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B 6. The variables x, y and z are linked by the two relationships f(x, y,z) = x+y+z-1 = 0, g(x, y, z) = x² - 2y² + 3z² - 2 = 0. Show that the differentials dx, dy, and dz satisfy Hence find dz in terms of x, y and z. If x = 1, find the two possible numerical values dx dy dz d.x dx of y and satisfying f = g = 0, and hence find the numerical values of and at these two points. f g = 0 defines two curves lying in the three-dimensional space xyz. Find the unit vectors tangent to these curves at the two points with x = 1. dy dx dx + dy+dz = 0, rd.r – 2ydy+3zd~ =0. and

B 6. The variables x, y and z are linked by the two relationships f(x, y,z) = x+y+z-1 = 0, g(x, y, z) = x² - 2y² + 3z² - 2 = 0. Show that the differentials dx, dy, and dz satisfy Hence find dz in terms of x, y and z. If x = 1, find the two possible numerical values dx dy dz d.x dx of y and satisfying f = g = 0, and hence find the numerical values of and at these two points. f g = 0 defines two curves lying in the three-dimensional space xyz. Find the unit vectors tangent to these curves at the two points with x = 1. dy dx dx + dy+dz = 0, rd.r – 2ydy+3zd~ =0. and

Advanced Engineering Mathematics
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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can you please provide explanations

B 6. The variables x, y and z are linked by the two relationships f(x, y, z)= x+y+z-1=0, g(x, y, z) = x² - 2y² + 32² - 2 = 0. Show that the differentials dx, dy, and dz satisfy dx + dy+dz = 0, dy Hence find and dx rdr – 2ydy +3zdz = 0. dz in terms of x, y and z. If x = 1, find the two possible numerical values dx dy dz of y and satisfying f = g = 0, and hence find the numerical values of and at these two dx dx points. f = g = 0 defines two curves lying in the three-dimensional space xyz. Find the unit vectors tangent to these curves at the two points with x = 1.
Transcribed Image Text:B 6. The variables x, y and z are linked by the two relationships f(x, y, z)= x+y+z-1=0, g(x, y, z) = x² - 2y² + 32² - 2 = 0. Show that the differentials dx, dy, and dz satisfy dx + dy+dz = 0, dy Hence find and dx rdr – 2ydy +3zdz = 0. dz in terms of x, y and z. If x = 1, find the two possible numerical values dx dy dz of y and satisfying f = g = 0, and hence find the numerical values of and at these two dx dx points. f = g = 0 defines two curves lying in the three-dimensional space xyz. Find the unit vectors tangent to these curves at the two points with x = 1.
Expert Solution
Step 1: Finding differentials

The two relationships are

f left parenthesis x comma y comma z right parenthesis equals x plus y plus z minus 1 equals 0

g left parenthesis x comma y comma z right parenthesis equals x squared minus 2 y squared plus 3 z squared minus 2 equals 0

Differentiating both equations, we get

d x plus d y plus d z equals 0 a n d 2 x d x minus 4 y d y plus 6 z d z equals 0 o r x d x minus 2 y d y plus 3 z d z equals 0

Hence

d x plus d y plus d z equals 0 a n d x d x minus 2 y d y plus 3 z d z equals 0


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