Find Values of Osculating Circle Tangent to a Parabola

  • #1
Ascendant0
57
11
Homework Statement
Find the values of h, k, and a that make
the circle (x - h)^2 + (y - k)^2 = a^2 tangent to the parabola
y = x^2 + 1 at the point (1, 2) and that also make the second
derivatives d^2y>dx^2 have the same value on both curves there
Relevant Equations
Included in my attempts at a solution
I'm at a loss as to how they got to certain steps in the solutions manual. Here's how far I got with this:

Since the circle is tangent to y = x^2 + 1, the slope at (1, 2) is going to be 2, as is the slope of the 2nd derivative of the circle, so then...

The derivative of the circle would be dy/dx = (x - h)/(y - k), and since dy/dx = 2, I can find that h = 5 - 2k

For the 2nd derivative, this is where the manual loses me. They skip like three steps as to how they simplified it, so I don't see how they get their answer. Doing the 2nd derivative, I get something sloppy with a k^2 and multiple answers, whereas in the book, they get a simple k = 9/2 from an equation they give no explanation how they got to

For the 2nd derivative, I have tried both making the (y - k)^(-1), so I can use the product rule, instead of what I feel is a sloppier quotient rule, but tried that one as well. Either way, it gets sloppy at this point, and the solution loses me:

d^2y/dx^2 = (x - h) (-1) (y - k)^(-2) (dy/dx) + (y - k)^(-1)

Of course I can simplify this, but whether the quotient or product rule for derivatives, I end up with a "k^2" as a part of my derivative, so I end up with two different answers. I haven't tried anything with either answer, as I really want to know how they got to their answer in the book, as I know there is a simpler way to do this, but I don't see how they did

In the book, they don't explain, they just go:

"Also, (x - h)^2 + (y - k)^2 = a^2 ----> y'' = (1 + (y')^2)/(k - y)" and give NO explanation how they got the 2nd derivative from it???

I've been racking my brain on this one, and it's driving me nuts. I need to figure out how to simplify that so I don't get a k^2 and two different answers, but I'm stuck. If someone could point me in the right direction here as to what I'm missing, it would be greatly appreciated. I can't accept not being able to do something like this. My brain goes crazy until I can :P
 
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  • #2
Ascendant0 said:
"Also, (x - h)^2 + (y - k)^2 = a^2 ----> y'' = (1 + (y')^2)/(k - y)" and give NO explanation how they got the 2nd derivative from it???
Simply differentiate the equation ##(x-h)^2+(y(x)-k)^2=a^2## twice with respect to ##x ## and then solve for ##y''(x)##.
 
  • #3
renormalize said:
Simply differentiate the equation ##(x-h)^2+(y(x)-k)^2=a^2## twice with respect to ##x ## and then solve for ##y''(x)##.
Well, the first time I do, I end up with the dy/dx = (x + h)/(y - k). That's fine, and what they have in the book

However, when I do the 2nd derivative, it makes it sloppy, and I end up with the "(y-k)^2" in the denominator, which is what's causing me to get two possible answers for k

I'm sure there is an easier way to do the 2nd derivative of the equation, but I'm just going by the quotient and product rule options shown so far in the textbook. This is one of the "advanced exercises," so I'm assuming it's a bit outside of what they've covered so far

I could just take x - h + (y-k)y' = 0 and do the derivative of that, but I'm not sure how without getting y' alone on one side of the equation? I can't figure out how they get from the first to the 2nd derivative without getting the k^2 in the denominator?
 
  • #4
Ascendant0 said:
However, when I do the 2nd derivative, it makes it sloppy, and I end up with the "(y-k)^2" in the denominator, which is what's causing me to get two possible answers for k
As I said, differentiate the circle equation twice, but don't solve for anything until you're done differentiating!
The first-derivative of the equation:$$\frac{d}{dx}\left(\left(x-h\right)^{2}+\left(y\left(x\right)-k\right)^{2}\right)=\frac{d}{dx}\left(a^{2}\right)$$is:$$2\left(x-h\right)+2\left(y\left(x\right)-k\right)y^{\prime}\left(x\right)=0$$Don't solve for ##y^{\prime}##, just differentiate again to get the second-derivative of the equation:$$2\frac{d}{dx}\left(x-h+y\left(x\right)y^{\prime}\left(x\right)-ky^{\prime}\left(x\right)\right)=0$$which evaluates to:$$2\left(1+\left(y^{\prime}\left(x\right)\right)^{2}+\left(y\left(x\right)-k\right)y^{\prime\prime}\left(x\right)\right)=0$$Now solve this last equation for ##y^{\prime\prime}## and you're done.
 
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  • #5
renormalize said:
As I said, differentiate the circle equation twice, but don't solve for anything until you're done differentiating!
The first-derivative of the equation:$$\frac{d}{dx}\left(\left(x-h\right)^{2}+\left(y\left(x\right)-k\right)^{2}\right)=\frac{d}{dx}\left(a^{2}\right)$$is:$$2\left(x-h\right)+2\left(y\left(x\right)-k\right)y^{\prime}\left(x\right)=0$$Don't solve for ##y^{\prime}##, just differentiate again to get the second-derivative of the equation:$$2\frac{d}{dx}\left(x-h+y\left(x\right)y^{\prime}\left(x\right)-ky^{\prime}\left(x\right)\right)=0$$which evaluates to:$$2\left(1+\left(y^{\prime}\left(x\right)\right)^{2}+\left(y\left(x\right)-k\right)y^{\prime\prime}\left(x\right)\right)=0$$Now solve this last equation for ##y^{\prime\prime}## and you're done.
Thank you for that. Now I see how that works and I get it. I wasn't sure how to do d/dx when it came to the y and y' parts of the equation. The x becoming "1" and the "h" becoming "0" made sense, but now I see with the other two parts, you can use the product rule on them and y' just becomes y'' when you take its derivative

I really appreciate your help with this. Now I can see how much easier this should've been than what I was making it out to be. Also, I really need to learn LaTeX...
 
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