Q. 37 In Exercises 37-42, use the part... [FREE SOLUTION]

Chapter 12: Q. 37 (page 944)

In Exercises 37-42, use the partial derivatives you found in Exercises 27-32 and the point 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).
Exercise 27, $(0, \frac{蟺}{2})$

Short Answer

Expert verified

Ans:

part (a). $x = t, y = \frac{蟺}{2}, z = \frac{蟺}{2} t$

part (b). $x = 0, y = \frac{蟺}{2} + t, z = 0$

part (c). $f_{x} (0, \frac{蟺}{2}) (x - 0) + f_{y} (0, \frac{蟺}{2}) (y - \frac{蟺}{2}) = z - f (0, \frac{蟺}{2}) \frac{蟺}{2} x = z$

Step by step solution

Step 1. Given information:

$(0, \frac{蟺}{2})$

Step 2. Differentiating the function:

Function $f = e^{x} \sin (x y)$

Differentiating f partially with respect to x

$f_{x} = e^{x} [\cos (x y)] y + [\sin (x y)] e^{x} = e^{x} [y \cos (x y) + \sin (x y)]$

Differentiating f partially with respect to y

$f_{y} = e^{x} [\cos (x y)] x = x e^{x} \cos (x y)$

Step 3. Solving part (a):

The equation of the line tangent to the surface defined by f in the x-direction is,

x = 0 + t, y = \frac{蟺}{2}, z = f (0, \frac{蟺}{2}) + f_{x} (0, \frac{蟺}{2}) t x = t, y = \frac{蟺}{2}, z = e^{0} \sin (0 (\frac{蟺}{2})) + e^{0} [\frac{蟺}{2} \cos (0 (\frac{蟺}{2})) + \sin (0 (\frac{蟺}{2}))] t x = t, y = \frac{蟺}{2}, z = \frac{蟺}{2} t

Step 4. Solving part (b):

The equation of the line tangent to the surface defined by f in the y-direction is,

$x = 0, y = \frac{蟺}{2} + t, z = f (0, \frac{蟺}{2}) + f_{y} (0, \frac{蟺}{2}) t x = 0, y = \frac{蟺}{2} + t, z = e^{0} \sin (0 (\frac{蟺}{2})) + 0 e^{0} \cos (0 (\frac{蟺}{2})) t x = 0, y = \frac{蟺}{2} + t, z = 0$

Step 5. Solving part (c):

The equation of the plane containing the lines in part (a) and part (b) is

$f_{x} (0, \frac{蟺}{2}) (x - 0) + f_{y} (0, \frac{蟺}{2}) (y - \frac{蟺}{2}) = z - f (0, \frac{蟺}{2}) e^{0} [\frac{蟺}{2} \cos (0 (\frac{蟺}{2})) + \sin (0 (\frac{蟺}{2}))] x + 0 e^{0} \cos (0 (\frac{蟺}{2})) (y - \frac{蟺}{2}) = z - e^{0} \sin (0 (\frac{蟺}{2})) \frac{蟺}{2} x = z$

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Short Answer

Step by step solution

Step 1. Given information:

Step 2. Differentiating the function:

Step 3. Solving part (a):

Step 4. Solving part (b):

Step 5. Solving part (c):

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