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.

- anonymous

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- schrodinger

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- anonymous

hi

- TuringTest

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?

- anonymous

ok.

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## More answers

- TuringTest

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

- anonymous

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

- anonymous

what do i sub for r value?

- TuringTest

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\[\nabla f\]?

- anonymous

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

- TuringTest

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

- TuringTest

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

- anonymous

yes

- TuringTest

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?

- anonymous

(a,b,c)?

- TuringTest

yes, and what is the gradient at that point ?

- anonymous

is it find the parametric eqn?

- TuringTest

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

- TuringTest

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

- anonymous

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

- TuringTest

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?

- anonymous

multiplied by 2

- TuringTest

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

- anonymous

sry, im nt sure

- TuringTest

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?

- anonymous

positive gradient

- anonymous

both pointing in the same dirn; parallel

- TuringTest

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

- TuringTest

|dw:1352129640330:dw|

- anonymous

the points are on the level surface

- TuringTest

|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?

- TuringTest

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

- anonymous

yes.

- TuringTest

good, then we're done :)

- anonymous

ihw do i answer the qn? mathematically written

- TuringTest

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.

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