Question

What is the equation of a tangent plane to the surface `z = xe^(x/y)` that passes through the origin?

Answer 1

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

`z = xe^(x/y)`

And so we define our surface function, `f`, by:

`f(x,y,z) = xe^(x/y) - z`

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. The partial derivatives are:

`(partial f)/(partial x) = (x)(1/y e^(x/y)) + (1)(e^(x/y)) = x/y e^(x/y) + e^(x/y)`

`(partial f)/(partial y) = xe^(x/y)x(-1/y^2) = -x^2/y^2e^(x/y)`

`(partial f)/(partial z) = -1`

And so:

`grad f = (x/y e^(x/y) + e^(x/y))hat(i) -(x^2/y^2e^(x/y))hat(j) -1hat(k)`

So for some particular point `P(a,b,c)` the normal vector to the surface is given by:

`grad f(a,b,c) = (a/b e^(a/b) + e^(a/b))hat(i) -(a^2/b^2e^(a/b))hat(j) -1hat(k)`

So the tangent plane to the surface `z` has this normal vector, `n=grad f(a,b,c)` and it also passes through the point `P(a,b,c)`. It will therefore have a vector equation of the form:

`vec r * vec n = vec p * vec n`

Where `vec r=((x),(y),(z))`; `vec n=grad f(a,b,c)`, is the normal vector and `p` is any point in the plane, in the case `P(a.b,c)`. Hence, the tangent plane equation at `P(a,b,c)` is:

`((x),(y),(z)) * ( (a/b e^(a/b) + e^(a/b)), (-a^2/b^2e^(a/b)), (-1) ) = ((a),(b),(c)) * ( (a/b e^(a/b) + e^(a/b)), (-a^2/b^2e^(a/b)), (-1) )`

`:. (a/b e^(a/b) + e^(a/b))x -a^2/b^2e^(a/b)y - z = (a/b e^(a/b) + e^(a/b))a -a^2/b^2e^(a/b)b - c`

`:. (a/b e^(a/b) + e^(a/b))x -a^2/b^2e^(a/b)y - z = a^2/b e^(a/b) + ae^(a/b) -a^2/be^(a/b) - c`

`:. (a/b e^(a/b) + e^(a/b))x -a^2/b^2e^(a/b)y - z = ae^(a/b) - c`

But `P(a,b,c)` also lies on the surface defined by the original equation, and so it satisfies, `z = xe^(x/y)`, so;

`c = ae^(a/b)`

Inserting this value of `c` into the above general equation of the tangent planes gives us:

`(a/b e^(a/b) + e^(a/b))x -a^2/b^2e^(a/b)y - z = ae^(a/b) - ae^(a/b)`

Hence the general equation of the tangent plane at some arbitrary point `P(a,b,c)` is given by:

`(a/b e^(a/b) + e^(a/b))x -a^2/b^2e^(a/b)y - z = 0`

It is clear that `(x,y,z)=(0,0,0)` satisfies this equation from which we conclude that every tangent plane to the surface `z = xe^(x/y)` passes through the origin. QED

The graph of the surface would confirm this: