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JH
Numerade Educator

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Problem 34 Hard Difficulty

Find the area of the region bounded by the hyperbola $ 9x^2 - 4y^2 = 36 $ and the line $ x = 3 $.

Answer

$\frac{9}{2} \sqrt{5}-6 \ln \frac{3+\sqrt{5}}{2}$

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Video Transcript

Let's find the area of the region bounded by the hyperbole. Nine. X squared minus four y squared equals thirty six in the line X equals three. So I'll go to the graphic calculator and second to verify this rest. But the hyperbole is here in blue and this is one has of the hyperbole there. So let's go to the graphic calculator. We have these wrapped already So in red this time it's in red is the hyperbole and we see the curve right here and then we see him blew the line X equals three and so I'll zoom in And this is the region that's bounded by the hyper bleh and the vertical line. So let me come back to the previous page. So this is the region that we're interested in So we want to go ahead and find that area. So we'LL just think of this as two separate curves and we'll do this by solving for Why So let's solve this equation for why we have nine x squared minus thirty six. He goes for wise where so divide by four and I take the square root thirty six and then the square the force too. So up here we have the plus radical and below the exact system bottom have This is the negative square root that we just found over here? No. So to find the area will just do the integral were it goes from two or three and then we'll do top curve minus bottom curve Seoul area will be the integral are bounce for X er two to three That's how curves plus query and then we're subtracting But the thing was attracting is the negative square because that's the bottom cruise the negative wide so we could actually just go ahead and combined these together we'LL get two of these radicals and then the two we'll be able to cancel those twos two or three nine x squared minus thirty six and actually here we could even go step further. Let's go. In fact, there are nine here, So if we pull out this nine, the squirt of that's a three and then we have X squared minus four. So he pulled out a nine and then the square nine was three. That's where this story is coming from. Now we're ready for a tricks up. Let's take extra B to see again. D X is to seek and data tan data and then we can go ahead and evaluate X squared minus four. So x squared is foresee can swear. Go ahead and pull the four outside. The radical you left was seeking score data minus one by the protagonist identity for tension and see, Can't that becomes two times the radical of tangents? Where and then we could write This is too tan data and since we're switching from extra data in the tricks off, let's just go ahead and not find that data values for two or three. These are not going to be data values from the unit circle. So in this case, when we go to the next step was shall do right now. So the next step is we replaced the radical with two ten, and we replaced the X with two Sikh and time Stan. So let's go to the next page. Hey, we have three then. Since we don't know what those new data values are for now, let's just call them and be We could deal with this. At the very end, the radical became too thin. Data and then the ex to see can't data Tanda defense. And so we could go ahead and we have four times three. Let's pull that out. We have tan square times. He can't. And for this type of integral, probably the best way to go here is toe rewrite tangent as seek and squared data minus one and then go ahead and multiply. The sea can't squared minus one time seeking so we can do that. And then we can break this up into two intervals. So actually, we could just leave it as one of the girls Doesn't matter. So after I distribute, can we get this? And actually, in this case, we could think about this. We probably should think about this is two hundred rules because they're so different. Some of right and Prentiss is because the twelve goes to both Now for this first in a rule here, you might have memorized this already. It's in your table. It's exercise that you would have seen in seven point three. But for this in general here, the best way to go about this one is to use integration, my parts. So here you will take you to be sick Aunt Ada, do you seek in time standing then Devi is the leftover which is sea can square and then the is tan data. So integration my part works well for this first integral. And after you do this for the first, integral will get So the next step in our chain of equations twelve and then the interval of sea can cubed that becomes one half parentheses. C can't tangent plus natural log of C camp LosAngeles sloppy here. And then for the second in a rule in a roll of Sikkim that's just who have a minus already and then natural log. She can't plus Tanja plus c. So now we could go ahead and simplify. So here, what can we combine? We have the natural log here. We also have it over here with the minus sign. But this first one that has the one half in the front. So when we combine these so first we have twelve and then one half so that we have a six seeking data. Santana and then we have one half natural log minus one natural log. So that's minus a half. And then we multiplied by swallow that gives us minus six and now we'LL go to the triangle, find to seek an intention in terms of X And then we could replace and be with our original bounce two and three. Let's go to the next page and we could even factor in the sixth year. So let's do that. But we have a six. And really, because we're doing a definitely a rule, I should be putting the A and B here, but we're going to change those anyways. So now this is the point where we will use our troops up. Our trips up was equivalent to see Can't data equals x over two. So strong that Rangel So we have our data down here in the bottom, right? Our angle X over two for C can't And then you could find aged by using protectorate Karen for a triangle. A square close to squared is X square. So that means ages the radical of X squared minus four And now we could find see Canon tangent. So C can we really know what that is? That's six over too. So it's right that in time's tangent, So tangents ancient over too. So we have the radical over to and then minus natural log. And then we just have the sum of these two so c can, except for two plus hand go ahead and absolute value there. And now, since we're back in the variable, it's we could switch from A to B back to our original bounds to win three. So this is why it wasn't necessary to find Andy. Now we just go ahead and played these numbers in. So we have a three over to plugging in three first and then we have nine minus four in the Radical. It's a five. There we get another two minus natural log, three halves plus radical five over two. And that's for plugging in three. And now it's plugged into, and when we plug into for X, we have to over too radical zero over, too. And then we also have minus natural log to over too less radical Cyril over too. So here, radical zero zero and then we have natural log of one and natural Aga one is also zero, so we could ignore these two terms and the last thing to do here is to just simplify it. So we have nine radical, five over, two minus and then distributing the six six natural log. You could lose the absolute value here because the thing inside this Ellen is positive. You could have those fractions together three plus through five all over, too. And there's our answer.