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anonymous
 5 years ago
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.
anonymous
 5 years ago
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.

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0oh im just simply salivating over this one lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0please take your time, thanks

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we take the point; and apply the gradient to it right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0but the initial form seems to be in a vector position equation...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm i am trying to find something in my book

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0(3u) i + (3u^2 +2v) j + (5v^2) k

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the derivative of a vector position is the tangent line .... if i recall correctly

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0tangent vector that is...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.03u = 6 3u^2 +2v = 8 5v^2 = 20 should give us values for u and v maybe?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0u = 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
 5 years ago
Best ResponseYou've already chosen the best response.0so the point is on the tangent plane itself, and not the original equation

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0then lets find a tangent vector that matches this contraption lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0Φ(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
 5 years ago
Best ResponseYou've already chosen the best response.0can we assume u' and v' = 1?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0im thinking yes ..... and im usually right about this unless im wrong lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0du/du = 1 and dv/dv = 1 is what im thinking...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0Φ'(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
 5 years ago
Best ResponseYou've already chosen the best response.0<3,6u+2,10v> <a, b , c >  <6, 8, 20> right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0thats 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
 5 years ago
Best ResponseYou've already chosen the best response.0its in the same plane; just have to account for rotation...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.010 means 20; i got no short term memory apparently lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the approach looks resonable

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0with 2 vectors on the plane we can cross product them to get the nomal right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0<3,6u+2,10v> <6, 8, 20>  < a, b, c> is an equivalent statement right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the cross product will give the normal i think

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0it will; but first we have to determine what the vectors should be :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0perhaps we can dot product this to an unknown vector to get an orthogonal in the plane to establish a good plave equation?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0<3,6u+2,10v> < r , s, t>  = 0

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0watchmath will surely let me know when my stupidity shows tho right ? :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0< 3, 6u+2, 10v > <6, 8, 20 >  <3, 6u+10, 10v+20> if my original thought is correct

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0lets cross these and see what we get... ;)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0(6u+2)(10v+20)  (8)(10v) [(3.20)  (6.10v)] (3.8)  (6(6u+2))

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0<(60uv 20v 120u +40) , (60  60v) ,(36u +36)> maybe ....

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if 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
 5 years ago
Best ResponseYou've already chosen the best response.0<(60uv 20v 120u +40+80v) , (60  60v) ,(36u +36)> <(60uv 60v 120u +40) , (60  60v) ,(36u +36)>

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0(60uv 60v 120u +40)(x6) +(60  60v)(y+8) +(36u +36)(z+20) = 0 is what I come up with.....

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0can we come up with a surface equation from the parametrics used? im sure we can....

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0x = 3u y = −3u^2 + 2v z = −5v^2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0u = 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
 5 years ago
Best ResponseYou've already chosen the best response.0iv often wondered if there can be a tangent plane to a curve..... it seems like trying;

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i spose if it can be done for a tangent line to a point; its possible for a plane on a curve :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0ok..... 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
 5 years ago
Best ResponseYou've already chosen the best response.03 6u 0 0 2 10v (60uv  2)i + (30v)j + (6)k <(60uv2),30v,6> is our normal to the plane whenever we determiine a u and v to input

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0to include the point (6,8,20) lets put together the plane equation (60uv2)(x6) + (30v)(y+8) + (6)(z+20) = 0

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0in 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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