Chapter 12: Q 41. (page 944)

In Exercises $37 - 42$ , use the partial derivatives of $f (x, y) = x \sin y$ and the point $(2, \frac{π}{3})$ 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 $(b)$ .

Short Answer

Expert verified

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

$x = 2 + t, y = \frac{π}{3}, z = \frac{3 \sqrt{3}}{2} t$

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

$x = 2, y = \frac{π}{3} + t, z = \sqrt{3} + t$

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

$3 \sqrt{3} x + 6 y - 6 z = 2 π$

Step by step solution

Step 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 $f (x, y) = x \sin y$ and point $(2, \frac{π}{3})$

So $a = 2, b = \frac{π}{3}, f (a, b) = \sqrt{3}, f_{x} (a, b) = \frac{\sqrt{3}}{2}$

Therefore, equation of tangent in $x$ direction is

$x = 2 + t, y = \frac{π}{3}, z = \frac{3 \sqrt{3}}{2} t$

Step 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 $f (x, y) = x \sin y$ and point $(2, \frac{π}{3})$

So $x = 2, y = \frac{π}{3}, f (a, b) = \sqrt{3}, f_{y} (a, b) = 1$

Therefore, equation of tangent in $y$ direction is

$x = 2, y = \frac{π}{3} + t, z = \sqrt{3} + t$

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

The equation of plane containing the given lines and point is

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

$\Rightarrow \frac{\sqrt{3}}{2} (x - 2) + 1 (y - \frac{π}{3}) = z - \sqrt{3}$

$\Rightarrow \frac{\sqrt{3}}{2} x + y - z = \frac{π}{3}$

role="math" localid="1650292265507" $\Rightarrow 3 \sqrt{3} x + 6 y - 6 z = 2 π$

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Short Answer

Step by step solution

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

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

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

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