that D1 was four cosine theta, so we can actually make a replacement here, a substitution. |
| [175.09s -> 180.02s] And then we can simplify this just a little bit by dividing numerator and denominator by 2. |
| [180.59s -> 191.41s] So we are left with an expression that represents the time required to go across the lake in the boat We're going to do the same thing for the time required to go along the arc length by foot |
| [191.44s -> 205.20s] We'll call that time T2. Recall that the distance of walking from point B to point C along the shore was 4 theta. And then the speed that the woman walks with was given to us as 4 miles per hour. |
| [205.58s -> 216.40s] Now of course this simplifies to just theta. So here is a representation of the time to walk along the arc length. The total time would just be these two times added together, so we can write that out. |
| [217.10s -> 231.66s] So here is the total time function. We basically have time as a function of the angle equal time 1 plus time 2. Now that we finally have a simplified equation in terms of just one variable, we can proceed in optimizing it. And to do that, we take the derivative. |
| [232.30s -> 245.87s] Now the derivative of 2 cosine theta will be negative 2 sine theta and then the derivative of theta will just be 1. We can then set the derivative equal to 0. We'll subtract 1. |
| [246.99s -> 254.67s] Then divide by negative 2 and so now we have the sine of theta equals 1 half. Now it turns out there are two solutions to this equation. We can have |
| [255.41s -> 266.88s] 30 degrees or 150 degrees Certainly 150 would be too much because if we had an angle that projected out to 150 degrees Then it would look something like this and that would be well |
| [266.88s -> 276.34s] outside of the bounds of even the circle. So we can confidently reject 150 degrees. We still have to make sure that this angle indeed creates a minimum amount of time. |
| [277.01s -> 287.50s] And to do that, what we have to do is test the time required at the end points of our interval and also at this critical number. So for example, |
| [287.50s -> 298.03s] We have the following interval here. The smallest that theta could be would be 0 degrees. The largest, as noted earlier, that it could be is 90 degrees. And then we have a critical number here at 30 degrees. |
| [298.35s -> 311.71s] Now to find out which one of these values leads to the absolutely minimum amount of time required to travel around the lake, we just plug them in for theta into the time equation. So for example, |
| [311.71s -> 323.50s] T of zero degrees or zero radians comes out to two hours. T of 30 degrees or pi over six comes out to approximately 2.25 hours. |
| [324.11s -> 332.32s] And T of 90 degrees, or pi over two radians, comes out to approximately 1.57 hours. Well, |
| [332.32s -> 340.14s] we're trying to minimize the time that it takes for her to reach her destination. So we can actually see that the correct answer, the correct angle, |
| [340.14s -> 350.69s] that we should have is the pi over two radians, or 90 degrees. Let's understand what this means in terms of the question by referring back to the picture. So in the original picture, this angle |
| [350.69s -> 365.07s] doesn't look like 90 degrees in fact it isn't 90 degrees in order to make it 90 degrees we would have to open it up a little bit so let's say we open it up about that much okay that's still not 90 degrees but we're getting closer so she would row her boat across this much of the lake and then she would get off and then walk the rest of the way |
| [365.07s -> 378.14s] to point C. Well, that wasn't the correct angle, right? This is still a little bit less than 90 degrees. So we'd have to open it up just a little bit more. So perhaps that would look like this if we try to make this angle 90 degrees. She would have to row just a little bit. |
| [378.14s -> 384.05s] across the lake get off the boat and then walk the remaining portion of the distance all the way to point c |
| [384.05s -> 398.35s] That angle right there would be the theta, but it's still not quite 90 degrees. The point is, as we open up the angle more and more to try to get to 90 degrees, she's essentially not going to have to go across the lake at all. She's simply going to walk around the perimeter. |
| [398.35s -> 409.68s] get to point C. And so the correct answer here is having the woman just walk around the perimeter of the lake in order to get to point C. She doesn't have to row at all. An interesting result. |
| [411.47s -> 420.56s] Thanks again for taking the time to watch this video. If you like it, please subscribe and feel free to send your own question to the following email address and stay tuned for |
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