2

Find the equation of the tangent plane to the surface $z=x^2+2y^3$ at the point $(1,1,3)$. I think that it is $z=2x+6y-5$. Is that right?

Zev Chonoles
  • 129,973
Maximilian1988
  • 1,323
  • it would be better if you showed us how you arrived at that answer – cats Feb 19 '13 at 06:00
  • 1
    The answer is correct. Presumably you arrived at it through the right calculation. – André Nicolas Feb 19 '13 at 06:24
  • oh thank god! I did it at the exam I don't even know how..my brain isn't working anymore...and something else...I was asked how f(x,Y)=3x^2+3y^2 look like, i had the following choices: a) hyperbolas, b) circles, c) parabolas d) straight lines... I chose circles, why? because I know that x^2+y^2 represents a circle, that was my guess, because honestly, I had no idea..so am I right? – Maximilian1988 Feb 19 '13 at 06:32
  • Cross-sections of $z=3x^2+3y^2$ parallel to the $x$-$y$ plane are expanding circles. As to $z=3x^2+3y^2$, it is a three-dimensional surface. – André Nicolas Feb 19 '13 at 06:43
  • yay! so I guess I got it correct...buuf thanks :) – Maximilian1988 Feb 19 '13 at 06:45

1 Answers1

4

The normal to the surface $x^2+2y^3-z=0$ is parallel the gradient: $(2x,6y^2,-1)$

At $(1,1,3)$, this is $(2,6,-1)$. The tangent plane is perpendicular to this vector. That is, $$ (2,6,-1)\cdot(x,y,z)=2x+6y-z=5\tag{1} $$ where $5$ is gotten by plugging $(1,1,3)$ into $2x+6y-z$. $(1)$ is the equation of the tangent plane.

robjohn
  • 345,667