anonymous
  • anonymous
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.
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
hi
TuringTest
  • 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
  • anonymous
ok.

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TuringTest
  • TuringTest
@mathstina what is the gradient vector for any point on the sphere?
anonymous
  • anonymous
what is point ? is it (0,0,0)?
anonymous
  • anonymous
what do i sub for r value?
TuringTest
  • 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
  • anonymous
2x i + 2y j + 2z k = r^2
TuringTest
  • TuringTest
r is a constant for a sphere, so this thingy should be =0
TuringTest
  • TuringTest
grad f=2x i + 2y j + 2z k = 0 still with me?
anonymous
  • anonymous
yes
TuringTest
  • 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
  • anonymous
(a,b,c)?
TuringTest
  • TuringTest
yes, and what is the gradient at that point ?
anonymous
  • anonymous
is it find the parametric eqn?
TuringTest
  • 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
  • TuringTest
a picture may help us visualize what we're doing|dw:1352129016463:dw|
anonymous
  • anonymous
2a i+2b j+2c k =0
TuringTest
  • 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
  • anonymous
multiplied by 2
TuringTest
  • TuringTest
yes, so what does that say bout the relationship between the two vectors?
anonymous
  • anonymous
sry, im nt sure
TuringTest
  • 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
  • anonymous
positive gradient
anonymous
  • anonymous
both pointing in the same dirn; parallel
TuringTest
  • TuringTest
yes, now look at what this means visually...
TuringTest
  • TuringTest
|dw:1352129640330:dw|
anonymous
  • anonymous
the points are on the level surface
TuringTest
  • 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
  • TuringTest
hence all normal lines to the surface pass through the origin*
anonymous
  • anonymous
yes.
TuringTest
  • TuringTest
good, then we're done :)
anonymous
  • anonymous
ihw do i answer the qn? mathematically written
TuringTest
  • 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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