Calculus

[nikuhalal]

I went through different website and asked for help but i did not get free site. I could not afford weekly fee for solving the questions.

Fortunatelly I got this site. It is great for me.

Could you please help me to solve this problem\


Consider the surface Z= 9-x^2-9y^2

a. if Z = f(x,y), describe what gradient (delta) f represents and how it is oriented with respect to the graph of z= 9-x^2-9y^2.
b. Treating the surface Z= 9-x^2-9y^2 as a level surface of a function g(x,y,z), describe what gradient ( delta) g represents and how it is oriented with respect to the graph of Z= 9-x^2-9y^2.
c. Find the points on Z= 9-x^2-9y^2 where the tangent plane is parallel to the plane 2x+4y-2z=1.

This is my first question in ths help site. So I kindly request you to solve the problem.

niku

 

[Deleted member 4993]

I went through different website and asked for help but i did not get free site. I could not afford weekly fee for solving the questions.

Fortunatelly I got this site. It is great for me.

Could you please help me to solve this problem\


Consider the surface Z= 9-x^2-9y^2

a. if Z = f(x,y), describe what gradient (delta) f represents and how it is oriented with respect to the graph of z= 9-x^2-9y^2.
b. Treating the surface Z= 9-x^2-9y^2 as a level surface of a function g(x,y,z), describe what gradient ( delta) g represents and how it is oriented with respect to the graph of Z= 9-x^2-9y^2.
c. Find the points on Z= 9-x^2-9y^2 where the tangent plane is parallel to the plane 2x+4y-2z=1.

This is my first question in ths help site. So I kindly request you to solve the problem.

niku

Please share with us your work, indicating exactly where you are stuck - so that we may know where to begin to help you.

 

[galactus]

c. Find the points on \(\displaystyle Z= 9-x^{2}-9y^{2}\) where the tangent plane is parallel to the plane \(\displaystyle 2x+4y-2z=1\).

\(\displaystyle \frac{{\partial}z}{{\partial}x}=-2x\)

\(\displaystyle \frac{{\partial}z}{{\partial}y}=-18y\)

So \(\displaystyle -2x_{0}i-18y_{0}j-k\) is normal to the surface at a point \(\displaystyle (x_{0},y_{0},z_{0})\) on the surface.

\(\displaystyle 2i+4j-2k\) is normal to the given plane.

The tangent plane and the given plane are parallel if their normals are parallel:

\(\displaystyle -2x_{0}=2\Rightarrow x_{0}=-1\)

\(\displaystyle -18y_{0}=4\Rightarrow y_{0}=\frac{-2}{9}\)

\(\displaystyle z=\frac{68}{9}\) at \(\displaystyle (-1,\frac{-2}{9})\)

The point is \(\displaystyle (-1,\frac{-2}{9},\frac{68}{9})\)