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Q 38.

Expert-verified
Calculus
Found in: Page 944
Calculus

Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

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Short Answer

In Exercises 37-42, use the partial derivatives of role="math" localid="1650186853142" gx,y=tan-1xy2 and the point role="math" localid="1650186870407" 1,0 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.

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

x=1+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=1, y=t, z=0

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

z=0

See the step by step solution

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,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=tan-1xy2 and point 1,0

So localid="1650290477299" a=1, b=0, fa,b=0, fxa,b=0

Therefore, equation of tangent in x direction is

x=1+t, y=0, z=0

Step 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=tan-1xy2 and point 1,0

So a=1, b=0, fa,b=0, fya,b=0

Therefore, equation of tangent in y direction is

role="math" localid="1650290595556" x=1, y=t, z=0

Step 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

0x-1+0y-0=z-0

z=0

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