Tangent Planes to Level Surfaces Examples 1

# Tangent Planes to Level Surfaces Examples 1

Recall from the Tangent Planes to Level Surfaces page that if we can represent a surface $S$ as a level surface of a three variable real-valued function $f(x, y, z) = k$, then the equation of the tangent plane at a point $P(x_0, y_0, z_0)$ on the surface can be calculated with the following formula:

(1)
\begin{align} \quad \quad \frac{\partial}{\partial x} f(x_0, y_0, z_0) (x - x_0) + \frac{\partial}{\partial y} f(x_0, y_0, z_0) (y - y_0) + \frac{\partial}{\partial x} f(x_0, y_0, z_0) (z - z_0) = 0 \end{align}

Recall that this formula is especially useful when the variable $z$ is not isolated in the equation of the surface. We are now going to look at some examples of calculating tangent planes using the formula above.

## Example 1

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

We note that $z^2 - x^2y = 3$ is the level surface to $w = f(x, y, z) = z^2 - x^2y$ at $k = 3$.

The partial derivatives of $f$ are $\frac{\partial w}{\partial x} = -2xy$, $\frac{\partial w}{\partial y} = -x^2$, and $\frac{\partial w}{\partial z} = 2z$. Therefore we have that $\frac{\partial}{\partial x} f(1, 1, 2) = -2$, $\frac{\partial}{\partial y} f(1, 1, 2) = -1$, and $\frac{\partial}{\partial z} f(1, 1, 2) = 4$. Applying these values to the formula above and we get that the tangent plane is given by:

(2)
\begin{equation} -2(x - 1) - 1(y - 1) + 4(z - 2) = 0 \end{equation} ## Example 2

Find the equation of the tangent plane to the surface $\cos z + x \sin y = 0$. at the point $\left ( \frac{\pi}{2}, 0, \frac{\pi}{2} \right )$.

We note that $\cos z + x \sin y = 0$ is the level surface to $w = f(x, y, z) = \cos z + x \sin y$ at $k = 0$.

The partial derivatives of $f$ are $\frac{\partial w}{\partial x} = \sin y$, $\frac{\partial w}{\partial y} = x \cos y$, and $\frac{\partial w}{\partial z} = -\sin z$. Therefore we have that $\frac{\partial}{\partial x} f \left ( \frac{\pi}{2}, 0, \frac{\pi}{2} \right ) = 0$, $\frac{\partial}{\partial y} f \left ( \frac{\pi}{2}, 0, \frac{\pi}{2} \right ) = \frac{\pi}{2}$, and $\frac{\partial}{\partial z} f \left ( \frac{\pi}{2}, 0, \frac{\pi}{2} \right ) = -1$. Applying these values to the formula above and we get that the tangent plane is given by:

(3)
\begin{align} \quad 0\left (x - \frac{\pi}{2} \right ) + \frac{\pi}{2} \left ( y - 0 \right ) -1 \left ( z - \frac{\pi}{2} \right ) = 0 \end{align} ## Example 3

Find the equation of the tangent plane to the surface $e^x + e^{yz} = 4$ at the point $(\ln (2), 2, \ln (\sqrt{2}))$.

We note that $e^x + e^{yz} = 4$ is the level surface to $w = f(x,y,z) = e^x + e^{yz}$ at $k = 4$.

The partial derivatives of $f$ are $\frac{\partial w}{\partial x} = e^x$, $\frac{\partial w}{\partial y} = ze^{yz}$, and $\frac{\partial w}{\partial z} = ye^{yz}$. Therefore we have that $\frac{\partial}{\partial x} f(\ln (2), 2, \ln (\sqrt{2})) = 2$, $\frac{\partial}{\partial y} f(\ln (2), 2, \ln (\sqrt{2})) = 2\ln (\sqrt{2}) = \ln (2)$, and $\frac{\partial}{\partial z} f(\ln (2), 2, \ln (\sqrt{2})) = 4$. Applying these values to the formula above and we get that the tangent plane is given by:

(4)
\begin{align} \quad 2(x - \ln (2)) + \ln (2) (y - 2) + 4(z - \ln (\sqrt{2})) = 0 \end{align} 