MATH1231 - Tangent Planes

## 1. Take partial derivative of x (∂z/∂x)

• This will be in terms of x and y, so sub in the points you have for them
• Use the point gradient formula with z and x, leaving y to its original value:
(1)
\begin{align} z – z_1 = \frac {∂z}{∂x}(x-x_1), y = y_1 \end{align}
• Set $λ = (x – x_1)$, and then put the equation in parametric vector form:
(2)
\begin{align} (x, y, z) = (x_1, y_1, z_1) + λ(1, 0, \frac {∂z}{∂x}) \end{align}

## 2. Take partial derivative of y (∂z/∂y)

• This will be in terms of x and y, so sub in the points you have for them.
• Use the point gradient formula with z and y, leaving x in its original form:
(3)
\begin{align} z – z_1 = \frac {∂z}{∂y}(y-y_1), x = x_1 \end{align}
• Set $μ = (y – y_1)$, and then put the equation in parametric vector form:
(4)
\begin{align} (x, y, z) = (x_1, y_1, z_1) + μ(0, 1, \frac {∂z}{∂y}) \end{align}

## 3. Finding the Tangent Plane

• Both of these lines lie in the tangent plane, and since the two vectors are non-parallel, the tangent plane to the surface at $(x_1, y_1, z_1)$ is equal to
(5)
\begin{align} (x, y, z) = (x_1, y_1, z_1) + λ(1, 0, \frac {∂z}{∂x}) + μ(0, 1, \frac {∂z}{∂y}) \end{align}

## 4. Finding the Normal

• We now need to find the normal to the surface at the point $(x_1, y_1, z_1)$
• This equals the cross product of $(0, 1, \frac {∂z}{∂y}) and (1, 0, \frac {∂z}{∂x})$
• So equals $(\frac {∂z}{∂x}, \frac {∂z}{∂y}, -1)$
• Hence tangent plane: $n \cdot (x – x_1, y – y_1, z – z_1) = 0$
page revision: 9, last edited: 12 Aug 2011 01:52