Question

(a) What is the equation of the tangent plane to the surface

z = 2xy + 3y^2 + 2x + y + 4

at the point P(-4,-3,60)?

(b) What are the parametric equations of the normal line at this point?

Answer 1

For (a), I get z = -12x -25y - 63

For (b),
I get that the constant are -4, -3, and 60 for x, y, and z respectively but, I don't know how to find the coefficients of t for x, y, and z and would appreciate it if someone could help me figure this out.

Answer 2

Basically, I get it all except the coefficients of t for part (b).

Answer 3

you might want to try part (a) again

Answer 4

My answer is correct. I checked.

Answer 5

did you type the question correctly....(-4,-3,60) is not a solution to the equation you gave above

Answer 6

Sorry,

z != 2xy + 3y^2 + 2x + y + 4
z = x^2 + 2xy + 3y^2 + 2x + y + 4

Answer 7

(I'm sleepy.)

Answer 8

do you know how to find the normal vector to the plane?

Answer 9

Is it -12 i - 25 j - 63k ?

Answer 10

and do you mean *a* normal vector?

Answer 11

no that is not a normal vector to the surface

Answer 12

So, I don't know. :(

Answer 13

\[a(x-x_0)+b(y-y_0)+c(z-z_0)=0\]

is a plane with normal vector \(\vec{n}=<a,b,c>\)

Answer 14

find the coefficients infront of the x,y,z variables

Answer 15

Oh wait! Is the normal vector: 12 i + 25 j + k ?

Answer 16

(I did that based on what you just said.)

Answer 17

yes

Answer 18

Oh so the coefficients of t are the coefficients of i, j, k.

Answer 19

I get it then :)
Thanks!

Answer 20

don't forget to include your original point (-4,-3,60)