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