A ship leaves port traveling southwest at a rate of 12 mi/hr. ...

Question

A ship leaves port traveling southwest at a rate of 12 mi/hr. At noon, the ship reaches its closest approach to a radar station, which is on the shore 1.5 mi from the port. If the ship maintains its speed and course, what is the rate of change of the tracking angle θ between the radar station and the ship at 1:30 P.M. (see figure)? (Hint: Use the Law of Sines.)

Accepted Answer

Welcome back, everyone. In this problem, a car departs from a town and travels southeast at a constant speed of 40 MPH. At 10 p.m., the car passes its closest point to a tracking station, which is located 3 miles east of the town. What is the rate of change of the angle theater between the tracking station and the car at 2:00 p.m. if the car continues on its path at the same speed? Use the law of science and assume that the path of the car is straight. Run the answer to 3 decimal places. Here we have a diagram of our car, the tone, and the tracking station. Now, if we're going to find the rate of change of the anger theater, OK, that means at some point we're trying to differentiate theater. Trying to differentiate theater with respect to time. OK, so let's look at our diagram, see what we can derive about theater, what we can figure out, and then try to somehow differentiate it with respect to time. Now what do we know about our angle? Well, in our diagram, OK, let's let alpha be our 3rd angle, OK, in the triangle that is formed, OK. Then, We can tell that our anger alpha is going to be equal to 180 degrees or pi minus the other two angles because we know all the angles will sum to pi radiance. So that's gonna be pi minus 45 degrees or 4th of pi minus theta, which equals 3/4 of pi minus theta, OK? This is uh how we could express our anger alpha in terms of theater. Now in our problem we were told that at 1 p.m. the car passes its closest point to a tracking station which is located 3 miles east of the town. So if we let the distance that our car has traveled from the town be S, OK, then that means at 1 p.m. Using what we know here, we know that at 1 p.m. this forms a right angle, so this is gonna be a right triangle and thus cause the cosine of 45 degrees is gonna be equal to the side adjacent to the angle, OK, which is S. Divided by the hypotenuse, which is 3. And in that case, that must mean then that S, the distance traveled at 1 p.m. is going to be 3 + 45, 3 route 2 divided by 2. Now at 2 p.m. OK. We want to find the rate of change of the angle theater, but what, what is going to be the distance at 2 p.m.? Well, no, remember we're traveling 40 MPH, so our distances is going to be 3 route 2 divided by 2 plus 40 multiplied by 1 because it's 1 hour. And when we were that old, it becomes 80 + 3 route 2 divided by 2. So this is the distance that will be traveled at 2 p.m. Now, since we have our distances here, we know what our distances are. How can we use that to help us to express the in terms of the distance? Because that will help us to find the rate of change. Well, that's where our law of science comes in, OK? I know by the law of science, OK. By the law of signs, recall, the law of signs basically tells us that the ratio of the sign of an angle to its opposite side in the triangle is the same all the way throughout. So in other words, the ratio of sine theta to s is going to be equal to the ratio of sine alpha to 3. So if we use that idea here, that means sine theta divided by S equals sine alpha, which we know is 3/4 of pi minus theta. Divided by 3. And know if we do some transposition here, then we can say that the sign of theta is going to be equal tos multiplied by the sign of 3/4 of pi minus theta. Divided by 3. Now we could take this a step further because if we recall one of the identities in trigonometry tells us that the sign of the difference between two angles, A and B or A minus B equals sign A B minus cost a b. OK. So if we apply that idea here, that means we could rewrite sine theater as a third of S multiplied by the sine of 3/4 of pi. theta minus costs 3/4 of pie. Signed the I know if we simplify here, OK. We know that the sign of 3/4 of pie is a half of root 2, so that's root 2. More divided by two cost theater, OK. And the, the, the cosine of 3/4 of pi equals negative route 2 divided by 2. So if we're subtracting that, that's going to become plus route 2 divided by 2, sorry, sine theta. And now we can factor out route 2 divided by 2. OK. They give us S route 2 divided by 6 multiplied by the cosine of theta plus the sine of theta. Now let's divide through by the cosign of theater, the simplifier expression. When we do that, we're gonna get the tangent of theater to be equal to a 6th of S root 2 multiplied by 1 + 10 theater, OK? And know if we were to group our like terms together. Then we should find that we can expand our term S root 2 divided by 6 across our bracket, carry the tangent term to the left hand side, and factor the tangent to get the tangent of theta multiplied by 1 minus S root 2 divided by 6 to be equal to S root 2 divided by 6. OK. Now where do we go from here? Well, no. If we were to divide through by our expression, then we should get the tangent of theater to be equal tos root 2 divided by 6 divided by 1 minuss root 2 divided by 6. I know when we simplify that we should get it to be equal to Sroot 2. Divided by 6 minuss root 2 if we expand the or if we make the expression in our denominator one term one fraction and thus if this is the tangent of theta then theta must be equal to the inverse tangent of S root 2 divided by 6 minus s root 2. Now before we continue, let's just make sure we're still all on the same page, OK? So we've done quite a bit of transposition here, but remember, essentially all we wanted to do, like we said at the start of our problem was to express theater, OK, in terms of S because we'll need that to help us figure out the rate of the change of the anger theater that is with respect to time. So now we have theater in terms of S. And because we have theater in terms of S, no we can differentiate theater with respect to the time, OK? So now, let me put that in red here. Differentiating theater. OK. With respect to the time. Then we're differentiating an inverse tangent, which means that the derivative. Is going to be equal to 1 divided by 1 plus root 2 divided by 6 minuss root 2 squad, that is 1 plus the tangents argument squared multiplied by the derivative of our argument. Now notice here we're going, that means we need to differentiates root 2 divided by 6 minus S root 2 and because it's a quotient, we're going to apply the quotient rule of differentiation. And when we do that, OK, we end up with 6 minus S root 2. Multiplied by route 2 DSDT where DSDT is the rate of change of a distance with respect to time minus. S root 2 multiplied by negative root 2 multiplied by. Again, our rate of change of a distance with respect to time. In other words, or speed, OK, or or velocity. Now in this case, all of this is going to be divided by 6 minus S root to all squared. Now here we can expand and go ahead and simplify and again I'll, I'll leave that to you in the interest of time, but when you do, you should find that our numerator simplifies to 6 root 2 multiplied by DSDT, OK, and our denominator will be 6 minus root 2 sorry, 6 minus root 2. Squared Plus 2 S 2. Now remember, we know what S and DSDT are. We know, like we said earlier that our car is traveling at a speed, the SDT or velocity of 40 MPH. So the SDT equals 40, and we said that at 2 p.m. or distances equals 80 + 32 divided by 2. So in that case then. This means that our derivative of theta with respect to time will be equal to 6 root 2 multiplied by 40. All divided, OK, by 6 minus S80 + 3 route 2 divided by 2 multiplied by route 2. Squared Plus 2 Squad, so that's 2 multiplied by 80 + 3 root 2 divided by 2 squared. I know when you solve here, if you were to put this into your calculator to 3 significant figures, you should find that it's approximately equal to 0.053 radiance per hour. So what does this mean? Well, no, basically, this tells us that our rate of change of the angleitheater between the tracking station and the car at 2 p.m. if it, if it continues on its path at the same speed, is going to be 0.053 radiance per hour. Thanks a lot for watching everyone. I hope this video helped.

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