Two boats leave a port at the same time, one traveling west at...

Question

Two boats leave a port at the same time, one traveling west at 20 mi/hr and the other traveling southwest ( 45° south of west) at 15 mi/hr. After 30 minutes, how far apart are the boats and at what rate is the distance between them changing? (Hint: Use the Law of Cosines.)

Accepted Answer

Hi, everyone. Let's take a look at this practice problem. This problem says two hikers start from the same point at the same time. One hiker walks due north at a speed of 5 kilometers per hour, while the other walks in the northwest direction, 45 degrees north of west, at a speed of 4 kilometers per hour. Find the distance between the hikers and the rate at which this distance is changing after 2 hours. Run the answers to two decimal places, and we're given a hint that says to use the law of cosines. We're given four possible choices as our answers. For choice A, we have 7.13 kilometers and 11.50 kilometers per hour. For choice B, we have 11.50 kilometers and 7.13 kilometers per hour. For choice C, we have 11.50 kilometers and 3.57 kilometers per hour. And for choice D, we have 7.13 kilometers and 3.57 kilometers per hour. Now, in our problem, we're told that one hiker walks due north at a speed of 5 kilometers per hour. So we'll draw that. As an arrow pointing north and will live the distance that that hiker has hiked as D1. And its speed will label as V1, which is going to be equal to 5 kilometers per hour. Now the other hiker walks in the northwest direction. So we'll draw an arrow that is going in the northwest direction. So that means that there's gonna be 45 degrees between these two paths. And what is the distance that that hiker Um, hikes as D2, and we'll label its speed as V2, which is going to be equal to 4 kilometers per hour. Now, the question was us to find the distance and the rate at which that distance is changing between the two hikers after 2 hours. So we'll label that distance as D. So the first thing we need to do is evaluate the values for D1 and D2. Now, D1, which is gonna be the distance that the hiker one has traveled, it's just going to be equal to their speed, which is going to be V1, multiplied by the time, which is T. So this is going to be equal to. B1, we have 5 kilometers. Per hour. And for tea, we have 2 hours. So when we multiply those two together, this is going to be equal to. 10 kilometers Likewise for D2, that is just going to be equal to the second higher speed, which is V2, multiplied by the time T, which is going to be equal to. 4 kilometers per hour. Multiplied by 2 hours. Which is going to be equal to 8 kilometers. So now we just need to determine the distance between the two hikers, and this is where we're going to use the law of cosines. So we call for the law of cosines that we have D2 is going to be equal to D12 plus D22 minus 2 multiplied by D1, multiplied by D2 multiplied by the cosine of 45 degrees since the angle between D1 and D2 is 45 degrees. So now we need to just solve for D, and to do that, we'll just take the square root of both sides, and we'll also substitute in our values for D1 and D2. So we'll have D is equal to the square root. Of the quantity for D1, we'll have 10 kilometers, and that's gonna be squared, then we'll have plus. 8 kilometers squared or D2 squad, you'll have minus 2 multiplied by 10 kilometers. Multiplied by 8 kilometers. Multiplied by cosine of 45 degrees. So we can evaluate this expression using our calculator, and this is going to be equal to. 7.13 kilometers. So here we've just rounded to two decimal places. So this is the first half of our problem answered. Now, the second half, we need to find the rate at which this distance is changing between them. So that means we're going to need to take a derivative with respect to time, and the quantity that we need to take that derivative of is going to be our law of cosines. So we'll have the derivative with respect to T. Of D2 is going to be equal to the derivative with respect to T of the quantity of D1 2 plus D2 squad. Minus 2 multiplied by D1, multiplied by D2, multiplied by the cosine of 45 degrees in quantity. So, when we take the derivative with respect to T of D2d, we get 2D multiplied by the derivative of D with respect to T, since we have to also apply our chain rule. And this is going to be equal to, on the right hand side, we'll have for the first term, the derivative with respect to T of D1 squared, and that's going to give me 2 multiplied by D1, multiplied by the derivative of D1 with respect to T. Then we'll have plus, for a second term, we'll have the derivative with respect to T of D2 squared. And so that's going to give me 2 multiplied by D2, multiplied by the derivative of D2 with respect to T. Then we'll have minus, and here we have two constants in the uh third term, that we can factor out, so it'll be 2 multiplied by the cosine of 45 degrees. And then we're going to have left is D1 multiplied by D2. So we'll need to take a time derivative of that. So here we're going to have to apply our product rule. So we'll have 2 multiplied by cosine of 45 degrees, multiplied by the quantity, and recall for the product rule that will have the derivative of the first term in our product, which is going to be the derivative of D1 with respect to T. This could get multiplied by the second terminal product, which is going to be D2. Then we'll have plus, the first terminal product, which is D1, multiplied by the derivative of the second term. So we'll have the derivative of D2 with respect to T. In quantity. So, I noticed here that I have a factor of 2 in every single term, so we can divide both sides of the equation by 2. And we now need to actually solve this equation for the derivative of D with respect to T. So to do that, um, I'll just need to divide both sides of the equation by D. In doing that, we will have the derivative of D with respect to T is equal to 1 divided by D multiplied by the quantity. And here I'll have D1, multiplied by the derivative of D1 with respect to T, but that there is just the speed of the first hiker, which is going to be V1. Then we'll have plus D2, multiplied by the derivative of D2 with respect to T, which again, is just going to be the speed of our second hiker, V2. Then we'll have minus the cosine of 45 degrees is 1 divided by the square root of 2. And that gets multiplied by the quantity of the derivative of D1 with respect to T, which is going to be V1, multiplied by D2. Plus D1 multiplied by V2 in quantity, since the derivative of D2 with respect to T is equal to V2. So now all we need to do is rearrange our equation on the right hand side. So this is going to be equal to 1 divided by D. Multiplying the quantity, and here we will collect um terms that are multiplying V1 and V2. So we'll have V1, multiplying the quantity of D1. Minus D2 divided by the square root of 2. In quantity, you'll have plus V2, multiplying the quantity of D2 minus D1 divided by the square root of 2 in quantity. So at this point we can substitute in values. So this is going to be equal to 1 divided by 7.13 kilometers for D. That gets multiplied by the quantity of V1, which is 5 kilometers per hour. That's multiplying the quantity of D1, which is 10 kilometers. Minus D2, which is 8 kilometers, and that is divided by the square root of 2. In quantity, plus V2, which is 4 kilometers per hour. That gets multiplied by the quantity of D2, which is 8 kilometers. Minus D1 divided by 22, so that's gonna be 10 kilometers divided by the square root of 2 in quantity. So, now we can evaluate this expression using our calculators, and when we do so, this, we find that this is going to be equal to. 3.57. Kilometers per hour. That is the second half of our problem answer. So now that we have both parts answered, we can compare that to our answer choices, and we see that the correct answer choice corresponds to answer choice D. So, I hope that this explanation has been useful, and I'll see you in the next video.

Two boats leave a port at the same time, one traveling west at 20 mi/hr and the other traveling southwest ( 45° south of west) at 15 mi/hr. After 30 minutes, how far apart are the boats and at what rate is the distance between them changing? (Hint: Use the Law of Cosines.)

Key Concepts

Law of Cosines

Relative Velocity

Trigonometric Functions

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