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Thus, the equation of the streamline passing through point (1 m, 2 m) is x² = Determine the derivative of the equation with respect to y. 6y(3y²)-(y³+5)(6) (бу? 18y -бу-30 36у? 2r4 dy 2rd dy 2y³–5 6y? dy 2y³–5 dy 12xy? Set the derivative d = 0 and rewrite the above equation. 2y³–5 0 = 12xy2 2у — 5 %3D 0 y = 1.357 m Determine the corresponding value of x. 5+y x = бу 5+(1.357 m) 6(1.357 m) x? : x = 0.960 m

Thus, the equation of the streamline passing through point (1 m, 2 m) is x² = Determine the derivative of the equation with respect to y. 6y(3y²)-(y³+5)(6) (бу? 18y -бу-30 36у? 2r4 dy 2rd dy 2y³–5 6y? dy 2y³–5 dy 12xy? Set the derivative d = 0 and rewrite the above equation. 2y³–5 0 = 12xy2 2у — 5 %3D 0 y = 1.357 m Determine the corresponding value of x. 5+y x = бу 5+(1.357 m) 6(1.357 m) x? : x = 0.960 m

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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5+y Thus, the equation of the streamline passing through point (1 m, 2 m) is x² бу Determine the derivative of the equation with respect to y. бу (3у?)-(у+5) (6) (бу? 18у -бу-30 - dx 2x dr dy 36y? dx 2x dy 2y³–5 бу? dx 2y-5 dy 12xy2 Set the derivative d = 0 and rewrite the above equation. 2y³–5 0 = 12xy2 2y – 5 = 0 y = 1.357 m Determine the corresponding value of x. 5+y x² бу 5+(1.357 m) x² = 6(1.357 m) x = 0.960 m
Transcribed Image Text:5+y Thus, the equation of the streamline passing through point (1 m, 2 m) is x² бу Determine the derivative of the equation with respect to y. бу (3у?)-(у+5) (6) (бу? 18у -бу-30 - dx 2x dr dy 36y? dx 2x dy 2y³–5 бу? dx 2y-5 dy 12xy2 Set the derivative d = 0 and rewrite the above equation. 2y³–5 0 = 12xy2 2y – 5 = 0 y = 1.357 m Determine the corresponding value of x. 5+y x² бу 5+(1.357 m) x² = 6(1.357 m) x = 0.960 m
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