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Parametnc equationsqiventanit}cot(t), 0 <t < 7 (a) Sketch the curve represented by the parmetrc cquaoons Use amtowys ncrease5Indicate the airectionthe cunve a...

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Parametnc equationsqiventanit}cot(t), 0 <t < 7 (a) Sketch the curve represented by the parmetrc cquaoons Use amtowys ncrease5Indicate the airectionthe cunve a5 (Find reccanqular-coordinale equatlon for the cuNvceliminating the parmeternere

parametnc equations qiven tanit} cot(t), 0 <t < 7 (a) Sketch the curve represented by the parmetrc cquaoons Use amtowys ncrease5 Indicate the airection the cunve a5 ( Find reccanqular-coordinale equatlon for the cuNvc eliminating the parmeter nere



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Use graphing technology to sketch the curve traced out by the given vector-valued function. $$\mathbf{r}(t)=\langle 8 \cos t+2 \cos 7 t, 8 \sin t+2 \sin 7 t\rangle$$

We are given parametric equations. They are X equals to plus five co sign data and y equals negative. Six plus four signed data. In part. They were asked to sketch the curve represented by these equations and to indicate the orientation of this curve. To do this, I'm going to make a table of values. Table will have three rows data X and Y Let's say to begin with zero and go in steps of higher or for all the way up to two pi. Yeah, Now, when theta is equal to zero, X is equal to two plus five times one or seven when data is equal to zero. Why is equal to negative six plus zero for negative six and they equals pi over four. X is two plus five times 2/2. So two plus five halves through to and why is equal to negative six plus four times route to over two negative six plus two or two? Yeah, when data is pi over, two X is equal to two plus zero or two, and why is equal to negative six plus four or negative, too, when data is equal to three pi over four x is equal to two plus five times negative 2/2. This is two minus five Hades route to and why is equal to negative six plus four times route to over two. This is native six plus two or two. When theta is equal to PI, X is equal to two plus five cents. Negative one or two minus five or negative three. And why is equal to negative six plus four times zero or negative six. When theta equals five pi over four X is equal to two plus five times negative route to over tombs. This is two minus five halves times route to and why is equal to negative six plus four times negative route to over two. This is negative. Six minus to route to when data is equal to three. Pi over two x is equal to two plus five times zero for two and why is equal to negative six plus four times negative one or negative 10. And when data is equal to seven, pi over four x is equal to two plus five and grew to over two or two plus five Hades route to and why is equal to negative six plus four times negative route to over two. This is negative. Six minus two Ruutu. Now I have 88 points to plot. Mhm. I'll let X range from negative three up to positive seven, and I'll let wide range from negative 10 up to negative two. Yeah, so in essence, I really let wide range from zero to negative. 10. Yeah, All right. So the point seven negative six it's located about here. 0.2 plus five halves. Route to well route to is about 14 is this is 5 10 7 about 35 plus two, which is about 5.5 and then negative six plus to route to Well, this is approximately negative. Six plus two times 1.4, which is negative. Six plus 2.8, which is negative. 3.2 to the point at about 5.5 negative 3.2. We also have a point at two negative two located here point at two minus five halves. Route to this is approximately two minus. What was this? Five halves of 1.4. So two minus 3.5, which is about negative 1.5 and then negative six plus two route to this is about negative. Six plus 2.8, which is we already saw. This was 3.2. So appointed about negative 1.5 negative. 3.2 then we have a point at negative three negative six located about here and then appoint at two minus five halves or two. Negative six minus two. Route to so to minus five. Hazard to is about negative. 1.5 and negative six minus two or two. This is about negative six minus two point eight, which is about negative. 8.8 pointed about negative 1.5 negative. 8.8 then we have a point at to negative 10. Okay, it's about here. And finally a point at two plus five. Hazard. Too negative six minus two or two or at about 5.5 negative. 8.8. And so the graph of our equation looks something like this, so it looks like an ellipse centered at approximately two negative six. The orientation is determined by the direction of increasing data. We see that as data increases from zero well X's decreasing and why is increasing so we start out at the 0.7 negative six and then we moved up into the left until I reached the point two whatever. And then we start moving down into the left and we continue in this way until we get back to our starting point. In part of these, were asked to eliminate the parameter and write the result in rectangular equation. This graph represents this curve and to adjust the domain if necessary. Ordinarily, to eliminate the parameter, we would try to solve one of the equations for that parameter. However, because we have triggered a metric functions here, solving for data is not simple. Instead, let's try to use trig identities such as Pythagorean Theorem. So if we sell for co sign data, we have chosen data is X minus 2/5, and likewise we have that Y plus 6/4 is signed data. Therefore, X minus 2/5 squared plus y plus 6/4 squared is equal to co sign square data plus sine squared data or once so X minus two squared over 25 plus y plus six square over 16 equals. One is the equation, and we've eliminated are parameter data. Notice that the graph of this equation is in fact an ellipse centered at the point to negative six. And so the graph of this equation represents the curb.

