A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

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

  • This Question is Closed
  1. JoannaBlackwelder
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @freckles

  2. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    have you found y'

  3. JoannaBlackwelder
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yeah, it is -x/y

  4. imqwerty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    this is a circle with the origin at center so any line tangent makes an angle of 90 with centr|dw:1443674243272:dw|

  5. JoannaBlackwelder
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yeah, I got that too, but I thought that it needed to be more "calculusy" and less conceptual

  6. JoannaBlackwelder
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So, that is an option if I can't think of another way, thanks @imqwerty

  7. imqwerty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    ok lets think that way :)

  8. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    \[(a,b) \text{ is on the circle means } a^2+b^2=r^2 \\ \text{ so } b^2=r^2-a^2 \\ b=\pm \sqrt{r^2-a^2} \text{ assume } b>0 \text{ so we have } (a,b)=(a,\sqrt{r^2-a^2}) \\ \text{ the tangent line at } (a,b) \text{ is } \\ y-\sqrt{r^2-a^2}=- \frac{a}{\sqrt{r^2-a^2}} (x-a) \\ y=\frac{-a}{\sqrt{r^2-a^2}}(x-a)+\sqrt{r^2-a^2} \\ \text{ now the line that is perpendicular \to this one going through } (a,\sqrt{r^2-a^2} ) \\ \text{ is } y-\sqrt{r^2-a^2}=\frac{\sqrt{r^2-a^2}}{a}(x-a) \\ y=\frac{\sqrt{r^2-a^2}}{a}(x-a)+\sqrt{r^2-a^2} \\ \text{ so we need to show } (0,0) \text{ is on this line }\] 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

  9. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    and also when b=0

  10. JoannaBlackwelder
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Wow, thanks!

  11. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    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

  12. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    and then I was about to find the equation of the normal line from there

  13. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    though we didn't really need calculus for this just algebra

  14. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    so I kind of like @imqwerty 's way :) less work

  15. imqwerty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    (: thanks

  16. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    |dw:1443675019274:dw|

  17. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    that was just using the one algebraic formula for finding slope

  18. imqwerty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    umm m thinking of something like partial derivative of the equation of circle x^2+y^2 :/ what will it give?

  19. JoannaBlackwelder
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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|

  20. JoannaBlackwelder
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So, the equation of the line has the y intercept as 0

  21. JoannaBlackwelder
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Which means the lines go through (0,0)

  22. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    partials give you something like this write the circle as level curve: \(\phi = x^2 + y^2 = const\) \(\large d \phi = \frac{\partial \phi}{\partial x} \ dx + \frac{\partial \phi}{\partial y} \ dy = <\frac{\partial \phi}{\partial x} , \frac{\partial \phi}{\partial y} >\bullet<dx,dy> = 0\) 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 \(\vec n \bullet \vec t = 0\)

  23. imqwerty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    thanks ^_^ @IrishBoy123 :)

  24. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.