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28. A fisherman hooks a trout and reels in his line at 4irsec Assume the tip of the fishing rod is 12 ft above the water and the fish is pulled horizontally directl...

Question

28. A fisherman hooks a trout and reels in his line at 4irsec Assume the tip of the fishing rod is 12 ft above the water and the fish is pulled horizontally directly towards the fisherman: Find the horizontal seep of the fish when it is 20 feet from the fisherman:

28. A fisherman hooks a trout and reels in his line at 4irsec Assume the tip of the fishing rod is 12 ft above the water and the fish is pulled horizontally directly towards the fisherman: Find the horizontal seep of the fish when it is 20 feet from the fisherman:



Answers

An angler hooks a trout and reels in his line at 4 in/s. Assume the tip of the fishing rod is 12 ft above the water and directly above the angler, and the fish is pulled horizontally directly toward the angler (see figure). Find the horizontal speed of the fish when it is $20 \mathrm{ft}$ from the angler. (FIGURE CANNOT COPY)

This problem. We have a fisherman who is standing longshore here. Okay? He has ah, hooked a trout, and he is reeling in the fishing line. Okay, The tip of his rod. Hey, his rod is perfectly vertical, and this is 12 feet tall. Okay, so we are going to label this, um, are for Rod. Okay, then here is our line, and our line is being reeled in. Hey, at a rate off four inches per second and the fish is being pulled this direction. So we're gonna call this F for fish. All right. So we know D L D t is negative. Four inches per second because it's the, um, length is decreasing. Okay. What we want to find is what is g f d t. When the fish is 20 feet from the shore. All right, so let's look here. Looks to me like we have a right triangle. So we're gonna work with the Pythagorean theorem. We have asked Squared plus R squared is equal to l squared. Okay. In this instance, we do have a constant the rod. The length of that raw never changes so we can go ahead and put 12 in for our huh? Now we're going to differentiate this equation with respect to time, so we're gonna have to f d F d t plus well, the derivative of a constant zero. Okay, so that doesn't even need to be there. So that's gonna be equal to two. L d l d t. All right, I'm gonna go ahead and cross off my choose because I'm just going to divide the entire equation by two. So I'm gonna substitute in what I know. So we want to know at 20 feet how fast the fish is being pulled in. So twenties we're gonna plug in for F, and we're solving for the f D. T el is Well, we gotta figure out l So let's come over here to the side and we'll do another path. Angry. And Vera, this time we're gonna put 20 in, we're at, and we're solving for l. All right, so let's look here. We're going to have if we swear 20 and square 12 and add them together. We have 544 and the square root of 544 is 23.32. So l is going to be 23.32 An de L D. T is negative four inches 1st 2nd but we're talking about feet for the distance of the fish. So let's convert this two feet per second. So four inches would be 1/3 of a foot, so negative one their word feet per second. All right, now we can go ahead and do some calculating. Here we have 23.32 times negative. 1/3 is gonna give us negative 7.77 Divide out the 20 from both sides and we have negative 0.3885 feet per second. And if you want that in inches, it would be 4.66 inches per second. Hey, that would be negative. And all the negative tells us is that the distance from the fish to the man is decreasing.

