Anonymous From: - Posts: - Votes: - |
I guys i am kind of new. I am in calculus right now, and i amnot sure if this problem is supposed to drive me crazy, but it isdoing a good job. If i could get any sort of help with it thatwould be lovely. And if you get help from anyone, could you please site yoursources. A circle of fixed radius is drawn centered at the point (1,0).(reacall that the equation for such a cirlce is (x-1)2 +y2 = 1). A second circle of changing radius is drawncentered at the origin. (if we call the radius of this secondcricle r the equation for this circle is x2 +y2 = r2). call the yintercept of this second circle P and the upper point ofintersection of these two circles Q. Finally let R be the xintercept of the line PQ. What happens to R as r shrinksto zero? In words, what are the linits of the coordinates of R asthe radius of the second crilce approaches zero? wish i could show the picture...but i can't..my computer isn'tvery nice |

Anonymous From: - Posts: - Votes: - |
This is actually sort of an interesting problem compared towhat most people ask on this board. The answer is 4. Here is howyou get it: 1) First solve for the x coordinate where the two circlesintersect solving for x in (x-1)2+y2-1 =x2+y2-r2which gives you x =(r^2)/2 Plug this into the equation on right hand sideand solve for the y coordinate where the two circles intersectwhich is found to be y = (r*Sqrt[4 - r^2])/2 2) Now youcan solve for the slope s of the straight line connecting thepoints P and R using the equation: (r*Sqrt[4 - r^2])/2 = r + (r^2*s)/2, so s = (-2 + Sqrt[4 -r^2])/r. 3) Now youhave the equation of the line PR which is y = r + ((-2 + Sqrt[4 -r^2])*x)/r. Set y = 0 in this equation and you can solve forthe value of the x intercept which is r^2/(2 - Sqrt[4 - r^2]). That's what we needed. Nowall you need to do is the take the limit of this as r goes to 0using L'Hospital's rule and you get the answer of4. Note thatthis solution depends of the fixed circle having a radius of 1which it what is has in the equation stated in the problem. If theradius of the fixed circle is changed to some other value, or somearbitrary number R, you'll need to work through the same logicagain. Goodluck! |

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