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Problem 41 Hard Difficulty

Two sides of a triangle have lengths $ 12 m $ and $ 15 m. $ The angle between them is increasing at a rate of $ 2^o/min. $ How fast is the length of the third side increasing when the angle between the side of fixed length is $ 60^o? $

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so whatever we're doing, like a related rate problem, the best thing to do is just go ahead and plug in early, try to draw a picture, and then plug in everything you have. So we have this triangle here with one side, like 12. 15, and the angle between them is increasing by a rate of two degrees per minute on, then. Over here, we have, like, our third side, so I'll just go ahead and call this third side. See, um, actually, before we do anything, um, whenever we are working with, like, calculus, we should convert everything from degrees to radiance. Um, because a lot of times, things will not work out if we do them in degrees. So I'm just going to come up here and convert this to radiance first. Before we go any further, eso remember to do that we would multiply by pi and divide by 1. 80 so it would be to over 1 80 times pi on, then. That would just be pi over 90. We have pie over 90 minutes. So that will be our units for this. Or I guess rads per minute will be it. Radiance per minute. And so this is very important for us to do that. So just kind of keep that in mind. Um, And then over here, I guess I should also convert this to, um Pi Third as well. So this is pie third. Yeah. Now let's see if we can find some way to relate all of these things. So we want to know about this third side, we have to side lengths of triangle. Um, and we have this angle here. Well, we can actually use the law of coastlines to relate everything here. So this is going to be C squared is equal to Hong. Call this A and B, And the reason why I'm using C is because we know the angle across from this. All right, So B C squared is equal to a squared plus B squared minus two a b, then cosine of the angle across from that which in this case, is going to be theta. Yeah, And now, since sides A and B, we're supposed to always be Constance. We could just go ahead and plug those in to this so we would have 12 squared, which is 1 44 so we have C squared is even toe 1 44 and then be square. So there'll be 15 square so to 25 and then two times 12 times 15 is 3. 60. And since data and see are going to be changing, we don't want to plug anything in for those at least quite yet. But let's go ahead and kind of simplify this down a little bit. So 1 44 plus 2 25 is 3 69. So this would be C squared is equal to 3. 69 minus 3. 60 co side they. So now we have this relationship between this angle theta and our side. See, now, over here, they want us to find how is the third side increasing at this angle? Um, so since we're trying to figure out how fast this is increasing so this is going to be D. C by D. T. So we want to find the derivative of sea with respect to time and then over here, Or actually I should say this here is actually telling us how data is changing. What perspective? Time. So we have D data by D. T. And at this point. If we just come down here and take the derivative of this, we can go ahead and implicitly differentiate each side. And that should give us so on the left we would have to see times and then we use change Rules of the derivative of sea with respect to time is just gonna be d C by d. T 3 69. That's just concepts that goes away. And then the derivative of cosine is negative sign. So that would be 3 60 sign data. But then we have to implicitly differentiate again, since data and time are not the same variable. So it be d data by D. T. Yeah. Um, And now, at this point, we can actually start plugging stuff in on. Then we'll see what we need after that, Um, because we've already implicitly differentiated or just took the derivative on each side. And so we're interested when fatal with 60 degrees or pie Third. And we were told that our change of theta is going to be a constant, um, pi over 90. Um, so let's just go out and plug those in over here. So actually, I'll divide each side by two first, so that would just give us 1 80 here. And so, um, we don't know what C is yet. You don't know what d C by DT is. That's what we're looking for That will have 1 80 times. So sign of pie from a sign of pie. Third, which is going to be Route three over to and then deflated by D t is supposed to be pi over 90. Okay, um, it's actually 90 and 1 80. These would simplify down to two. And then actually, those twos can't stop with each other. So if we were to rearrange everything now, I'll divide over by C So we get that our change of C prospective time is going to be just Route three over route three times pi over. See? So we need to go ahead and figure out what is he going to be? And then from there we can go ahead and get an estimation for this. So we know our relationship between c released how to find C is given by this equation right here. And so again, remember, we're interested when Seita is equal to pi third. So at that point pie third or co sign is going to be one half. So that would be 3 69 minus 1 80. So 69 minus 1 80 is 1 89 and then we just take the square root on each side. So we get C is route 1 89. So then we come up here and plug that it. All right, so this would be route three pi over group 1 89. Um, and the units with this should be eso the units for our side. We're in meters, and then time we're in minutes. So this would be meters per minute and then actually, let's just go ahead and plug all these and then see what we get s so we could get, like, an estimation of it. So, fruit me times. Hi. I have been all over route 1 89 and this is going to give us something around zero point. Um, e lost it zero point 395 m per minute. So if you want the exact answer, it's up there. And if you want this approximate answer, it's there. But again, remember something that's really important. When we did this was when we're doing calculus stuff, any kind of degree. You always convert it to radiance unless they for some reason, tell you not to. Because the entire way we have, like these derivatives set up, it's always in terms of radiance for, like, sine cosine and all of that. So, yeah, just kind of keep in mind we always use radiance, because otherwise at this point, we would have just got a completely different answer.

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James Stewart

Calculus: Early Transcendentals

Grace He

Heather Zimmers

Kristen Karbon

Joseph Lentino

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Grace He

Heather Zimmers

Kristen Karbon

Joseph Lentino

James StewartCalculus: Early Transcendentals

Grace He

Heather Zimmers

Kristen Karbon

Joseph Lentino

James Stewart

Calculus: Early Transcendentals