Finding a Tangent Plane on a Surface

We have just defined what a tangent plane to a surface $S$ at the point on the surface is. We will now go about finding equations for these tangent planes similarly to how we found equations of tangent lines for points on single variable functions.

Let $z = f(x, y)$ be a function that generates the surface $S$ and let $P(x_0, y_0, z_0)$ be a point on $S$, and suppose that we want to find the tangent plane to this point. Clearly, the tangent plane to $P$ passes through $P$ and so this plane has an equation in the form of (by letting $-\frac{A}{C} = a$ and $-\frac{B}{C} = b$):

(1)

\begin{align} A(x - x_0) + B(y - y_0) + C(z - z_0) = 0 \\ Ax - Ax_0 + By - By_0 + Cz - Cz_0 = 0 \\ Cz - Cz_0 = -Ax + Ax_0 - By + By_0 \\ C(z - z_0) = -Ax + Ax_0 - By + By_0 \\ z - z_0 = \frac{-A}{C}x + \frac{A}{C}x - \frac{B}{C}y + \frac{B}{C}y_0 \\ z - z_0 = a(x - x_0) + b(y - y_0) \end{align}

So $z - z_0 = a(x - x_0) + b(y - y_0)$ is the equation of the tangent plane at $P$. If we intersect the tangent plane at $P$ with the plane $y = y_0$ then the intersection is:

(2)

\begin{align} z - z_0 = a(x - x_0) + b(y_0 - y_0) \\ z - z_0 = a(x - x_0) \end{align}

We recognize this equation as the point-slope form of the tangent line of the curve of intersection of $S$ with $y = y_0$ at the point $P$. Recall from the Partial Derivatives page that the slope of this line is given by the partial derivative $f_x (x_0, y_0) = \frac{\partial}{\partial x} f(x_0, y_0)$, and so $a (x - x_0) = f_x (x_0, y_0) (x - x_0)$.

Similarly, if we intersect the tangent plane at $P$ with the plane $x = x_0$ then the intersection is:

(3)

\begin{align} z - z_0 = a(x_0 - x_0) + b(y - y_0) \\ z - z_0 = b(y - y_0) \end{align}

Once again, we recognize this equation as the point-slope form of the tangent line of the curve of intersection of $S$ with $x = x_0$ at the point $P$. The slope of this line is given by the partial derivative $f_y (x_0, y_0) = \frac{\partial}{\partial y} f(x_0, y_0)$, and so $b(y - y_0) = f_y (x_0, y_0) (y - y_0)$.

Substituting these into the $z - z_0 = a(x - x_0) + b(y - y_0)$ and the equation of the tangent plane at $P$ is given by the following formula:

(4)

\begin{align} z - z_0 = a (x - x_0) + b (y - y_0) \\ z - z_0 = f_x (x_0, y_0) (x - x_0) + f_y (x_0, y_0) (y - y_0) \end{align}

Definition: If $z = f(x, y)$ is a two variable real-valued function that generates the surface $S$, then the Equation of the Tangent Plane at $P(x_0, y_0, z_0)$ is given by the formula $z - z_0 = f_x (x_0, y_0) (x - x_0) + f_y (x_0, y_0) (y - y_0)$ provided that $f$ has continuous partial derivatives.

We will now look at some examples of finding tangent planes to points on surfaces.

Example 1

Find the equation of the tangent plane to the surface $f(x,y) = x^2y^3 + 4xy^5$ at the point $(1, 1, 5)$.

We note that $f_x (x, y) = 2xy^3 + 4y^5$ and so $f_x (1, 1) = 6$. Also, we note that $f_y (x, y) = 3x^2y^2 + 20xy^4$ and so $f_y (1, 1) = 23$. Therefore the equation of the tangent plane at $(1, 1, 5)$ is:

(5)

\begin{align} z - 5 = f_x (1, 1) (x - 1) + f_y (1, 1) (y - 1) \\ z - 5 = 6(x - 1) + 23(y - 1) \end{align}

Example 2

**Find the equation of the tangent plane to the paraboloid $z = 2x^2 + 3y^2$ at the point $(1, 2, 14)$.

We note that $f_x (x, y) = 4x$ and so $f_x (1, 2) = 4$. Also, we note that $f_y (x, y) = 6y$ and so $f_y (1, 2) = 12$. Therefore the equation of the tangent plane at $(1, 2, 14)$ is:

(6)

\begin{align} z - 14 = f_x (1, 2) (x - 1) + f_y (1, 2) (y - 2) \\ z - 14 = 4(x - 1) + 12(y - 2) \end{align}