Let's go ahead and draw the situation. And so were the fisherman's head, located here at point H H for head and height, 1.55 meters above the ground. We're above the water in this case, and we know the trail is some distance. A lot of the water. Let's say the traps here. So this is T and this distance here we're told this 0.45 years. Now we're told the distance between the fisherman said, and the trout is 3.5 meters, and so the total mom's gonna to know here is 3.5 years. But let's say that the distance here, if we draw this down, there's going to be summer fraction here. I didn't draw that good enough, do you? So there's gonna be a refraction right there. And so let's say after the refraction, the distance war zone to the trail. This here is distance D. That means if I draw this down this distance, here is D. Now this angle here is St II writes the incident angle in this year, a state a T. It's the transmitted angle, and so I hope my pictures not getting too sloppy here. But if this is distance D in the total of 3.5 in this distance right here must be 3.5 my honesty. And so now we're ready to apply geometry and sells long. Let's apply Snell's law first. We know says that in one scientist Guy is equal to into science data T. In this case, we know that in one is one and into is 1.33 And so Snow's long eyes an equation that relates the incident angle to the transplant angle at this point. Now let's use on the country extend the starting lineup. And then if this is 1.55 then this side over here is also 1.55 And so that tells us that the tangent of the incident angle I'm just reading it off from this top trying here. Is he going to offset over adjacent? So it's 3.5 over 1.55 Oh, uh, what's uh I messed up here, so it's actually 3.5 minus t. So this 3.5 extends past this and it goes all the way to the trial. But really the opposite is just this part here, which is the 3.5 minus D. So they were you OK, fixed it. Tangent of the incident angle is 3.5, minus the distance D. So it's this length over the 155 Okay, that makes more sense now. And let's do the same thing for the transmitted angle. So Tangent of fate A T. So now I'm looking at this straggle down here and then the opposite of this is D over the adjacent, which is point for five. I really apologize about the picture here. It's kind of messy at this point. We have three equations and three unknowns. We have cells along. And then these two facts from geometry or I should say, trigonometry. The three unknowns are they don't high tea. Andy. In Solving this system of equations is actually not very easy. It's quite hard, actually. I recommend using some new miracle software like Mathematica were something else in there. And when you use that, you give that immunity, which is our cure unknown, right? It's the angle from the vertical that the trout sees head of the fisherman. We get that this is equal to 42 0.2 degrees. You can sell for the other unknowns but problems not asking for them. So I'm not going to present them here, and that is the solution.

Given that we are pulling in a fish with on a dock that's 10 ft high. We're interested at the moment that this length of fishing line is 25 ft and we know that it's being reeled in. We'll go ahead and call that Z that green line that's being reeled in that negative one her second because the length is decreasing, which is why it's negative. So that's really important. And then we could set up first, just a relationship between the angle Fada so we could do sign, which is opposite tennis Fixed. That's a constant Z is going to change that green fishing line length and then another way to rewrite that is just 10 times. See to the negative one. It's downstairs. So then let's take the derivative of both sides So derivative of Sign is co sign. But we're differentiating with respect to time, so it's gonna be the data over DT. Then we're gonna move the negative one in front, so that becomes a negative 10 z to the negative, too. Mhm, the negative two. And then all of that times dizzy 80. But the good news is we know what thes e V T is just like every related great problem. We have to know one rate to solve for the other one. But that's a negative one. Let's just go and replace that with a negative one the leave the units off for just a minute. We also know that Z at this instant is 25. And so let's just rewrite that as all divided by 25 squared is the same thing is Z to the negative, too Negative. 10 stays up top and then we're going to divide the co sign over to the other side. So that's going to go in the denominator cosign of guns, Bread data. Okay, so at this point, we've now gotten rid of because I'm there and now this will allow us to solve or the data duty. The only thing we need to know now is what data is. But we could set up a relationship Sign of data is equal to 10 over 25 and then move the sign over. That would be an inverse sign or take the universe sign of both sides. And that is what we're going to substitute birth data there, so it's gonna be negative times a negative. So these will actually turn both into positive now, which makes sense because the angle should be, uh, increasing. It should be getting bigger here, so that makes sense. Why it's positive. And so 10 divided by 25 squared cosine of signed in verse 10/25 and that should give us the data. BT is equal to zero 0.0 17 and then it keeps going 45 or for six if we ran. Radiance is the unit for data divided by time, which is, and seconds so radium for one second. Now, if you wanted to convert that, of course you could do that times, um, 1 80 over pie. So let's do that over time. And that basically tells us the angle is changing by that one degree. Every second there is the same idea. Okay, so that's it.

40 question first rodeo diagram. So it is Strangle a B C A. B C is He is 50 feet and angle is This is 35 degree. We need to find the A B that is equal to D. So for the triangle, ABC cause 35 degree is equal to 50 way d 35. Cost 35 is equal to 0.819 That is equal to 50 by the so value of become so Toby 50 by 0.819 So we get the well US 61.5 feet.


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