- #1

beneakin

- 13

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1. Homework Statement

http://www.nellilevental.com/calc1.jpg

2. Homework Equations

C1 = \left(x - 1\right) ^{2} + y^{2} = 1

C2 = x^{2} + y^{2} = r^{2}

3. 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

C1_{upper} = \sqrt{1-(x-1)^{2}} = C2_{upper} = \sqrt{r^{2}-x^{2}}

\sqrt{1-(x-1)^{2}} = \sqrt{r^{2}-x^{2}}

1-(x-1)^{2} = r^{2}-x^{2}

1-(x^{2}-2x+1) = r^{2}-x^{2}

2x = r^{2}

x = \frac{r^{2}}{2}

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:

(\frac{r^{2}}{2}, \frac{\sqrt{4r^{2}-r^{4}}}{2})

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:

Y = X(\frac{\sqrt{4r^{2}-r^{4}}-2r}{r^{2}})+r

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:

X=\frac{-r^{3}}{\sqrt{4r^{2}-r^{4}}-2r}

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

dynamicsolo

Jan11-08, 01:53 PM

the book is James Stewart Single Variable Calculus 6e, ch 2.3 exercise 60

Heh, I recognized this problem right away, since that's the book we use where I am... (BTW, it's exercise 62.)

C1 = \left(x - 1\right) ^{2} + y^{2} = 1

C2 = x^{2} + y^{2} = r^{2}

I see no problems with your derivation of the x-intercept.

I was going to suggest a tip here on solving for the upper intersection point of the two circles. If you expand the first equation as

(x^{2} - 2x + 1) + y^{2} = 1, giving

x^{2} - 2x + y^{2} = 0 ,

you can immediately substitute the second equation into this to obtain

r^{2} - 2x = 0 .

In any event, looking at your result for the intercept

X=\frac{-r^{3}}{\sqrt{4r^{2}-r^{4}}-2r} ,

the limit as x->0 gives you zero in the denominator and an indeterminate ratio 0/0 for the limit.

Given that you're in Section 2.3 of Stewart, you're past the point where you're allowed to just graph the function; you'll have to actually solve this using a technique you've learned. To make life a bit easier, factor out an "r" from the numerator and denominator to get

\frac{-r^{2}}{\sqrt{4 - r^{2}} - 2}

and using the "conjugate factor" method by multiplying the numerator and denominator by \sqrt{4 - r^{2}} + 2 . This will help you get rid of the troublesome denominator and give you something easy to take the limit of for x->0.

Having taught "Calc-One" recently, I happen to have the solutions manual for Stewart here and find that there is also a nice completely geometric solution given. But you're learning about limits, so go to it... (The limit is indeed X = 4 , which does seem counterintuitive.)

ice109

Jan11-08, 03:26 PM

can someone give a clear geometric argument for why this is true?

mda

Jan11-08, 04:15 PM

can someone give a clear geometric argument for why this is true?

Once you have the intersection point Q, it is a case of constructing similar triangles.

Q is approximately (1/2 r^2, r).

Taking angle POQ we bisect and intersect with PQR, calling this point S say.

S is approximately (1/4 r^2, r).

OS is perpendicular to PQR, so OSP and RPO are similar triangles.

Therefore OR = OS OP/PS = r r / (1/4 r^2) = 4.

iceman713

Jan11-08, 04:34 PM

Heh, I recognized this problem right away, since that's the book we use where I am... (BTW, it's exercise 62.)

I see no problems with your derivation of the x-intercept.

I was going to suggest a tip here on solving for the upper intersection point of the two circles. If you expand the first equation as

(x^{2} - 2x + 1) + y^{2} = 1, giving

x^{2} - 2x + y^{2} = 0 ,

you can immediately substitute the second equation into this to obtain

r^{2} - 2x = 0 .

In any event, looking at your result for the intercept

X=\frac{-r^{3}}{\sqrt{4r^{2}-r^{4}}-2r} ,

the limit as x->0 gives you zero in the denominator and an indeterminate ratio 0/0 for the limit.

Given that you're in Section 2.3 of Stewart, you're past the point where you're allowed to just graph the function; you'll have to actually solve this using a technique you've learned. To make life a bit easier, factor out an "r" from the numerator and denominator to get

\frac{-r^{2}}{\sqrt{4 - r^{2}} - 2}

and using the "conjugate factor" method by multiplying the numerator and denominator by \sqrt{4 - r^{2}} + 2 . This will help you get rid of the troublesome denominator and give you something easy to take the limit of for x->0.

Having taught "Calc-One" recently, I happen to have the solutions manual for Stewart here and find that there is also a nice completely geometric solution given. But you're learning about limits, so go to it... (The limit is indeed X = 4 , which does seem counterintuitive.)

Look at the picture, it is indeed problem 60.

And I did completely forget that I wasn't allowed to just graph it, as I type this I'm trying to force my eyes to not look at your post so I won't cheat.

Nice tip too, thanks.

Edit: whoooo, I did it

dynamicsolo

Jan11-08, 08:00 PM

Look at the picture, it is indeed problem 60.

I'll have to go back and check the Fifth Edition of Stewart, then. In the Sixth Edition, it is #62 both in the textbook and the solutions manual.

And I did completely forget that I wasn't allowed to just graph it...

I made the comment I did about you're being in Section 2.3 also because there is another technique for dealing with limits which lead to indeterminate ratios like this one, but you won't see it until Chapter 4. (Although, frankly, the "conjugate factor" method is faster for something like this than l'Hopital's Rule is...)

