მს For the function (x, y) = √√x²y², we have that 8 (0,0) = 0 and (0,0) = 0. From this, it follows that is differentiable at the origin. Hence the linearisation (tangent plane approximation) of at the point (0, 0) is given by: მს მს z - (0,0) = -(0, 0) (x −0) + -(0, 0) (y - 0) dy Ox z = 0+ 0(x0) + 0(y − 0) = 0. Therefore, from the above computation, it follows that the xy-plane (z = 0) is the tangent plane to at the origin. State whether you agree or disagree with the foregoing argument and claim. Give a brief reason for your answer.
მს For the function (x, y) = √√x²y², we have that 8 (0,0) = 0 and (0,0) = 0. From this, it follows that is differentiable at the origin. Hence the linearisation (tangent plane approximation) of at the point (0, 0) is given by: მს მს z - (0,0) = -(0, 0) (x −0) + -(0, 0) (y - 0) dy Ox z = 0+ 0(x0) + 0(y − 0) = 0. Therefore, from the above computation, it follows that the xy-plane (z = 0) is the tangent plane to at the origin. State whether you agree or disagree with the foregoing argument and claim. Give a brief reason for your answer.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:For the function v(x, y) = /x²y², we have that (0,0) = 0 and (0,0) = 0.
dy
From this, it follows that is differentiable at the origin. Hence the linearisation
(tangent plane approximation) of at the point (0,0) is given by:
2 – Þ(0,0)
(0,0)(x – 0) +
(0,0)(y – 0)
ду
0 + 0(x – 0) + 0(y – 0) = 0.
Therefore, from the above computation, it follows that the xy-plane (z = 0) is
%3D
the tangent plane to at the origin.
State whether you agree or disagree with the foregoing argument and claim.
Give a brief reason for your answer.
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