1.
Determine the area bounded by the x-axis, y-axis, y=12, and perimeter of the conic section. The equation of the conic section is provided below. Equation : (x-15)^2 – r^2 = -(y-12)^2 Where r is the positive square root value of D rounded to the nearest whole number Use the equations on the next slide to determine D
3.
First of all, we must understand what the question is asking for. We must find the area, which is bounded by these four things: x-axis y-axis y=12 perimeter of (x-15)^2 – r^2 = -(y-12)^2
4.
The x-axis is the horizontal axis. The y-axis is the vertical axis. The equation y=12 is just a simple linear equation where y is Always going to be equal to 12 no matter what value x is. As for the perimeter of the equation, we must first identify the Shape of this conic section. X-axis Y-axis Y=12
5.
<ul><li>Because we see r^2, we might say that this equation is for a circle. Let’s go through the requirements of a circular equation to prove that this equation is in fact a circle. </li></ul><ul><li>A number not multiplied by x or y is positive when isolated from the rest of the terms </li></ul><ul><li>(x-15)^2 + (y-12)^2 = r^2 </li></ul><ul><li>TRUE since r^2 is positive when isolated </li></ul><ul><li>Both x and y are to the exponent 2 </li></ul><ul><li>(x-15)^2 + (y-12)^2 = r^2 </li></ul><ul><li>TRUE </li></ul><ul><li>The factors of compressions/stretches for x and y are the same </li></ul><ul><li>(x-15)^2 + (y-12)^2 = r^2 </li></ul><ul><li>TRUE since the factors are only 1 </li></ul><ul><li>Eg. 5x^2 + 5y^2 = r^2 where the factors are 5 </li></ul><ul><li>Both x and y are positive when isolated from a positive number </li></ul><ul><li>(x-15)^2 + (y-12)^2 = r^2 </li></ul><ul><li>TRUE since everything is positive when r^2 is isolated </li></ul>
6.
Given by our proofs, it is safe to say that the equation we have is a circular one. We must obtain the perimeter (circumference) of this circle because it is necessary to find the area that which it helps bound. We do not have sufficient information though to construct this circle. However, we can obtain sufficient information if we work out the four equations because of this specific piece of information: Equation : (x-15)^2 – r^2 = -(y-12)^2 Or (x-15)^2 + (y-12)^2 = r^2 Where r is the positive square root value of D rounded to the nearest whole number So let us begin by analyzing the 4 given equations
7.
Here are the four equations. The first equation consists of only one variable, A. The second equation consists of three variables, A, B, and C. The third equation consists of four variables, A, B, C, and D. The fourth equation consists of four variables, A, B, C, and D. (e is not a variable) Now we can just think of ways to solve these equations. We shall start with the first equation.
8.
Let’s go through the first equation. Step 1 We have to simply everything as much as possible to obtain a. If you look at the base, there is the letter to a certain exponent. This should be familiar, otherwise, just look at Euler’s Identity, which is shown below. If you transpose the one, e^[i(pi)] will equal -1. So now we can just substitute what we have in the equation for -1. The base of the logarithm should now be (-2)(-1) which equals 2.
9.
The 7/2 is already simplified, so we leave that alone and proceed to simplifying the exponent (right hand side of equation. Step 2 By looking at this, you should easily identify certain trigonometric identities. The denominator is not pleasant looking, so we can expand it into (a+1)(a-1) You should notice that there is only one term in the numerator. Everything in the numerator is being multiplied.
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Step 3 You can find (a+1) and (a-1) in the numerator, which can be reduced to 1 from (a+1)(a-1) in the denominator. There is sin(a) and csc(a) which both reduce each other to 1 because csc(a)=1/sin(a) and 1/csc(a)=sin(a). That leaves us with sec(a)-tan(a), which amazingly also equals 1. As a final result, the right hand side should be 2a(1) which equals 2a. Thus, providing us with this equation. The Base is 2, the argument (power) is 7/2 and the exponent is 2a. Let’s convert this logarithm into its exponential form because our calculator can only calculate logarithms with the base of 10. We should obtain this.
