Question

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.

Answer 1

oh im just simply salivating over this one lol

Answer 2

please take your time, thanks

Answer 3

we take the point; and apply the gradient to it right?

Answer 4

but the initial form seems to be in a vector position equation...

Answer 5

hmm i am trying to find something in my book

Answer 6

(3u) i + (3u^2 +2v) j + (-5v^2) k

Answer 7

the derivative of a vector position is the tangent line .... if i recall correctly

Answer 8

tangent vector that is...

Answer 9

i believe so

Answer 10

3u = 6

3u^2 +2v = -8

-5v^2 = -20

should give us values for u and v maybe?

Answer 11

it's worth a try

Answer 12

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..

Answer 13

so the point is on the tangent plane itself, and not the original equation

Answer 14

yes

Answer 15

then lets find a tangent vector that matches this contraption lol

Answer 16

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

Answer 17

can we assume u' and v' = 1?

Answer 18

i don't know

Answer 19

im thinking yes ..... and im usually right about this unless im wrong lol

Answer 20

du/du = 1 and dv/dv = 1 is what im thinking...

Answer 21

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....

Answer 22

<3,-6u+2,-10v>
<a, b , c >
-----------------
<6, -8, -20> right?

Answer 23

you mean add them?

Answer 24

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..

Answer 25

its in the same plane; just have to account for rotation...

Answer 26

-10 means -20; i got no short term memory apparently lol

Answer 27

the approach looks resonable

Answer 28

with 2 vectors on the plane we can cross product them to get the nomal right?

Answer 29

<3,-6u+2,-10v>
-<6, -8, -20>
-----------------
< a, b, c> is an equivalent statement right?

Answer 30

the cross product will give the normal i think

Answer 31

it will; but first we have to determine what the vectors should be :)

Answer 32

perhaps we can dot product this to an unknown vector to get an orthogonal in the plane to establish a good plave equation?

Answer 33

<3,-6u+2,-10v>
< r , s, t>
---------------
= 0

Answer 34

watchmath will surely let me know when my stupidity shows tho right ? :)

Answer 35

< 3, -6u+2, -10v >
<-6, 8, 20 >
---------------------
<-3, -6u+10, -10v+20> if my original thought is correct

Answer 36

lets cross these and see what we get... ;)

Answer 37

(-6u+2)(-10v+20) - (8)(-10v)

-[(3.20) - (-6.-10v)]

(3.8) - (-6(-6u+2))

Answer 38

<(60uv -20v -120u +40) , (-60 - 60v) ,(-36u +36)>

maybe ....

Answer 39

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

Answer 40

<(60uv -20v -120u +40+80v) , (-60 - 60v) ,(-36u +36)>

<(60uv -60v -120u +40) , (-60 - 60v) ,(-36u +36)>

Answer 41

(60uv -60v -120u +40)(x-6)

+(-60 - 60v)(y+8)

+(-36u +36)(z+20)

= 0 is what I come up with.....

Answer 42

can we come up with a surface equation from the parametrics used? im sure we can....

Answer 43

x = 3u
y = −3u^2 + 2v
z = −5v^2

Answer 44

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

Answer 45

iv often wondered if there can be a tangent plane to a curve..... it seems like trying;

Answer 46

i spose if it can be done for a tangent line to a point; its possible for a plane on a curve :)

Answer 47

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

Answer 48

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

Answer 49

to include the point (6,-8,-20) lets put together the plane equation

(60uv-2)(x-6) + (30v)(y+8) + (6)(z+20) = 0

Answer 50

in order for the coeff of x = 240, then

60uv -2 = 240

60uv = 242

uv = 242/60 if i did it right