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Find the area of the region bounded between the curve 7(t) (tcos t,sin t) for 0 < t < T, and the line segment [rom (~t,0) to (0,0)....

Question

Find the area of the region bounded between the curve 7(t) (tcos t,sin t) for 0 < t < T, and the line segment [rom (~t,0) to (0,0).

Find the area of the region bounded between the curve 7(t) (tcos t,sin t) for 0 < t < T, and the line segment [rom (~t,0) to (0,0).



Answers

Find the area of the region bounded by hypocycloid $\mathbf{r}(t)=\cos ^{3}(t) \mathbf{i}+\sin ^{3}(t) \mathbf{j} .$ The curve is parameterized by $t \in[0,2 \pi].$

We wish to find the area under the curve. Why is equal to one over coastline square T. And this is between he is equal to zero and T. Is equal to pi over two. So to find the area under the curve, we need to go ahead and find the interrupt. So what we're looking for is the integral from zero to pi over two of our curve. So one over coastline square T. Can be written as second squared T. So now this is an improper integral because when we try to put in pi over two we get um 1/0, right? Because co sign of high over two is zero. So now this is one divided by zero. So that means the first thing we need to do is we need to replace that with a dummy variable. So we're going to say the limit as a approaches. Hi over two of the integral from zero to a. Of seeking square T. And now we can go ahead and find the integral. So the anti derivative of second squared is tangent. So therefore this is equal to the limit as a approaches high over to of tangent. Okay, evaluated from zero to A. So this is equal to limit. As a approaches pi over to put in um substitute in our limits of integration. And you should get this is tangent but a minus tangent of zero. So 10 0. Well, that's just zero and tangent of A as a approaches high over two is going towards infinity. So that means this is divergent. So what this means is the area under this curve between zero and pi over two is divergent. So it doesn't have a limit, it's going towards infinity.

Hello. You are pretty good. Purse are probably number 26 here. Out off. The is gone by. They endure wired minus the square. More like minus two squared. We d my Nesler. Come on. So are off. Yeah, this except where you can find us. Do you mind us? Thank you. OK, but my one minus the flip. You are minus stayed a square comma minus stupid. So a DEA media Because half moved there. Lex de vie minus. Why your DX off? Who did? Why Nesler do one. What Really X value exercise P d one minus the square off. Three minus. Thank you. OK, device wine A stupid minus by e f Learn Linus de squid The exists Linus. Three days square, big. So this can be expanded us. You have even like it's got great spent about Subkoff. Well, in a okay, I'm next plentiful minus two Dinner T minus two days square again. Plus to be rest Two for minus goodbye. Minus three d square plus three days square minus 50 square plus thief square plus mine off. Three dealers to for digging. Is it run by? Do you, Linus, learn to one through the years, therefore minus treaty Arrested for my steel is too full minus two p square rested Its got teeth square. That's true. He's okay, my Nesler interdicting. It's all this one we will get like, minus. Do you know theist if I But if I minus to take you by three plus deep minus alert, we will get minus eight by 50 cents Expedia. But this paid but with me. Thank you.

So for this problem we are calculating the area bounded by and I'm just gonna do a quick sketch here. E. To the negative X. Which is E. To the X. But reflected over the Y. Axis. So that's why it's exponential decay. Another way to think about each the negative X. Is it's the same thing as one over E. To the X. Um And then the line through the 10.1 and one comma one over ease one comma one over E. Oh okay. So I actually might be too zoomed out here to really see what's going on. Um But it looks like that line is above the exponential function between zero and 1. Um But what we can do is just go ahead and start finding the area. If we end up with a negative answer, that just means that the curve was actually above the line. But seeing as the fact that the line and the curve both cross 01 and 11 over E. They both go through those points um the curve looks to be below that line. Um So to find the area between the curves that is simply going to be the integral from 0 to 1. Because those are our bounds we're going from X equals 02 X equals one. And then it's going to be upper minus lower. So the upper line. So I'm gonna put an L. There for now. It's gonna be the line minus E. To the negative X. Dx. But what I need is the line. So the line you can find the slope is going to be y tu minus Y. One. Mhm Over X 2. -11. So that's one over E minus one. Or one minus E. Over E. So that's our slope. Um And then we can just use um let's just use slope intercept form oh mm X plus B. And you can see that because it goes through +01 that your B. Value is just one. So um let's go ahead and plug that into. My integral. So the integral from 01 of one minus E. Over E times X plus one minus E. To the negative X. Dx. And we're gonna go ahead and keep calculate this. Um And a girl which we can do by hand. So the anti derivative of one minus E over E. To the X. So that's X. To the first power. The anti derivative will be X. Squared. So we have to take our constant divide by two. And then it be times X squared. Because if you think about multiplying the two in front to use the power role that twos will cancel. And you'll be left with one minus E over E times X. To the first. Um plus the anti directive one is X. And then negative E. To the negative X. The anti derivative would be positive E. To the negative X. Because the derivative of negative X is negative one which we can see here. And the derivative of V. Two something is still lead to the something. Um And then we're going from 0 to 1. So once I have my anti derivative I can plug in one, plug in zero and subtract. So if I plug in one I get um one minus E. One minus E over to E times one squared plus one plus E. To the negative one is one over E. Um minus if you plug in zero, the X square become zero. The X becomes zero but each of the negative zero is not zero. It's one Because anything to the zero power is one. Um And they were going to simplify here, so are ones cancel. And so I get one minus E over to E. Plus get a common denominator to over to E. So I get one plus two, which is three minus E all over to E. And I did get a positive number because E is smaller than three. So this is a positive number, which means I was correct about the line being above the curve.

Sachs is equal to Oh Science T why is sequel to you to the party? He is between serial on DDE high over too So we can scratch a graph of this girl one on one. The curves like this So we need to compute the area of this part. So first we computed area the whole part this part then minus one Here The area of this squire is equal to one So the area of this hope heart is equal. Chu Absolutely you of integral from zero to high over too. Why is e t on did curative of acts Connective science t Okay, which is the co two integral from zero to pi over too. Yeah, you too. Sign of tea. Hey! Of integration by pars we have this is equal to sign T e t from zero to high over too minus to go from zero to high over too. You do thi co scientist t you see which is equal to two pi over too minus use integration. My parts again we have This is equal to coast science t you do you? I'm zero to high over shoo and plus into girl from zero to high over too. T sign of tea. Did he? This part is he hurt you next you want you're this part in same. I had this part. So we have integral from zero to high over too. You could be a sign of heat heat. This is equal to you. Too high over too. My last one over two. So the area off this part A rear. It's equal to you too, Pai. Over to us. One over two, minus a rear of this part. Which is you go to one. So the answer is into high over too minus one over two, which is about 1.90 foul.


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