11.
Now, we can convert both sides of the equation into logarithmic form with 10 as the base. This will allow us to isolate a and use our calculators. Okay, the base is not “2.” It is 10, so do not get fooled. A logarithm is an exponent, so what does it mean when an exponent is being multiplied? Well, it obviously means the exponent has its own exponent. In this case, 2a is the exponent, so we multiply it by Log2. Now we can isolate a by dividing both sides of the equation by 2 and Log2. That is the last step.
12.
Now it is time to look at the second equation we were given. Check this beastly equation out. It’s going to require knowledge on integrals and differentials. Just joking, it obviously does not, but I wish. Since this equation is extremely long, we will simplify the equation in a manner that goes through sections of the equation. We shall begin this part. We can’t do anything with a. As for the next term, we see the exponent -1. Because the (-1/3) is to the exponent -1, we just obtain the reciprocal of it. Negative exponents just make switch bases into their reciprocal form. For example, 10^-3 = (1^3)/(10^3).
13.
Now let’s do this part of the equation, which is actually an exponent. This part is perhaps the most difficult one out of this whole problem. J/K!!! Okay, keep in mind that this is an exponent. We see cos(-a), but it not cos(a) which is 1/sec(a). By this, I was thinking of reducing sec(a) and 1/sec(a).This is the basic cosine graph and not the actual cosine operation we see in the equation. As you can see, it says cos(-a). The (-) simply flips the graph horizontally. If we flip this graph, it is exactly the same as it was before, so we can say that Cos(-a) = Cos(a)
14.
This gives us this new equation, which simplifies into this by multiplying cos(a) and sec(a) together. We are not done simplifying. Through the most primitive Pythagorean Identity: We can say that ( 1-cos^2(a) ) = sin^2(a). Because it is in front of (-), the negative symbol, we now have –sin^2(a). The whole thing should equal 1 since sin^2(a) - sin^2(a) = 0 leaving 1 by itself. 1*1=1, so the exponent is just 1.
15.
We still have a chunk left to simplify. Our current equation should be this: This is the numerator of the fraction in this equation. I see a lot of trigonometric identities. It is kind of hard to explain the reasoning behind manipulating identities. You just have to use your head and base your luck on chance!!! You can solve this stuff in myriads of ways, so I’ll just take you guys through a concise simplification of the numerator.
16.
The positive symbols next to the radicals (square root symbols) indicate that the square root value is positive. I aligned everything so that every vertical column has the same value. Notice that I input unnecessary brackets in cos(c+∏). The only puzzling thing should be –cos(c). Let’s take a look at a graph of the basic cosine function. F(c) = cos(c)
17.
F(c) = cos(c) Now let’s view –cos(c) and cos(∏+c) cos(∏+c) – cos(c) Alrighty then, –cos(c) and cos(∏+c) are exactly the same thing. To obtain –cos(c), cos(c) was just flipped vertically, not horizontally. The value of parameter a is (-1). As for cos(∏+c) , cos(c) was just shifted to the left by ∏.
19.
So “1” is our numerator, now let’s find the denominator. We can see that Euler’s Identity is involved in this, as well as factorial notation. How about we factor out the a!b!c! ! Hold up! This is Euler’s Identity which equals ZERO. The denominator is zero then, so all our hard work finding the numerator has been wasted. Oh well, +0 is what we found for this section of the equation.
20.
Our current equation should be this: It is the same as it was previously. This is the last section of the equation. We can’t do anything with it, so let’s obtain our final equation. This is our final equation. We still have 2 more to go though.
21.
Here are all of our equations. We had four equations earlier, but the first one was just used to isolate A. Based on the looks of these three equations, we can apply “systems of equations.”
22.
