Solution to Quiz 9: Derivatives and Tangent Planes - Prof. Naoki Saito, Quizzes of Calculus

The solutions to quiz 9 questions, including finding the derivative of a function at a given point in a specific direction and determining the equation for the tangent plane on a surface at a given point. Useful for students in advanced calculus or mathematics courses.

Typology: Quizzes

Pre 2010

Uploaded on 12/09/2010

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Quiz 9 Solution
Problem 1 (5 points): Find the derivative of the function f(x;y ; z) = ln(1 +
xyz)at P0(2;3;1) in the direction A=๎˜€i+j
grad (f) = fxi+fyj+fzk=yz
1+xyz i+xz
1+xyz j+xy
1+xyz k
Thus the directional derivative in that direction Ais given by:
grad (f)๎˜๎˜€i+j
j๎˜€i+jj= grad (f)๎˜๎˜€๎˜€p2i+p2j๎˜=๎˜€p2yz
1+xyz +p2xz
1+xyz
Problem 2 (5 points): Find the equation for the tangent plane on the given
surface y2๎˜€2xy ๎˜€z= 1 at point P0(1;4;7).
Take f(z; y ; z) = y2๎˜€2xy ๎˜€z๎˜€1. Then:
fx=๎˜€2y
fy= 2y๎˜€2x
fz=๎˜€1
Hence the tangent plane is given by:
fx(1;4;7) (x๎˜€1) + fy(1;4;7) (y๎˜€4) + fz(1;4;7) (z๎˜€7) = 0
๎˜€8 (x๎˜€1) + 6 (y๎˜€4) ๎˜€(z๎˜€7) = 0
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Quiz 9 Solution

Problem 1 (5 points): Find the derivative of the function f (x; y; z) = ln(1 +

xyz) at P 0 (2; 3 ; 1 ) in the direction A = i + j

grad (f ) = fxi + fy j + fz k = yz 1+xyz i + xz 1+xyz j + xy 1+xyz k

Thus the directional derivative in that direction A is given by:

grad (f )  i+j ji+jj = grad (f ) 

p 2 i +

p 2 j

p 2 yz 1+xyz

p 2 xz 1+xyz

Problem 2 (5 points): Find the equation for the tangent plane on the given

surface y 2 2 xy z = 1 at point P 0 (1; 4 ; 7).

Take f (z; y; z) = y 2 2 xy z 1. Then:

fx = 2 y fy = 2y 2 x

fz = 1

Hence the tangent plane is given by: fx (1; 4 ; 7) (x 1) + fy (1; 4 ; 7) (y 4) + fz (1; 4 ; 7) (z 7) = 0

8 (x 1) + 6 (y 4) (z 7) = 0