Question

What is the equation of the tangent plane to the graph of `g(x,y)= e^{-y}sin(x)+e^{-2y}` at `(0,0,1)`...? Thanks :)

Answer 1

`pi: qquad - x + 2y + z = 1`

Explanation:

You're looking for the normal vector for the tangent plane.

There are at least 2 ways to do this:

1)

For function:

• `z = g(x,y)= e^{-y}sin(x)+e^{-2y}`

Create level surface S:

• `S(x,y,z) = e^{-y}sin(x)+e^{-2y} - z`

It's normal is:

• `nabla S = ((e^(-y)cos x),(- e^(-y) sin x - 2 e^(- 2y)),(-1))`

At `(0,0,1)` that evaluates as `mathbf n = (( 1),( - 2 ),(-1))`

The plane is then: `(mathbf r - mathbf r_o)* mathbf n = 0` where you can use `mathbf r_o = langle 0,0,1 rangle`

So

• `pi: qquad - x + 2y + z = 1`

2)

Or you can write the surface as being parameterised in x and y:

`mathbf r = ((x),(y),(e^{-y}sin(x)+e^{-2y} ))`

.... with tangent vectors:

• `mathbf r_x = ((1),(0),(e^{-y}cos(x) ))`

• `mathbf r_y = ((0),(1),(- e^{-y}sin(x)- 2 e^{-2y} ))`

These evaluate as `langle 1,0, 1 rangle` and `langle 0,1, -2 rangle`

Their cross product is:

`det ((hat x, hat y , hat z),(1,0,1),(0,1,-2)) = langle -1, 2, 1 rangle = mathbf n`

[Same vector, just pointing in the other direction.]

Answer 2

• `pi: qquad - x + 2y + z = 1`

Explanation:

You're looking for the normal vector for the tangent plane.

There are at least 2 ways to do this:

1)

For function:

• `z = g(x,y)= e^{-y}sin(x)+e^{-2y}`

Create level surface S:

• `S(x,y,z) = e^{-y}sin(x)+e^{-2y} - z`

From the directional derivative, it's normal is:

• `nabla S = ((e^(-y)cos x),(- e^(-y) sin x - 2 e^(- 2y)),(-1))`

At `(0,0,1)` that evaluates as `mathbf n = (( 1),( - 2 ),(-1))`

The plane is then: `(mathbf r - mathbf r_o)* mathbf n = 0` where you can use `mathbf r_o = langle 0,0,1 rangle`

So

• `pi: qquad - x + 2y + z = 1`

2)

Or you can write the surface as being parameterised in x and y:

`mathbf r = ((x),(y),(e^{-y}sin(x)+e^{-2y} ))`

.... with tangent vectors:

• `mathbf r_x = ((1),(0),(e^{-y}cos(x) ))`

• `mathbf r_y = ((0),(1),(- e^{-y}sin(x)- 2 e^{-2y} ))`

These evaluate as `langle 1,0, 1 rangle` and `langle 0,1, -2 rangle`

Their cross product is:

`det ((hat x, hat y , hat z),(1,0,1),(0,1,-2)) = langle -1, 2, 1 rangle = mathbf n`

[Same vector, just pointing in the other direction.]