The argument that mda gives above is related to the alternate solution offered in the manual; there, the solver used constructions built on the point diametrically opposite to P on C2...

http://www.nellilevental.com/calc1.jpg

2. Homework Equations

C1 = \left(x - 1\right) ^{2} + y^{2} = 1

C2 = x^{2} + y^{2} = r^{2}

3. 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

C1_{upper} = \sqrt{1-(x-1)^{2}} = C2_{upper} = \sqrt{r^{2}-x^{2}}

\sqrt{1-(x-1)^{2}} = \sqrt{r^{2}-x^{2}}

1-(x-1)^{2} = r^{2}-x^{2}

1-(x^{2}-2x+1) = r^{2}-x^{2}

2x = r^{2}

x = \frac{r^{2}}{2}

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:

(\frac{r^{2}}{2}, \frac{\sqrt{4r^{2}-r^{4}}}{2})

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:

Y = X(\frac{\sqrt{4r^{2}-r^{4}}-2r}{r^{2}})+r

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:

X=\frac{-r^{3}}{\sqrt{4r^{2}-r^{4}}-2r}

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

dynamicsolo

Jan11-08, 01:53 PM

the book is James Stewart Single Variable Calculus 6e, ch 2.3 exercise 60

Heh, I recognized this problem right away, since that's the book we use where I am... (BTW, it's exercise 62.)

C1 = \left(x - 1\right) ^{2} + y^{2} = 1

C2 = x^{2} + y^{2} = r^{2}

I see no problems with your derivation of the x-intercept.

I was going to suggest a tip here on solving for the upper intersection point of the two circles. If you expand the first equation as

(x^{2} - 2x + 1) + y^{2} = 1, giving

x^{2} - 2x + y^{2} = 0 ,

you can immediately substitute the second equation into this to obtain

r^{2} - 2x = 0 .

In any event, looking at your result for the intercept

X=\frac{-r^{3}}{\sqrt{4r^{2}-r^{4}}-2r} ,

the limit as x->0 gives you zero in the denominator and an indeterminate ratio 0/0 for the limit.

Given that you're in Section 2.3 of Stewart, you're past the point where you're allowed to just graph the function; you'll have to actually solve this using a technique you've learned. To make life a bit easier, factor out an "r" from the numerator and denominator to get

\frac{-r^{2}}{\sqrt{4 - r^{2}} - 2}

and using the "conjugate factor" method by multiplying the numerator and denominator by \sqrt{4 - r^{2}} + 2 . This will help you get rid of the troublesome denominator and give you something easy to take the limit of for x->0.

Having taught "Calc-One" recently, I happen to have the solutions manual for Stewart here and find that there is also a nice completely geometric solution given. But you're learning about limits, so go to it... (The limit is indeed X = 4 , which does seem counterintuitive.)

ice109

Jan11-08, 03:26 PM

can someone give a clear geometric argument for why this is true?

mda

Jan11-08, 04:15 PM

can someone give a clear geometric argument for why this is true?

Once you have the intersection point Q, it is a case of constructing similar triangles.

Q is approximately (1/2 r^2, r).

Taking angle POQ we bisect and intersect with PQR, calling this point S say.

S is approximately (1/4 r^2, r).

OS is perpendicular to PQR, so OSP and RPO are similar triangles.

Therefore OR = OS OP/PS = r r / (1/4 r^2) = 4.

iceman713

Jan11-08, 04:34 PM

Heh, I recognized this problem right away, since that's the book we use where I am... (BTW, it's exercise 62.)

I see no problems with your derivation of the x-intercept.

I was going to suggest a tip here on solving for the upper intersection point of the two circles. If you expand the first equation as

(x^{2} - 2x + 1) + y^{2} = 1, giving

x^{2} - 2x + y^{2} = 0 ,

you can immediately substitute the second equation into this to obtain

r^{2} - 2x = 0 .

In any event, looking at your result for the intercept

X=\frac{-r^{3}}{\sqrt{4r^{2}-r^{4}}-2r} ,

the limit as x->0 gives you zero in the denominator and an indeterminate ratio 0/0 for the limit.

Given that you're in Section 2.3 of Stewart, you're past the point where you're allowed to just graph the function; you'll have to actually solve this using a technique you've learned. To make life a bit easier, factor out an "r" from the numerator and denominator to get

\frac{-r^{2}}{\sqrt{4 - r^{2}} - 2}

and using the "conjugate factor" method by multiplying the numerator and denominator by \sqrt{4 - r^{2}} + 2 . This will help you get rid of the troublesome denominator and give you something easy to take the limit of for x->0.

Having taught "Calc-One" recently, I happen to have the solutions manual for Stewart here and find that there is also a nice completely geometric solution given. But you're learning about limits, so go to it... (The limit is indeed X = 4 , which does seem counterintuitive.)

Look at the picture, it is indeed problem 60.

And I did completely forget that I wasn't allowed to just graph it, as I type this I'm trying to force my eyes to not look at your post so I won't cheat.

Nice tip too, thanks.

Edit: whoooo, I did it

dynamicsolo

Jan11-08, 08:00 PM

Look at the picture, it is indeed problem 60.

I'll have to go back and check the Fifth Edition of Stewart, then. In the Sixth Edition, it is #62 both in the textbook and the solutions manual.

And I did completely forget that I wasn't allowed to just graph it...

I made the comment I did about you're being in Section 2.3 also because there is another technique for dealing with limits which lead to indeterminate ratios like this one, but you won't see it until Chapter 4. (Although, frankly, the "conjugate factor" method is faster for something like this than l'Hopital's Rule is...)

The argument that mda gives above is related to the alternate solution offered in the manual; there, the solver used constructions built on the point diametrically opposite to P on C2...

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