Question

# Find the planes tangent to the following surfaces at the indicated points: a) x^2+2y^2+3xz=10, at the point (1,2,\frac{1}{3}) b) y^2-x^2=3, at the point (1,2,8) c) xyz=1, at the point (1,1,1)

Functions
Find the planes tangent to the following surfaces at the indicated points:
a) $$x^2+2y^2+3xz=10$$, at the point ($$1,2,\frac{1}{3}$$)
b) $$y^2-x^2=3$$, at the point (1,2,8)
c) $$xyz=1$$, at the point (1,1,1)

2021-05-05
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:
$$x+y+z=3$$