Q. 66

Question

Use the first-order partial derivatives of the functions in Exercises $65$ and $66$ to find the equation of the hyperplane tangent to the graph of the function at the indicated point P. Note that these are the same functions as in Exercises $53$ and $54$ .

f (x, y, z) = \frac{x}{y^{2} + z^{2}}, P = (4, 0, 2)

Step-by-Step Solution

Verified

Answer

The solution to the equation $f (x, y, z) = \frac{x}{y^{2} + z^{2}}, P = (4, 0, 2)$ is $x - 4 z - 4 w = - 8$

1Step1:Given data

$f (x, y, z) = \frac{x}{y^{2} + z^{2}} at point P = (x_{0}, y_{0}, z_{0}) = (4, 0, 2)$

$f_{x} (x_{0}, y_{0}, z_{0}) (x - x_{0}) + f_{y} (x_{0}, y_{0}, z_{0}) (y - y_{0}) + f_{z} (x_{0}, y_{0}, z_{0}) (z - z_{0}) = w - f (x_{0}, y_{0}, z_{0})$

$f_{x} (4, 0, 2) (x - 4) + f_{y} (4, 0, 2) (y - 0) + f_{z} (4, 0, 2) (z - 2) = w - f (4, 0, 2)$ $(1)$

2Step 2:Find fx

$f_{x} (x_{0}, y_{0}, z_{0}) = {\frac{d}{d x} f (x, y, z)|}_{(x_{0}, y_{0}, z_{0})} = {\frac{d}{d x} \frac{x}{y^{2} + z^{2}}|}_{(x_{0}, y_{0}, z_{0})}$

$f_{x} (4, 0, 2) = {\frac{1}{y^{2} + z^{2}} \frac{d}{d x} (x)|}_{(4, 0, 2)}$

$f_{x} (4, 0, 2) = {\frac{1}{y^{2} + z^{2}} \times 1|}_{(4, 0, 2)}$

$f_{x} (4, 0, 2) = {\frac{1}{0^{2} + 2^{2}}|}_{(4, 0, 2)}$

$f_{x} (4, 0, 2) = \frac{1}{4}$ $(2)$

3Step 3: Find fy

$f_{y} (x_{0}, y_{0}, z_{0}) = {\frac{d}{d y} f (x, y, z)|}_{(x_{0}, y_{0}, z_{0})} = {\frac{d}{d y} \frac{x}{y^{2} + z^{2}}|}_{(x_{0}, y_{0}, z_{0})}$

$f_{y} (x_{0}, y_{0}, z_{0}) = {x \frac{d}{d y} (\frac{1}{y^{2} + z^{2}})|}_{(x_{0}, y_{0}, z_{0})} = x \frac{d}{d y} {({(y^{2} + z^{2})}^{- 1})}_{(x_{0}, y_{0}, z_{0})}$

$f_{y} (x_{0}, y_{0}, z_{0}) = {x \frac{d}{d y} ({(y^{2} + z^{2})}^{- 1}) \frac{d}{d y} (y^{2} + z^{2})|}_{(x_{0}, y_{0}, z_{0})}$

$f_{y} (x_{0}, y_{0}, z_{0}) = - {\frac{2 x y}{{(y^{2} + z^{2})}^{2}}|}_{(x_{0}, y_{0}, z_{0})}$

$f_{y} (4, 0, 2) = - {\frac{2 x y}{{(y^{2} + z^{2})}^{2}}|}_{(4, 0, 2)} = - \frac{2 \times 4 \times 0}{{(0^{2} + 2^{2})}^{2}}$

$f_{y} (4, 0, 2) = 0$ $(3)$

4Step 4:Find fz

$f_{z} (x_{0}, y_{0}, z_{0}) = {\frac{d}{d z} f (x, y, z)|}_{(x_{0}, y_{0}, z_{0})} = {\frac{d}{d z} \frac{x}{y^{2} + z^{2}}|}_{(x_{0} - y_{0}, z_{0})}$

$f_{z} (x_{0}, y_{0}, z_{0}) = {x \frac{d}{d z} (\frac{1}{y^{2} + z^{2}})|}_{(x_{0}, y_{0}, z_{0})} = x \frac{d}{d z} {({(y^{2} + z^{2})}^{- 1})}_{(x_{0}, y_{0}, z_{0})}$

$f_{z} (x_{0}, y_{0}, z_{0}) = {x \frac{d}{d z} ({(y^{2} + z^{2})}^{- 1}) \frac{d}{d z} (y^{2} + z^{2})|}_{(x_{0}, y_{0}, z_{0})}$

$f_{z} (x_{0}, y_{0}, z_{0}) = {x (- \frac{1}{{(y^{2} + z^{2})}^{2}}) \cdot 2 z|}_{(x_{0}, y_{0}, z_{0})} = - {\frac{2 x z}{{(y^{2} + z^{2})}^{2}}|}_{(x_{0}, y_{0}, z_{0})}$

$f_{y} (4, 0, 2) = - {\frac{2 x y}{{(y^{2} + z^{2})}^{2}}|}_{(4, 0, 2)} = - \frac{2 \times 4 \times 2}{{(0^{2} + 2^{2})}^{2}}$

$f_{z} (4, 0, 2) = - \frac{16}{16}$

$f_{z} (4, 0, 2) = - 1$ $(4)$

5Step 5: Find(fx,fy,fz)

$f (x_{0}, y_{0}, z_{0}) = {\frac{x}{y^{2} + z^{2}}|}_{(x_{0}, y_{0}, z_{0})}$

$f (x_{0}, y_{0}, z_{0}) = f (4, 0, 2) = \frac{4}{0^{2} + 2^{2}}$

$f (x_{0}, y_{0}, z_{0}) = 1$ $(5)$

6Step6:Solution

Substituting Equation $f_{x} (4, 0, 2) = \frac{1}{4}$ , $f_{y} (4, 0, 2) = 0$ , $f_{z} (4, 0, 2) = - 1$ and $f (x_{0}, y_{0}, z_{0}) = 1$ in $f_{x} (4, 0, 2) (x - 4) + f_{y} (4, 0, 2) (y - 0) + f_{z} (4, 0, 2) (z - 2) = w - f (4, 0, 2)$ get