anonymous
  • anonymous
An equation of the tangent plane to the parametrized surface Φ(u,v) = (3u, −3u2 + 2v, −5v2) at the point (6, −8, −20) is (in the variables x, y, z). To normalize the answer, make sure your coefficient of x is 240. = 0.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • amistre64
oh im just simply salivating over this one lol
anonymous
  • anonymous
please take your time, thanks
amistre64
  • amistre64
we take the point; and apply the gradient to it right?

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amistre64
  • amistre64
but the initial form seems to be in a vector position equation...
anonymous
  • anonymous
hmm i am trying to find something in my book
amistre64
  • amistre64
(3u) i + (3u^2 +2v) j + (-5v^2) k
amistre64
  • amistre64
the derivative of a vector position is the tangent line .... if i recall correctly
amistre64
  • amistre64
tangent vector that is...
anonymous
  • anonymous
i believe so
amistre64
  • amistre64
3u = 6 3u^2 +2v = -8 -5v^2 = -20 should give us values for u and v maybe?
anonymous
  • anonymous
it's worth a try
amistre64
  • amistre64
u = 2 and v = 2, -2 3(2^2) + 2v = -8 12 + 2v = -8 2v = -20 v = -10 doesnt seem to fit the point in question..
amistre64
  • amistre64
so the point is on the tangent plane itself, and not the original equation
anonymous
  • anonymous
yes
amistre64
  • amistre64
then lets find a tangent vector that matches this contraption lol
amistre64
  • amistre64
Φ(u,v) = (3u, −3u^2 + 2v, −5v^2) Φ'(u,v) = (3, m', −10v) m = −3u^2 + 2v m' = -6u u' + 2 v' if i see it right
amistre64
  • amistre64
can we assume u' and v' = 1?
anonymous
  • anonymous
i don't know
amistre64
  • amistre64
im thinking yes ..... and im usually right about this unless im wrong lol
amistre64
  • amistre64
du/du = 1 and dv/dv = 1 is what im thinking...
amistre64
  • amistre64
Φ'(u,v) = (3, (-6u +2), −10v) would be our tangent vector; now lets point that to our point and see the vector we get....
amistre64
  • amistre64
<3,-6u+2,-10v> ----------------- <6, -8, -20> right?
anonymous
  • anonymous
you mean add them?
amistre64
  • amistre64
thats my thought; we have a vector going from whereever the point on the curve and to point it to the point outside of it we have to add something to it to get to there..
amistre64
  • amistre64
its in the same plane; just have to account for rotation...
amistre64
  • amistre64
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amistre64
  • amistre64
-10 means -20; i got no short term memory apparently lol
anonymous
  • anonymous
the approach looks resonable
amistre64
  • amistre64
with 2 vectors on the plane we can cross product them to get the nomal right?
amistre64
  • amistre64
<3,-6u+2,-10v> -<6, -8, -20> ----------------- < a, b, c> is an equivalent statement right?
anonymous
  • anonymous
the cross product will give the normal i think
amistre64
  • amistre64
it will; but first we have to determine what the vectors should be :)
amistre64
  • amistre64
perhaps we can dot product this to an unknown vector to get an orthogonal in the plane to establish a good plave equation?
amistre64
  • amistre64
<3,-6u+2,-10v> < r , s, t> --------------- = 0
amistre64
  • amistre64
watchmath will surely let me know when my stupidity shows tho right ? :)
amistre64
  • amistre64
< 3, -6u+2, -10v > <-6, 8, 20 > --------------------- <-3, -6u+10, -10v+20> if my original thought is correct
amistre64
  • amistre64
lets cross these and see what we get... ;)
amistre64
  • amistre64
(-6u+2)(-10v+20) - (8)(-10v) -[(3.20) - (-6.-10v)] (3.8) - (-6(-6u+2))
amistre64
  • amistre64
<(60uv -20v -120u +40) , (-60 - 60v) ,(-36u +36)> maybe ....
amistre64
  • amistre64
if we had a surface instead of a vector function; that would be easier i think lol ... I gotta wonder if I interpreted that first part correctly
amistre64
  • amistre64
<(60uv -20v -120u +40+80v) , (-60 - 60v) ,(-36u +36)> <(60uv -60v -120u +40) , (-60 - 60v) ,(-36u +36)>
amistre64
  • amistre64
(60uv -60v -120u +40)(x-6) +(-60 - 60v)(y+8) +(-36u +36)(z+20) = 0 is what I come up with.....
amistre64
  • amistre64
can we come up with a surface equation from the parametrics used? im sure we can....
amistre64
  • amistre64
x = 3u y = −3u^2 + 2v z = −5v^2
amistre64
  • amistre64
u = 3/x y = -27/x^2 + 2v y + 27/x^2 = 2v v = (y + 27)/ 2x^2 gonna have to research this one some more maybe lol
amistre64
  • amistre64
iv often wondered if there can be a tangent plane to a curve..... it seems like trying;
amistre64
  • amistre64
i spose if it can be done for a tangent line to a point; its possible for a plane on a curve :)
amistre64
  • amistre64
ok..... the derivatives for df/du and df/dv give us the vectors for our plane at the point (x,y,z) f(u,v)=<3u, −3u^2 + 2v, −5v^2> df/du = <3, −6u, 0> and df/dv=<0, 2, −10v> ; we cross these to get a normal to the plane at f(u,v) then
amistre64
  • amistre64
3 -6u 0 0 2 -10v (60uv - 2)i + (30v)j + (6)k <(60uv-2),30v,6> is our normal to the plane whenever we determiine a u and v to input
amistre64
  • amistre64
to include the point (6,-8,-20) lets put together the plane equation (60uv-2)(x-6) + (30v)(y+8) + (6)(z+20) = 0
amistre64
  • amistre64
in order for the coeff of x = 240, then 60uv -2 = 240 60uv = 242 uv = 242/60 if i did it right

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