Okay. Forgiven Parametric equations. These are X equals one plus co sign data and why equals one plus two signed data in part. They were asked to sketch the curve represented by these Parametric equations and indicate the orientation of the curve. To do this, I'll make a table of values. Yes, Data X and Y, as the rose begin with data equals zero and I'll go up to data equals two pi. So have data, then pi over two and pie, then three pi over two. You know what? Let's be a little more granular. Yeah. Instead, I'm going to do pi over four, then pi over two three pi over four pie five pi over four three pi over two and seven pi over four. After that, the values will repeat themselves. Okay, So when data is equal to zero, X is equal to one plus one or two and why is equal to one plus zero or one and data equals pi over four. X is equal to one plus route to over two. And why is equal to one plus two times through to over two or one plus route to and data equals pi over two X is equal to one plus zero or one, and why is equal to one plus two or three, and they d equals three. Pi over four X is equal to one minus with 2/2. Why is equal to one classroom to, and that equals Pi X is equal to one minus one or zero. Lie is equal to one plus zero or one data equals five. Pirates four X is equal to one minus route to over two. And why is equal to one plus two times negative route to over two. This is one minus route to, and data is equal to three. Pi over two X is equal to one plus zero or one. Why is equal to one minus two or negative one? And when data is equal to seven, fire before X is equal to one plus 2/2. And why is equal to one minus route to mhm? With these eight points, I'll plot the graph, so I'll let X range from I get zero up to two, and I'll let y range from negative one up to three. Mhm. So the first point to one it's located here. The next point one plus route to over to this is about one plus 0.7 or 1.7 and then one plus route to is about one plus 1.4 or 2.4. We have 1.72 point four located about here. Mhm. The next point is 13 located about here. Next point is one minus. Route to over to this is about one minus 10.7 or 0.3 and one plus route to, which is about 2.4. And we have 01 located out here than one minus route to over two and one minus or two. This is about 0.3 and one minus 1.4 is about negative 0.4. So this is located out here. Next, we have one negative one about here and then one plus route to over 21 minus route to this is again approximately 1.7 and 0.3. Sorry. Negative point negative 0.4. And then we return to 21 And so the graph looks something like this. It's a little bit rough, but this looks like the graph of a circle centered at 11 with a radius of one. The orientation is in the direction of increasing data. Well, we see that starting at 30 equals zero. As data increases X decreases in, Why increases? So starting at the point to one, we move up and to the left. And then once you reach the 0.13 we start moving down into the left until I reached the point 01 When we start moving down into the right until we reach the 10.1 negative one. When we start moving up into the right until we reach to one and we repeat in part B were asked to eliminate the parameter and write the result in rectangular equation whose graph represents this curve, adjusting the domain if necessary. Ordinarily, we want to solve one of these equations for data. However, because we have to go metric functions, this would be hard to do. Instead, let's try using an identity like the Pythagorean theorem. So if we solve for coastline data, see, the X minus one equals co sign data solving for sine theta. We see that why minus one over to equal sign data and therefore X minus one squared plus y minus one over to squared is equal to co sign square data plus sine square data, which is one and so we have X minus one squared plus why minus one squared over, two squared, which is four equals one. So I actually made a mistake before. If we really looked at the axes with the same scale, it would look more like an ellipse. And in fact, the graph of this equation is an ellipse centered at 11 and so the graph represents the curve.

