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JoannaBlackwelder
 one year ago
show that the normal line at any point on the circle x^2+y^2=r^2 passes through the origin
JoannaBlackwelder
 one year ago
show that the normal line at any point on the circle x^2+y^2=r^2 passes through the origin

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JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, it is x/y

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2this is a circle with the origin at center so any line tangent makes an angle of 90 with centrdw:1443674243272:dw

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, I got that too, but I thought that it needed to be more "calculusy" and less conceptual

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.0So, that is an option if I can't think of another way, thanks @imqwerty

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2ok lets think that way :)

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[(a,b) \text{ is on the circle means } a^2+b^2=r^2 \\ \text{ so } b^2=r^2a^2 \\ b=\pm \sqrt{r^2a^2} \text{ assume } b>0 \text{ so we have } (a,b)=(a,\sqrt{r^2a^2}) \\ \text{ the tangent line at } (a,b) \text{ is } \\ y\sqrt{r^2a^2}= \frac{a}{\sqrt{r^2a^2}} (xa) \\ y=\frac{a}{\sqrt{r^2a^2}}(xa)+\sqrt{r^2a^2} \\ \text{ now the line that is perpendicular \to this one going through } (a,\sqrt{r^2a^2} ) \\ \text{ is } y\sqrt{r^2a^2}=\frac{\sqrt{r^2a^2}}{a}(xa) \\ y=\frac{\sqrt{r^2a^2}}{a}(xa)+\sqrt{r^2a^2} \\ \text{ so we need to show } (0,0) \text{ is on this line }\] 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

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.0Wow, thanks!

freckles
 one year ago
Best ResponseYou've already chosen the best response.3notice 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
 one year ago
Best ResponseYou've already chosen the best response.3and then I was about to find the equation of the normal line from there

freckles
 one year ago
Best ResponseYou've already chosen the best response.3though we didn't really need calculus for this just algebra

freckles
 one year ago
Best ResponseYou've already chosen the best response.3so I kind of like @imqwerty 's way :) less work

freckles
 one year ago
Best ResponseYou've already chosen the best response.3dw:1443675019274:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.3that was just using the one algebraic formula for finding slope

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2umm m thinking of something like partial derivative of the equation of circle x^2+y^2 :/ what will it give?

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.0I 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/adw:1443676370689:dw

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.0So, the equation of the line has the y intercept as 0

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.0Which means the lines go through (0,0)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0partials give you something like this write the circle as level curve: \(\phi = x^2 + y^2 = const\) \(\large d \phi = \frac{\partial \phi}{\partial x} \ dx + \frac{\partial \phi}{\partial y} \ dy = <\frac{\partial \phi}{\partial x} , \frac{\partial \phi}{\partial y} >\bullet<dx,dy> = 0\) because \(d (const.) = 0\) but this also \(\implies \vec n_{x,y} = <\frac{\partial \phi}{\partial x} , \frac{\partial \phi}{\partial y} > \equiv<x,y>\) for point P on circle at generalised point (a,b), \(\vec{OP} = <a,b> \), \(\vec {n_P} = <a,b>\) ie \(\vec{OP} \parallel \vec{n_P}\) so \(\vec {n_P}\) goes through O. which is all kinda like implicit diff in reverse because \(2yy' + 2x = 0, \; y' = \frac{x}{y}, \; \vec t = <y,x> \implies \vec n = <x,y> \) as \(\vec n \bullet \vec t = 0\)

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2thanks ^_^ @IrishBoy123 :)
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