Let \(f:\mathbb{R}^n\to\mathbb{R}\) be a differentiable function. Recal that the tangent plane of the surface consisting of points such that f(x)=k for some constant k at the point \(x_0\) is given with:

\(\triangledown f(x_0)\cdot(x-x_0)=0\) (1)

a) Here we set \(f(x,y,z)=x^2+2y^2+3xz,x_0=(1,2,\frac{1}{3})\) and k=10. Note that \(f(x_0)=k\).

Now we can calculate:

\(\triangledown f(x)=(2x+3z,4y,3x)\Rightarrow\triangledown f(1,2,\frac{1}{3})=(3,8,3)\)

Using (1) we easily get:

\(0=(3,8,3)\cdot(x-1,y-2,z-\frac{1}{3})=3(x-1)+8(y-2)+3(z-\frac{1}{3})\)

Thus, the tangent plane equation is:

\(3x+8y+3z=20\)

b) Here we set \(f(x,y,z)=y^2-x^2, x_0=(1,2,8)\) and k=3. Note that \(f(x_0)=k\)

Now we can calculate:

\(\triangledown f(x)=(-2x,2y,0)\Rightarrow\triangledown f(1,2,8)=(-2,4,0)\)

Using (1) we easily get:

\(0=(-2,4,0)\cdot(x-1,y-2,z-8)=-2(x-1)+4(y-2)+0(z-8)\)

Thus, the tangent plane equation is:

c) Here we set \(f(x,y,z)=xyz,x_0=(1,1,1)\) and k=1. Note that \(f(x_0)=k\). Now we can calculate:

\(\triangledown f(x)=(yz,xz,xy)\Rightarrow\triangledown f(1,1,1)=(1,1,1)\)

Using (1) we easily get:

\(0=(1,1,1)\cdot(x-1,y-1,z-1)=1(x-1)+1(y-1)+1(z-1)\)

Thus, the tangent plane equation is:

\(x+y+z=3\)

\(\triangledown f(x_0)\cdot(x-x_0)=0\) (1)

a) Here we set \(f(x,y,z)=x^2+2y^2+3xz,x_0=(1,2,\frac{1}{3})\) and k=10. Note that \(f(x_0)=k\).

Now we can calculate:

\(\triangledown f(x)=(2x+3z,4y,3x)\Rightarrow\triangledown f(1,2,\frac{1}{3})=(3,8,3)\)

Using (1) we easily get:

\(0=(3,8,3)\cdot(x-1,y-2,z-\frac{1}{3})=3(x-1)+8(y-2)+3(z-\frac{1}{3})\)

Thus, the tangent plane equation is:

\(3x+8y+3z=20\)

b) Here we set \(f(x,y,z)=y^2-x^2, x_0=(1,2,8)\) and k=3. Note that \(f(x_0)=k\)

Now we can calculate:

\(\triangledown f(x)=(-2x,2y,0)\Rightarrow\triangledown f(1,2,8)=(-2,4,0)\)

Using (1) we easily get:

\(0=(-2,4,0)\cdot(x-1,y-2,z-8)=-2(x-1)+4(y-2)+0(z-8)\)

Thus, the tangent plane equation is:

c) Here we set \(f(x,y,z)=xyz,x_0=(1,1,1)\) and k=1. Note that \(f(x_0)=k\). Now we can calculate:

\(\triangledown f(x)=(yz,xz,xy)\Rightarrow\triangledown f(1,1,1)=(1,1,1)\)

Using (1) we easily get:

\(0=(1,1,1)\cdot(x-1,y-1,z-1)=1(x-1)+1(y-1)+1(z-1)\)

Thus, the tangent plane equation is:

\(x+y+z=3\)