In Exercises 71–76, eliminate the parameter and graph the plane c...

Question

In Exercises 71–76, eliminate the parameter and graph the plane curve represented by the parametric equations. Use arrows to show the orientation of each plane curve.

x = 3 + 2 cos t, y = 1+2 sin t; 0 ≤ t < 2π

Accepted Answer

Write the corresponding rectangular equation for the following parametric equation by eliminating T draw a graph of the plane curve using a rectangular equation indicate direction of the curve using errors. You have X equals one plus cosine T Y equals five plus seven ST on the interval 0 to 2 pi for T were also given a court in the plane to Grafton. Now to solve this, we first need to solve in terms of tea, we have X equals one plus seven plus nine T. First, we can subtract one, get X minus one equals seven cos nine T and then divide both sides by seven, we get X minus one divided by seven equals cosine T. Now we will leave this as it is for. Now, we also have our Y versus Y equals five plus seven sine T subtracting five to get Y minus five equals seven sine T. Finally dividing by seven to get Y minus five divided by seven equals sine T. Now, one of our trick properties we know is that if we have square of T plus sine squared T, this is equals to one. So we can make use of this identity, we will get X minus one divided by seven squared plus Y minus five, divided by seven squared equals one. We will get X minus one squared divided by plus Y minus five squared divided by 49 equals one. And finally multiplying both sides by 49 we get X minus one squared plus Y minus five squared equals 49. Now, this equation is in rectangular and this is actually the equation of a circle. So our graph will be a circle. Now, we just need to draw a graph by finding some points. Let's make a table txny for T we will use zero pi divided by two pi and three pi divided by two. You know X is one plus seven plus nine, Y is five plus 7 70. Now, we can plug in our values plugged in T 04 X. We will get eight that we can continue for every one of these angles getting one negative six and one same for what you will get. 5 12, 5 and negative two. Now, we have all points we can use in our graph. We have a point at 85 1st over eight and the X and we go up five 1 12. So we have over one, up 12 now 65, so little less six of five and finally one negative two. Now I did mention this graph was a circle based on our equation. So we will just make a circle between these four points in this direction goes counterclockwise. We can draw arrows in the counterclockwise direction. Ok? I hope they help you solve the problem. Thanks for watching. Goodbye.

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In Exercises 71–76, eliminate the parameter and graph the plane curve represented by the parametric equations. Use arrows to show the orientation of each plane curve. x = 3 + 2 cos t, y = 1+2 sin t; 0 ≤ t < 2π