I assigned each of the equations numbers to make things easier. When doing systems of equations, you stack two or more lines in a manner that is similar to addition. However, when the lines are added, at least one variable should be removed. The next slide will show this. Oh and you must use 2 combinations of 2 equations where there is one common equation within each combination. By this, I mean 1,2 and 1,3 go together, 2,1 and 2,3 go together, and 3,1 and 3,2 go together. I’m going to do 1,2 and 1,3. 1 2 3
23.
1,2 (equations 1 and 2) It is basically addition, except there are columns assigned for each term where you can’t carry numbers. The left side shows the original equations and the right side shows the new equations. I decided to get rid of b because I felt like it. In order to get rid of b, one line must consist of (+b) while the other consists of (-b). For the first line, the factor I multiplied everything by to achieve -6b was 2. For the second line, the factor I multiplied everything by to achieve +6b was -3.
24.
1,3 (equations 1 and 3) B has to be removed since it was removed earlier. The factor for the first line is 5 and the factor for the second line is 3. Now we use 1,2 and 1,3, except their next equations. The top factor is just 1 and bottom one is (-1). We have to remove another variable. I chose d. We shouldn’t choose “a” because we found “a” earlier. Although, we could have plugged in “a” earlier with our value we found, but that is just another method to solve things. Hopefully you understand that or learned from that. This just means that we would have one less variable to worry about.
25.
Now we add the 2 lines, and obtain a new equation! These are our original three equations. a = 0.903677461 It is substitution and more isolation time !
26.
Let’s Find C a = 0.903677461 We use this equation, which was our last equation obtained from using systems of equations, Because it only has the variable, which we need to find. We already have “a.” Now let’s substitute “a” in. If you do everything properly, then: c = 24.51190959 Let’s Find B We use this equation Because we already have “a” and “c.” We don’t have “d,” so the other equations are useless. If you do everything properly, then: b = 39.9877418
27.
We now have three values. Let’s find D a = 0.903677461 We use any 2 of the equations to find D. If we use both, the resultant value of D will be the same. Let’s try one of them out. Note that if you use both of the equations, then it makes sure that your values are correct. We shall use this equation. c = 24.51190959 b = 39.9877418 (7)(0.903677461) + (2)(39.9877418) + 24.51190959 = D D = 149.1378318
28.
Woohoo! Pat yourself on the back! All we did was find d, which is used to find what the question is asking for. Woo, what a long question this is! This takes mighty long! You deserve a certificate for getting this far.
29.
This is what the question stated earlier: D = 149.1378318 R is the positive square root value of D rounded to the nearest whole number. We round the square root value and not D itself. So basically, the square root of D = 12.2474. When that is rounded to the nearest whole number, we obtain 12. Therefore, r=12 Let’s plug that into our circular equation (x-15)^2 + (y-12)^2 = 144 So now, let’s graph this relation, not function!
30.
The vertex is at (15,12). The radius is 12. Based on those two bits of information, we can easily construct this circle below. (x-15)^2 + (y-12)^2 = 144
31.
The vertex is at (15,12). The radius is 12. Based on those two bits of information, we can easily construct this circle below. Great! We have the circle. The question wants us to find the area bounded by this circle, the x-axis, the y-axis, and line y=12. The green is the area we must find. We can solve this several ways.
32.
Analyze the diagram first Here is my way to solve it. I see that there is a rectangle formed, so I obtain the area of that. After that, I subtract ¼ of the area of the circle from the rectangular area. You can see that ¼ of the circle occupies an area of the rectangle. The rectangle has a base of 15 units and height of 12 units, which equals the radius of the circle. You can also see that the radius touching and perpendicular to the x-axis is what creates the rectangle. That, along with y=12 creates the rectangular shape.
33.
Analyze the diagram first Area of rectangle = 15 units x 12 units = 180 units^2 Area of circle = ∏r^2 = ∏(12 units)^2 = 452.3893421 units^2 Bounded Area = Area of rectangle - (¼) Area of circle = 180 units^2 – (¼)(452.3893421 units^2) = 66.9027 units^2 Therefore, the area bounded is 66.9027 units^2.
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