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iceman713
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Homework Statement
http://www.nellilevental.com/calc1.jpg
Homework Equations
C1 = [tex]\left(x - 1\right) ^{2} + y^{2} = 1[/tex]
C2 = [tex]x^{2} + y^{2} = r^{2}[/tex]
The Attempt at a Solution
Initially I thought that the "value" of R was just going to increase without bound since the slope of the line was going to get more and more parallel to the X-axis. The skeptic in me doesn't think the answer would actually be that obvious though. So this is what I did:
I found the expressions for both "upper halves" of the circles, and set them equal to each other to find the general form of the point of intersection Q
[tex]C1_{upper} = \sqrt{1-(x-1)^{2}} = C2_{upper} = \sqrt{r^{2}-x^{2}} [/tex]
[tex]\sqrt{1-(x-1)^{2}} = \sqrt{r^{2}-x^{2}}[/tex]
[tex]1-(x-1)^{2} = r^{2}-x^{2} [/tex]
[tex]1-(x^{2}-2x+1) = r^{2}-x^{2}[/tex]
[tex]2x = r^{2}[/tex]
[tex]x = \frac{r^{2}}{2}[/tex]
so that gives me the x-coordinate of Q as a function of r, and I just plugged that back into C2upper to get the y-coordinate which gives me the coordinates of Q as:
[tex](\frac{r^{2}}{2}, \frac{\sqrt{4r^{2}-r^{4}}}{2})[/tex]
I then used that and the initial point of (0,r) to find the slope of the line PR, and since I already had the y-intercept I got this as the equation of the line PR:
[tex]Y = X(\frac{\sqrt{4r^{2}-r^{4}}-2r}{r^{2}})+r[/tex]
THEN, I set that equal to 0 and solved for X to get an expression for the X value of R(the x-intercept of the line) as a function of the radius of the initial shrinking circle and got:
[tex]X=\frac{-r^{3}}{\sqrt{4r^{2}-r^{4}}-2r}[/tex]
Now, if I graph that, it seems to work, at r = 2, the x value is 2 which is correct, and it seems like as r->0+ the limit of R is 4
At this point I just have no idea whether I'm right or not, 4 seems awfully small compared to the initial infinity I thought it would approach, but what the **** do I know, and it's an even problem so I feel like I'm SOL, which is why I come here
the book is James Stewart Single Variable Calculus 6e, ch 2.3 exercise 60
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