Question

How do you find the equations for the tangent plane to the surface `x=y(2z-3)` through `(4, 4, 2)`?

Answer 1

`x-y-8z =-16`

Explanation:

First we rearrange the equation of the surface into the form `f(x,y,z)=0`

`x=y(2z-3)`
`:. x=2yz-3y`
`:. x+3y-2yz = 0`

And so we have our function:

`f(x,y,z) = x+3y-2yz`

In order to find the normal at any particular point in vector space we use the Del, or gradient operator:

`grad f(x,y,z) = (partial f)/(partial x) hat(i) + (partial f)/(partial y) hat(j) + (partial f)/(partial z) hat(k)`

remember when partially differentiating that we differentiate wrt the variable in question whilst treating the other variables as constant. And so:

`grad f = ((partial)/(partial x) (x+3y-2yz))hat(i) +`
`" " ((partial)/(partial y) (x+3y-2yz))hat(j) +`
`" " ((partial)/(partial z) (x+3y-2yz))hat(k)`
`" "= (1)hat(i) + (3-2z)hat(j) + (-2y)hat(k)`
`" "= hat(i) + (3-2z)hat(j) -(2y)hat(k)`

So for the particular point `(4,4,2)` the normal vector to the surface is given by:

`grad f(4,4,2) = hat(i) + (3-2*2)hat(j) -(2*4)hat(k)`
`" " = hat(i) -hat(j) -8hat(k)`
`" " = ( (1), (-1),(-8) )`

So the tangent plane to the surface `x=y(2z-3)` has this normal vector and it also passes though the point `(4,4,2)`. It will therefore have a vector equation of the form:

`vec r * vec n = vec a * vec n`

Where `vec r=((x),(y),(z))`; `vec n` is the normal, and `a` is any point in the plane

Hence, the tangent plane equation is:

`((x),(y),(z)) * ( (1), (-1),(-8) ) = ((4),(4),(2)) * ( (1), (-1),(-8) )`
`:. (x)(1) + (y)(-1) + (z)(-8) = (4)(1) + (4)(-1) + (2)(-8)`
`:. x-y-8z = 4-4-16`
`:. x-y-8z =-16`

We can confirm this graphically: Here is the surface with the normal vector:

and here is the surface with the tangent plane and the normal vector: