What is going on here? How is this tangent equation being derived in the sol'n manual?

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allegansveritatem

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The following is a (small) portion of a solution from a solutions manual keyed to a much renowned calculus text--author James Stewart. The whole problem is complex and not really relevant to report here. What I want to know is: How is this tangent equation being derived. I can't for the life of me figure out the algebraic machinations involved. For one thing, they seem to be saying that x times x sub o can be written( x sub o) squared. But these are two different values of x! It's impossible to write them as one of them squared. Can anyone comment on this point and just on how they are deriving this equation. I mean I don't see how they are getting from = to = here at all. I include my attempt at coming out with the tangent equation.
solutionprob1.PNG
This is what I did ( it is pretty nasty looking, I know ):
solutionprob2.PNG
 
allegansveritatem said:
The following is a (small) portion of a solution from a solutions manual keyed to a much renowned calculus text--author James Stewart. The whole problem is complex and not really relevant to report here. What I want to know is: How is this tangent equation being derived. I can't for the life of me figure out the algebraic machinations involved. For one thing, they seem to be saying that x times x sub o can be written( x sub o) squared. But these are two different values of x! It's impossible to write them as one of them squared. Can anyone comment on this point and just on how they are deriving this equation. I mean I don't see how they are getting from = to = here at all. I include my attempt at coming out with the tangent equation.
View attachment 36331
This is what I did ( it is pretty nasty looking, I know ):
View attachment 36332
Click to expand...
As always, step by step:
[imath]y - y_0 = - \dfrac{4}{9} \dfrac{x_0}{y_0} (x - x_0)[/imath]

[imath](9y_0)(y - y_0) = -4x_0 (x - x_0)[/imath]

[imath]9y_0 y - 9y_0^2 = -4x_0 x + 4 x_0^2[/imath]

[imath]4x_0 x + 9y_0 y = 4 x_0^2 + 9y_0^2[/imath]

-Dan
 
topsquark said:
As always, step by step:
[imath]y - y_0 = - \dfrac{4}{9} \dfrac{x_0}{y_0} (x - x_0)[/imath]

[imath](9y_0)(y - y_0) = -4x_0 (x - x_0)[/imath]

[imath]9y_0 y - 9y_0^2 = -4x_0 x + 4 x_0^2[/imath]

[imath]4x_0 x + 9y_0 y = 4 x_0^2 + 9y_0^2[/imath]

-Dan
Click to expand...
I am sorry that I posted that second image--it was completely wrong. What happened is that I had worked out the tangent equation a couple of times but it was so messy that I quickly rewrote it so that it would look more presentable and in the process left out steps and made a complete hash of things.
Anyway, what you present here is pretty much what I came up with, but it is not what seems to be given in the solutions manual. they seems to be saying that 4(x sub 0)^2 plus 4( y sub o)^2 is the equation of the tangent complete. Is it? If so...how?
 
It is NOT stating that 4x0x + 9y0y simplifies to 4(x0) 2 + 9(y0) 2.
Just like in x+3x = 5x, it doesn't say that x+3x is always 5x. For the record, x+3x=5x implies that 4x=5x or x=0. If x+3x was identically 5x, then it would be true for ALL x. This equation is called a conditional equation--because equality holds only when x=0
 
allegansveritatem said:
I am sorry that I posted that second image--it was completely wrong. What happened is that I had worked out the tangent equation a couple of times but it was so messy that I quickly rewrote it so that it would look more presentable and in the process left out steps and made a complete hash of things.
Anyway, what you present here is pretty much what I came up with, but it is not what seems to be given in the solutions manual. they seems to be saying that 4(x sub 0)^2 plus 4( y sub o)^2 is the equation of the tangent complete. Is it? If so...how?
Click to expand...
Steven G said it, but as you have pretty much repeated your confusion from your OP, I think it needs to be said again: the equation the text derived is a function y=y(x), the tangent line to a given point [imath](x_0,y_0)[/imath] on the ellipse. It is no different than saying that it has the form y=mx+b... it's just written in a different format.

-Dan
 
Steven G said:
It is NOT stating that 4x0x + 9y0y simplifies to 4(x0) 2 + 9(y0) 2.
Just like in x+3x = 5x, it doesn't say that x+3x is always 5x. For the record, x+3x=5x implies that 4x=5x or x=0. If x+3x was identically 5x, then it would be true for ALL x. This equation is called a conditional equation--because equality holds only when x=0
Click to expand...
Yes, I see that now. I have misread the solution . This sometimes happens to me. I misread something and from then on that misreading undermines the whole enterprise.
 
topsquark said:
Steven G said it, but as you have pretty much repeated your confusion from your OP, I think it needs to be said again: the equation the text derived is a function y=y(x), the tangent line to a given point [imath](x_0,y_0)[/imath] on the ellipse. It is no different than saying that it has the form y=mx+b... it's just written in a different format.

-Dan
Click to expand...
I recognize now that I read the solution too fast. In other words, I misread it. Speed reading, if it works anywhere (highly debatable) , doesn't in mathematics.
 
allegansveritatem said:
I recognize now that I read the solution too fast. In other words, I misread it. Speed reading, if it works anywhere (highly debatable) , doesn't in mathematics.
Click to expand...
I can read a "leisure" book in three days. (I've done some in one.) It can take me a whole day to do a single page in my textbooks, depending on how "dense" the material is.

I sympathize!

-Dan
 
topsquark said:
I can read a "leisure" book in three days. (I've done some in one.) It can take me a whole day to do a single page in my textbooks, depending on how "dense" the material is.

I sympathize!

-Dan
Click to expand...
I can easily beat that. It can take me a whole day to understand a single line in my textbooks, especially when it says as easily seen.
 
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