Q 42.

Question

In Exercises 37-42, use the partial derivatives of gx,y=xcosy and the point 1,π 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.

Step-by-Step Solution

Verified
Answer

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

x=1+t, y=π, z=-1-t

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

x=1, y=π+t, z=-1

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

x+z=0

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

The line tangent to the surface at a,b,fa,b in the x direction is given by the parametric equation

x=a+t, y=b, z=fa,b+fxa,bt

Now we have function gx,y=xcosy and point 1,π

So a=1, b=π, fa,b=-1, fxa,b=-1

Therefore, equation of tangent in x direction is

x=1+t, y=π, z=-1-t

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

The line tangent to the surface at a,b,fa,b in the y direction is given by the parametric equation

x=a, y=b+t, z=fa,b+fya,bt

Now we have function gx,y=xcosy and point 1,π

So a=1, b=π, fa,b=-1, fya,b=0

Therefore, equation of tangent in y direction is

x=1, y=π+t, z=-1

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

The equation of plane containing the given lines and point is 

fxa,bx-a+fya,by-b=z-fa,b

-1x-1+0y-π=z+1

-x-z=0

x+z=0