You're given Parametric equations. These are X equals co sign of data. Why equals to sign of tooth data? Yeah. In part A were asked to sketch the curve represented by these Parametric equations and indicate the orientation of this curve. Mhm. Yes. I'm going to make a table of values Table will have three rows Data X and Y Mhm. Yeah. So all that data range from zero of two. Let's see higher four and pi over two, three pair before and pie and five pi over four, then three pi over two, then 7.3 or four. So we'll have eight points. Yeah. Now when data equals zero, X is equal to one and why is equal to zero when data is pi over four X is equal to Route two of her too. And why is equal to two times the sine of pi over two, which is two when data equals pi over two X is equal to zero y is equal to two times the sine of pi which is zero when data equals three pi over four x is equal to negative route to over two and why is he too two times the sine of three pi over two. This is negative. Two. When theta equals pi X is equal to negative one, and why is equal to two times the sine of two pi, which is zero and theta is five. Pi over four x is equal to negative route to over two and why is equal to two times the sign of five pi over two. This is the same as two times a sign of prior to or two when data is equal to three. Pi over two x is equal to zero. Why is equal to two times the sine of three pi, which is zero? Finally, when fate is equal to 75 or four X is equal to route to over two and why is equal to two times the sine of seven pi over two? This is the same as two times the sine of three pi over two, which is negative. Two. Once we reach two pi, we began all over again, so I'll let X range from negative one negative two, I guess, to positive two and walked negative to the positive, too. Route two of her, too to well route to, is about 1.4 is This is about 0.72 It's located about here. Next point is 00 located here, and we have negative route to over to This is about negative 0.7, negative two located about here. The next point is negative. 10 Okay, it's about here. Then we have negative route to over 22 located about here. Then 00 just here again. Finally grew to over two negative two located about here. So we began here, then removed here, and we moved to here, then here and here. Then here and here. Then here and then here. So we get what looks like the graph with two lobes. Now, the orientation of this graph is in the direction of increasing data. So if we begin with data equals zero, you see that as data increases, X is decreasing. And why increases so beginning at the 0.10 we move up into the left. Then once we reach route 2/2 two, we start moving down until left until we go through the origin. Continue moving down into the left until we reached the point negative route to over two negative two when we start moving up into the left until I reached the point. Negative one zero. When we start moving up into the right, encourage the point negative route to over 22 We start moving down into the right until he passed through the origin and continue moving down into the right until you reach negative route to over. Sorry. Route to over two. Negative two when we start moving up into the left. Sorry. Up into the right. And then once we reach 10 we start over all over again. Okay, then in part B, we were asked to eliminate the parameter and write the resulting rectangular equation whose graph represents this curve adjusting the domain if necessary. So ordinarily, what we would do is try to solve one of these equations for data. However, since both equations have trig and metric functions, this doesn't seem like a good idea. Instead, let's try to use identities. Excuse me. So what do we know about this sign of tooth data? Or we can leave that as it is. But if X is equal to co sign data, well, we know that co sign of tooth data this is CoSine squared, data minus sine squared data. Or this is to co sign square data minus one. And looking at Parametric equations. This is two X squared minus one. So we have that two X squared minus one squared plus y squared. This is equal to cosine squared of two theta plus sine squared of tooth data. And since the arguments are the same for co sine squared in science squared by the path agree and theorem, this is equal one. And so we've eliminated are variable data. And so the equation is two x squared minus one squared plus y squared equals one. This is not an equation you're probably familiar with. It's not a comic section, but if you do plug this into I'm sorry I made a mistake here. This should be plus why over two squared equals science. Great data. This is because why over two is equal to sign of tooth data, and so we have our final equation is y squared over four. But if you do plug this equation into a graph, you will get a two lobed graph like we have up here in part A. And therefore, the graph of this equation represents the curve

Were given parametric equations. These are X equals to co sign data. Why is equal to three signed data In part A. We are asked to sketch the curve represented by these equations and to indicate the orientation. To do this and going to make a table of values, the table will have three rows data thanks. And why? Select data range from zero and then hi over to hi three pi over two and two pi. So in beta zero X is to y zero and data is tired for two X zero and Y is three and data is pie X is negative. Two y is zero when data is three pi over two x zero and why is negative? Three. Yeah, we're very equals two pi. This is actually the same as the first point Now applied these points I'll let x range from negative to deposited to. In fact, I'll let both axes range from negative three. Positive three. The point 20 lies in the curve as the 03 negative 20 zero negative three and back to zero. So we keep going in a cycle. And so the graft looks something like this And so we see that the graph is something like in the lips. The orientation goes in the direction of increasing data. You start at day equals zero, starting at the point 20 We see that as data increases, X is decreasing and why it increased, removed up and to the left. And then once we reach the 0.3 we start moving down into the left and go on until we get back to the point 20 where we start over again. All right, then in part B were asked to eliminate the parameter and write the result in rectangular equation, whose craft represents this curve to eliminate the parameter. It doesn't seem wise to try to solve for data like we do in some other problems. Instead, let's try to use the Pythagorean theorem. So we know that except for two is equal to co sign of data and that y over three is equal to the sign of data and therefore it follows the X over two weird plus y over three squared equals co sine squared data plus sine squared data which bypassed, agree and theorem is one and therefore X squared over four plus y squared over nine is one. And this is the equation of an ellipse with a vertical major axis of length six and a minor axis of length to centered at the origin. And therefore, the graph of this equation represents this curve.


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