Slope of tangent line (DE)

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Here's the problem:

detrmine DE giving the slope of the tangent line at point (x,y) for the given family of curves:

[tex] 2cy = x^2 - c^2 [/tex]

Here's what I have:

slope of this curve:

[tex] y' = \frac{x}{c} [/tex]

So, then the slope of tangent line is negative inverse:

[tex] y' = \m \frac{-c}{x} [/tex]

so now I need c, but I don't see how to pull it out from the original equation!

Thanks for any help.
 
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  • #2
AKG
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The negative reciprocal would be the slope of the perpendicular line. The slope of the tangent line would just be x/c. What you want to do now is express c in terms of x and y (use the quadratic formula).
 
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AKG said:
What you want to do now is express c in terms of x and y (use the quadratic formula).
That is precisely the problem.
and I'm not seeing how I can do that, even with quadratic formula.
 
  • #4
AKG
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[tex] 2cy = x^2 - c^2 [/tex]

This is a quadratic equation in c.
 
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AKG said:
[tex] 2cy = x^2 - c^2 [/tex]

This is a quadratic equation in c.
Are you talking about [tex] (a + b)^2[/tex]? but there's y.
RHS has [tex] (a + b) (a - b)[/tex] but I don't know how that's helping me.
 
  • #6
AKG
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A quadratic equation in x takes the form ax² + bx + c = 0, and the quadratic formula allows you to find the two complex solutions to this equation:

[tex]x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
 
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AKG said:
A quadratic equation in x takes the form ax² + bx + c = 0, and the quadratic formula allows you to find the two complex solutions to this equation:

[tex]x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
Oh, yeah, complex... I didn't think of them being here.
Thanks a lot.
 
  • #8
AKG
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Well your solutions for this quadratic equation will be real, but real numbers are complex numbers.
 
  • #9
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English terminology was the hardest part here: quadratic equation. It's called somth. else in my language and does't translate word for word, so I made a wrong association.
Thanks for your help, that is really like hischool algebra, I feel dumb.... :redface:
 
  • #10
HallsofIvy
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EvLer said:
Here's the problem:

detrmine DE giving the slope of the tangent line at point (x,y) for the given family of curves:

[tex] 2cy = x^2 - c^2 [/tex]

Here's what I have:

slope of this curve:

[tex] y' = \frac{x}{c} [/tex]
You are completely off base here. You need to eliminate the constant c, getting a differential equation that does not involve c. Yes, 2cy'= 2x. Now, [tex]c^2+ 2cy= x^2[/tex] so [tex]c^2+ 2cy+ y^2= (c+y)^2= x^2+ y^2[/tex]. That is: [tex]c+ y= \pm\sqrt{x^2+ y^2}[/tex]
and [tex]c= -y+ \sqrt{x^2+ y^2}[/tex] or [tex] c= -y- \sqrt{x^2+ y^2}[/tex]. The slope satisfies [tex](-y+ \sqrt{x^2+ y^2})y'= x[/tex] or [tex](-y- \sqrt{x^2+ y^2}= x[/tex].

So, then the slope of tangent line is negative inverse:

[tex] y' = \m \frac{-c}{x} [/tex]
No, the slope of the tangent line is the derivative. You are thinking of the slope of the normal line.

so now I need c, but I don't see how to pull it out from the original equation!

Thanks for any help.
 
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