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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
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

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TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
@mathstina what is the gradient vector for any point on the sphere?
 one year ago

mathstina Group TitleBest ResponseYou've already chosen the best response.1
what is point ? is it (0,0,0)?
 one year ago

mathstina Group TitleBest ResponseYou've already chosen the best response.1
what do i sub for r value?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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\[\nabla f\]?
 one year ago

mathstina Group TitleBest ResponseYou've already chosen the best response.1
2x i + 2y j + 2z k = r^2
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
r is a constant for a sphere, so this thingy should be =0
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
grad f=2x i + 2y j + 2z k = 0 still with me?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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?
 one year ago

mathstina Group TitleBest ResponseYou've already chosen the best response.1
(a,b,c)?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
yes, and what is the gradient at that point ?
 one year ago

mathstina Group TitleBest ResponseYou've already chosen the best response.1
is it find the parametric eqn?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
? no you just need the two vectors to solve this\[P(a,b,c)=\langle a,b,c\rangle\]\[\nabla f(a,b,c)=?\]
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
a picture may help us visualize what we're doingdw:1352129016463:dw
 one year ago

mathstina Group TitleBest ResponseYou've already chosen the best response.1
2a i+2b j+2c k =0
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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?
 one year ago

mathstina Group TitleBest ResponseYou've already chosen the best response.1
multiplied by 2
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
yes, so what does that say bout the relationship between the two vectors?
 one year ago

mathstina Group TitleBest ResponseYou've already chosen the best response.1
sry, im nt sure
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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?
 one year ago

mathstina Group TitleBest ResponseYou've already chosen the best response.1
positive gradient
 one year ago

mathstina Group TitleBest ResponseYou've already chosen the best response.1
both pointing in the same dirn; parallel
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
yes, now look at what this means visually...
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
dw:1352129640330:dw
 one year ago

mathstina Group TitleBest ResponseYou've already chosen the best response.1
the points are on the level surface
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
dw:1352129723349:dwsince 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?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
hence all normal lines to the surface pass through the origin*
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
good, then we're done :)
 one year ago

mathstina Group TitleBest ResponseYou've already chosen the best response.1
ihw do i answer the qn? mathematically written
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago
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