Q 40.

In Exercises $37 - 42$ , use the partial derivatives of $g (x, y) = \frac{\ln (y^{2} + 1)}{x}$ and the point $(3, 0)$ specified to

$(a)$ find the equation of the line tangent to the surface defined by the function in the $x$ direction,

$(b)$ find the equation of the line tangent to the surface defined by the function in the $y$ direction, and

$(c)$ find the equation of the plane containing the lines you found in parts $(a)$ and $(b)$ .

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Answer

Part $(a)$ , The equation of the line tangent to the surface defined by the function in the $x$ direction is

$x = 3 + t, y = 0, z = 0$

Part $(b)$ , The equation of the line tangent to the surface defined by the function in the $y$ direction is

$x = 3, y = t, z = 0$

Part $(c)$ , The equation of the plane containing the lines you found in parts $(a)$ and $(b)$ is

$z = 0$

1Step 1. Explanation of part a , Finding equation of tangent in x direction

The line tangent to the surface at $(a, b, f (a, b))$ in the $x$ direction is given by the parametric equation

$x = a + t, y = b, z = f (a, b) + f_{x} (a, b) t$

Now we have function $g (x, y) = \frac{\ln (y^{2} + 1)}{x}$ and point $(3, 0)$

So $a = 3, b = 0, f (a, b) = 0, f_{x} (a, b) = 0$

Therefore, equation of tangent in $x$ direction is

$x = 3 + t, y = 0, z = 0$

2Step 2. Explanation of part b , Finding equation of tangent in y direction

The line tangent to the surface at $(a, b,, f (a, b))$ in the $y$ direction is given by the parametric equation

$x = a, y = b + t, z = f (a, b) + f_{y} (a, b) t$

Now we have function $g (x, y) = \frac{\ln (y^{2} + 1)}{x}$ and point $(3, 0)$

So $a = 3, b = 0, f (a, b) = 0, f_{y} (a, b) = 0$

Therefore, equation of tangent in $y$ direction is

$x = 3, y = t, z = 0$

3Step 3. Explanation of part c , Finding equation of plane

The equation of plane containing the given lines and point is

$f_{x} (a, b) (x - a) + f_{y} (a, b) (y - b) = z - f (a, b)$

$\Rightarrow 0 (x - 3) + 0 (y - 0) = z - 0$

$\Rightarrow z = 0$