• s3a

(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?

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  • s3a

(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?

Mathematics
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  • s3a
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.
  • s3a
Basically, I get it all except the coefficients of t for part (b).
you might want to try part (a) again

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  • s3a
My answer is correct. I checked.
did you type the question correctly....(-4,-3,60) is not a solution to the equation you gave above
  • s3a
Sorry, z != 2xy + 3y^2 + 2x + y + 4 z = x^2 + 2xy + 3y^2 + 2x + y + 4
  • s3a
(I'm sleepy.)
do you know how to find the normal vector to the plane?
  • s3a
Is it -12 i - 25 j - 63k ?
  • s3a
and do you mean *a* normal vector?
no that is not a normal vector to the surface
  • s3a
So, I don't know. :(
\[a(x-x_0)+b(y-y_0)+c(z-z_0)=0\] is a plane with normal vector \(\vec{n}=\)
find the coefficients infront of the x,y,z variables
  • s3a
Oh wait! Is the normal vector: 12 i + 25 j + k ?
  • s3a
(I did that based on what you just said.)
yes
  • s3a
Oh so the coefficients of t are the coefficients of i, j, k.
  • s3a
I get it then :) Thanks!
don't forget to include your original point (-4,-3,60)

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