JoannaBlackwelder
  • JoannaBlackwelder
show that the normal line at any point on the circle x^2+y^2=r^2 passes through the origin
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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JoannaBlackwelder
  • JoannaBlackwelder
@freckles
freckles
  • freckles
have you found y'
JoannaBlackwelder
  • JoannaBlackwelder
Yeah, it is -x/y

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imqwerty
  • imqwerty
this is a circle with the origin at center so any line tangent makes an angle of 90 with centr|dw:1443674243272:dw|
JoannaBlackwelder
  • JoannaBlackwelder
Yeah, I got that too, but I thought that it needed to be more "calculusy" and less conceptual
JoannaBlackwelder
  • JoannaBlackwelder
So, that is an option if I can't think of another way, thanks @imqwerty
imqwerty
  • imqwerty
ok lets think that way :)
freckles
  • freckles
\[(a,b) \text{ is on the circle means } a^2+b^2=r^2 \\ \text{ so } b^2=r^2-a^2 \\ b=\pm \sqrt{r^2-a^2} \text{ assume } b>0 \text{ so we have } (a,b)=(a,\sqrt{r^2-a^2}) \\ \text{ the tangent line at } (a,b) \text{ is } \\ y-\sqrt{r^2-a^2}=- \frac{a}{\sqrt{r^2-a^2}} (x-a) \\ y=\frac{-a}{\sqrt{r^2-a^2}}(x-a)+\sqrt{r^2-a^2} \\ \text{ now the line that is perpendicular \to this one going through } (a,\sqrt{r^2-a^2} ) \\ \text{ is } y-\sqrt{r^2-a^2}=\frac{\sqrt{r^2-a^2}}{a}(x-a) \\ y=\frac{\sqrt{r^2-a^2}}{a}(x-a)+\sqrt{r^2-a^2} \\ \text{ so we need to show } (0,0) \text{ is on this line }\] just plug in (0,0) in the normal line there to verify then you can also do the same for the other assumption when b<0
freckles
  • freckles
and also when b=0
JoannaBlackwelder
  • JoannaBlackwelder
Wow, thanks!
freckles
  • freckles
notice I didn't need to actually find the tangent line I just needed the slope of the tangent line and the slope of the normal line would be the opposite reciprocal of that
freckles
  • freckles
and then I was about to find the equation of the normal line from there
freckles
  • freckles
though we didn't really need calculus for this just algebra
freckles
  • freckles
so I kind of like @imqwerty 's way :) less work
imqwerty
  • imqwerty
(: thanks
freckles
  • freckles
|dw:1443675019274:dw|
freckles
  • freckles
that was just using the one algebraic formula for finding slope
imqwerty
  • imqwerty
umm m thinking of something like partial derivative of the equation of circle x^2+y^2 :/ what will it give?
JoannaBlackwelder
  • JoannaBlackwelder
I found another way. :-) Since we got the slope of the tangent line to be -x/y, the slope of the normal line is y/x. Using the point (a.b), the new slope is b/a|dw:1443676370689:dw|
JoannaBlackwelder
  • JoannaBlackwelder
So, the equation of the line has the y intercept as 0
JoannaBlackwelder
  • JoannaBlackwelder
Which means the lines go through (0,0)
IrishBoy123
  • IrishBoy123
partials give you something like this write the circle as level curve: \(\phi = x^2 + y^2 = const\) \(\large d \phi = \frac{\partial \phi}{\partial x} \ dx + \frac{\partial \phi}{\partial y} \ dy = <\frac{\partial \phi}{\partial x} , \frac{\partial \phi}{\partial y} >\bullet = 0\) because \(d (const.) = 0\) but this also \(\implies \vec n_{x,y} = <\frac{\partial \phi}{\partial x} , \frac{\partial \phi}{\partial y} > \equiv\) for point P on circle at generalised point (a,b), \(\vec{OP} = \), \(\vec {n_P} = \) ie \(\vec{OP} \parallel \vec{n_P}\) so \(\vec {n_P}\) goes through O. which is all kinda like implicit diff in reverse because \(2yy' + 2x = 0, \; y' = -\frac{x}{y}, \; \vec t = <-y,x> \implies \vec n = \) as \(\vec n \bullet \vec t = 0\)
imqwerty
  • imqwerty
thanks ^_^ @IrishBoy123 :)

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