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

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- amistre64

oh im just simply salivating over this one lol

- anonymous

please take your time, thanks

- amistre64

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- amistre64

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

- anonymous

hmm i am trying to find something in my book

- amistre64

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

- amistre64

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

- amistre64

tangent vector that is...

- anonymous

i believe so

- amistre64

3u = 6
3u^2 +2v = -8
-5v^2 = -20
should give us values for u and v maybe?

- anonymous

it's worth a try

- 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

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

- anonymous

yes

- amistre64

then lets find a tangent vector that matches this contraption lol

- 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

can we assume u' and v' = 1?

- anonymous

i don't know

- amistre64

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

- amistre64

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

- 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

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

- anonymous

you mean add them?

- 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

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

- amistre64

##### 1 Attachment

- amistre64

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

- anonymous

the approach looks resonable

- amistre64

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

- amistre64

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

- anonymous

the cross product will give the normal i think

- amistre64

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

- 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

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

- amistre64

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

- amistre64

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

- amistre64

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

- amistre64

(-6u+2)(-10v+20) - (8)(-10v)
-[(3.20) - (-6.-10v)]
(3.8) - (-6(-6u+2))

- amistre64

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

- 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

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

- amistre64

(60uv -60v -120u +40)(x-6)
+(-60 - 60v)(y+8)
+(-36u +36)(z+20)
= 0 is what I come up with.....

- amistre64

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

- amistre64

x = 3u
y = −3u^2 + 2v
z = −5v^2

- 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

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

- amistre64

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

- 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

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

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

in order for the coeff of x = 240, then
60uv -2 = 240
60uv = 242
uv = 242/60 if i did it right

Looking for something else?

Not the answer you are looking for? Search for more explanations.