41–44. Intersection points and area Find all the inters...

Question

41–44. Intersection points and area Find all the intersection points of the following curves. Find the area of the entire region that lies within both curves

r = 1 + sin θ and r = 1 + cos θ

Accepted Answer

Hello. In this video, we want to find the intersection points of the given polar curves. The polar curves given to us is R is equal to 3 + 3 sine of theta, and the second polar curve given to us is R is equal to 3 + 3 cosine of theta. Now, in this case here, in order to find the points of intersection, recall that the points of intersection between any two curves are the points where the curves are equal to each other. So in order to start finding the points of intersection, the first thing we want to do is we want to go ahead and set the equations equal to each other. That is going to give us 3 + 3 sin of theta. Equal to 3 + 3 cosine of theta. Now since we have a constant of 3 on both sides of this equation, we can start simplifying this by subtracting 3 to both sides. That will leave us with 3 sine of theta equal to 3 cosine of theta. We can simplify the coefficients of 3 by dividing both sides of the equation by 3. Again, that will leave us with sine of theta equal to cosine of theta. Now we can further simplify this equation by dividing both sides of the equation by cosine of theta. If we do this, the right hand side of the equation will equal to one, and the left hand side will be sine of theta divided by cosine of theta, but that matches the identity for tangent, so the left hand side can be rewritten as tangent of theta. Now the question here is what angles of theta allow the tangent function to be equal to one. Well, if we take a look at the unit circle, tangent of theta is equal to 1 whenever sine and cosine have the same value, and there's only two angles on the unit circle where sine and cosine are equivalent to each other. The first angle is that theta is equal to pi divided by 4. And the second angle is theta is equal to 5 pi divided by 4. But keep in mind that the original problem did not give us a restriction for the value for the domain of the functions. So, what we want to do is we want to go ahead and restrict our solutions between 0 and 2 pi. But we also have to recall that tangent has a period of just pi. So the general solution for this equation is going to be theta is equal to multiples of pi divided by 4, or in other words, theta is equal to pi divided by 4 plus pik. So, what we need to do now is we need to go ahead and evaluate the values of i at the solutions of pi divided by 4 and phi pi divided by 4. So, let's go ahead and start with the first solution when theta is equal to pi divided by 4. Now recall that the general form of a polar point or a coordinate on the polar axis is going to be R, theta. So what we want to do is we want to solve for the radius r when theta is equal to pi divided by 4. Now, if we take this value of theta and plug this back into the first polar equation, we get R is equal to 3 + 3 sin of pi divided by 4. Now sin of pi divided by 4 is going to simplify to be square root 2 divided by 2. So we have 3 plus 3 multiplied by square root of 2 divided by 2. Further simplifying this is going to be 3 + 3 square root of 2 divided by 2. And this will be the value of R. So what this means is that the first intersection point for the polar curves is going to be located at 3 + 3 square root of 2 divided by 2, pi divided by 4. This will be the first solution to the problem. Now we need to find the 2nd solution, and again, as previously stated, the 2nd solution will exist when pi is equal to 5 pi divided by 4. Now, if we were to plug in this value of 5 pi divided by 4 into the first polar equation, we get R is equal to 3 + 3 sin of 5 pi divided by 4. The sin of phi pi divided by 4 simplifies to square root 2 divided by 2. So we have 3 + 3 multiplied by the negative square root of 2 divided by 2. And this will further simplify to be 3 minus 3 square root of 2 divided by 2. And so what this means is that the second point of intersection is going to be located at 3 minus 32 root of 2 divided by 25 divided by 4, and this is going to be the second solution to the given problem. So in total, there these are going to be the two points of intersection between the two polar curves, and the overall solution to this problem is going to be option A. So I hope this video helps you in understanding on how to approach this problem, and we will go ahead and see you all in the next video.

41–44. Intersection points and area Find all the intersection points of the following curves. Find the area of the entire region that lies within both curvesr = 1 + sin θ and r = 1 + cos θ

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41–44. Intersection points and area Find all the intersection points of the following curves. Find the area of the entire region that lies within both curves

r = 1 + sin θ and r = 1 + cos θ