Finding a Tangent Plane on a Surface Examples 2

# Finding a Tangent Plane on a Surface Examples 2

Recall from the Finding a Tangent Plane on a Surface page that if $z = f(x, y)$ is a two variable real-valued function, then the equation of the tangent plane at a point $P(x_0, y_0)$ on the surface is given by the following formula:

(1)
\begin{align} \quad z - z_0 = f_x (x_0, y_0) (x - x_0) + f_y (x_0, y_0) \end{align}

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

## Example 1

Find the equation of the tangent plane that passes through point generated by $(2, \pi) \in D(f)$ on the surface $f(x, y) = \sin ( \cos (xy))$.

The partial derivatives of $f$ are $f_x(x, y) = -y\sin (xy) \cos (\cos (xy))$ and $f_y(x, y) = -x \sin (xy) \cos ( \cos (xy))$ and so $f_x(2, \pi) = -\pi \sin (2\pi) \cos (\cos (2 \pi)) = 0$ and $f_y (2 \pi) = -2 \sin (2 \pi) \cos ( \cos (2 \pi)) = 0$. Furthermore we have that $f(2, \pi) = \sin ( \cos (2\pi)) = \sin (1)$. Therefore the tangent plane that passes through the point $(2, \pi, \sin(1))$ is given by the following formula:

(2)
\begin{align} \quad z - \sin(1) = 0 \end{align} ## Example 2

Find the equation of the tangent plane that passes through point generated by $(1, 2) \in D(f)$ on the surface $f(x, y) = e^x 2^y$.

The partial derivatives of $f$ are $f_x(x, y) = e^x 2^y$ and $f_y(x, y) = \ln (2) e^x 2^y$ and so $f_x(1, 2) = 4e$ and $f_y(1, 2) = \ln (16) e$. Furthermore we have that $f(1, 2) = 4e$. Therefore the tangent plane that passes through the point $(1, 2, 4e)$ is given by the formula:

(3)
\begin{align} \quad z - 4e = 4e(x - 1) + \ln(16)e(y - 2) \end{align} ## Example 3

Find the equation of the tangent plane that passes through point generated by $(-2, 1) \in D(f)$ on the surface $f(x, y) = -2^x xy$.

The partial derivatives of $f$ are $f_x(x, y) = -y(2^x + \ln (2) 2^x x)$ and $f_y (x, y) = -2^x x$ and so $f_x(-2, 1) = -\left ( \frac{1}{4} - \frac{\ln (2)}{2} \right )$ and $f_y(-2, 1) = \frac{1}{2}$. Furthermore we have that $f(-2, 1) = \frac{1}{2}$. Therefore the tangent plane that passes through the point $\left ( -2, 1, \frac{1}{2} \right )$ is given by the formula:

(4)
\begin{align} \quad z - \frac{1}{2} = -\left ( \frac{1}{4} - \frac{\ln (2)}{2} \right ) (x + 2) + \frac{1}{2}(y - 1) \end{align} 