## mathstina Group Title Show that every normal line to the sphere x^2+y^2+z^2=r^2,where r is the radius,passes through the centre of the sphere. one year ago one year ago

1. mathstina Group Title

hi

2. TuringTest Group Title

label a point on the surface of the sphere with the position vector$\vec P=\langle a,b,c\rangle$what is the gradient at that point?

3. mathstina Group Title

ok.

4. TuringTest Group Title

@mathstina what is the gradient vector for any point on the sphere?

5. mathstina Group Title

what is point ? is it (0,0,0)?

6. mathstina Group Title

what do i sub for r value?

7. TuringTest Group Title

we aren't going top specify the point yet, because we want to prove the statement for the general case the function for the surface of the sphere is$f(x,y,z)=x^2+y^2+z^2=r^2$what is$\nabla f$?

8. mathstina Group Title

2x i + 2y j + 2z k = r^2

9. TuringTest Group Title

r is a constant for a sphere, so this thingy should be =0

10. TuringTest Group Title

grad f=2x i + 2y j + 2z k = 0 still with me?

11. mathstina Group Title

yes

12. TuringTest Group Title

okay, now let the point be some x=a, y=b, and z=c that satisfies the function f and so is on the sphere what is the position vector at that point?

13. mathstina Group Title

(a,b,c)?

14. TuringTest Group Title

yes, and what is the gradient at that point ?

15. mathstina Group Title

is it find the parametric eqn?

16. TuringTest Group Title

? no you just need the two vectors to solve this$P(a,b,c)=\langle a,b,c\rangle$$\nabla f(a,b,c)=?$

17. TuringTest Group Title

a picture may help us visualize what we're doing|dw:1352129016463:dw|

18. mathstina Group Title

2a i+2b j+2c k =0

19. TuringTest Group Title

yes, but be consistent in your notation. Write *all* vectors, including the position vector, the same way now compare the position vector and the gradient vector. What do you see?

20. mathstina Group Title

multiplied by 2

21. TuringTest Group Title

yes, so what does that say bout the relationship between the two vectors?

22. mathstina Group Title

sry, im nt sure

23. TuringTest Group Title

the scalar multiple of any vector points in the *same direction* as the original the gradient vector is a scalar multiple of the position vector, hence what can we say about their relative orientations?

24. mathstina Group Title

25. mathstina Group Title

both pointing in the same dirn; parallel

26. TuringTest Group Title

yes, now look at what this means visually...

27. TuringTest Group Title

|dw:1352129640330:dw|

28. mathstina Group Title

the points are on the level surface

29. TuringTest Group Title

|dw:1352129723349:dw|since the position vector and grad f are parallel for any point on the sphere, they are collinear. Since gradf is normal to the sphere, so must be the line hence, we can draw a line that contains both. Since the position vector passes through the origin, we know that any such line will also pass through the origin. Hence all normal lines to the surface. Make sense?

30. TuringTest Group Title

hence all normal lines to the surface pass through the origin*

31. mathstina Group Title

yes.

32. TuringTest Group Title

good, then we're done :)

33. mathstina Group Title

ihw do i answer the qn? mathematically written

34. TuringTest Group Title

let P be a point on the sphere$\vec P=\langle a,b,c\rangle$then$\nabla f=2\vec P\implies \nabla f\parallel\vec P$hence they are collinear. since the line containing P passes through the origin, and any line that contains P must also contain grad f, this means that so does any line normal to the sphere passes though the origin. QED I don't think any more math symbols are necessary, sometimes words are important in proofs.