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PLOT THE GRAPH FOR THE GIVEN CURVES AND CONDITIONS BELOW Need to find out the differential equation that represents the circles with center on the line y=x and passes through the origin. Standard equation of the circle is, (x-h)² + (y-k)² = r² As it passes through the origin, Put x=0, y=0 (0h)² + (0-k)² = r² h² +k² = r² Put h=k, as x=y Now, r² = 2h² (x-h)² + (y-h)² = 2h² x² + h² − 2xh + y² + h² − 2yh = 2h² (Since, (a - b)² = a² + b² - 2ab) x² + y² - h (2x + 2y) = 0 (x²+x²) = 2h x+y Now differentiate it with respect to x, (x+y)(2x+2yy)-(x²+y²)(1+y) = 0 (x+y)² Quotient rule is used: (#)' = -u', (x²) 12 dx On simplification, 2x² + 2xyy' + 2xy + 2y²y¹ − x² − x²y² − y² - y²y¹ = 0 y' = (−x² + y² − 2xy)/ (−x² + y² + 2xy) - (x²-y²+2xy) x²-y²-2xy = x² - y² + y2xy + 2y²x² - y² + 2xy = 0 - 1²). y (−x² + y² + 2xy) = −x² + y² − 2xy y' = This is the nx"-1

PLOT THE GRAPH FOR THE GIVEN CURVES AND CONDITIONS BELOW Need to find out the differential equation that represents the circles with center on the line y=x and passes through the origin. Standard equation of the circle is, (x-h)² + (y-k)² = r² As it passes through the origin, Put x=0, y=0 (0h)² + (0-k)² = r² h² +k² = r² Put h=k, as x=y Now, r² = 2h² (x-h)² + (y-h)² = 2h² x² + h² − 2xh + y² + h² − 2yh = 2h² (Since, (a - b)² = a² + b² - 2ab) x² + y² - h (2x + 2y) = 0 (x²+x²) = 2h x+y Now differentiate it with respect to x, (x+y)(2x+2yy)-(x²+y²)(1+y) = 0 (x+y)² Quotient rule is used: (#)' = -u', (x²) 12 dx On simplification, 2x² + 2xyy' + 2xy + 2y²y¹ − x² − x²y² − y² - y²y¹ = 0 y' = (−x² + y² − 2xy)/ (−x² + y² + 2xy) - (x²-y²+2xy) x²-y²-2xy = x² - y² + y2xy + 2y²x² - y² + 2xy = 0 - 1²). y (−x² + y² + 2xy) = −x² + y² − 2xy y' = This is the nx"-1

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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PLOT THE GRAPH FOR THE GIVEN CURVES AND CONDITIONS BELOW Need to find out the differential equation that represents the circles with center on the line y=x and passes through the origin. Standard equation of the circle is, (x-h)² + (y-k)² = r² As it passes through the origin, Put x=0, y=0 (0h)² + (0-k)² = r² h² +k² = r² Put h=k, as x=y Now, r² = 2h² (x-h)² + (y-h)² = 2h² x² + h² − 2xh + y² + h² − 2yh = 2h² (Since, (a - b)² = a² + b² - 2ab) x² + y² - h (2x + 2y) = 0 (x²+x²) = 2h x+y Now differentiate it with respect to x, (x+y)(2x+2yy)-(x²+y²)(1+y) = 0 (x+y)² Quotient rule is used: (#)' = -u', (x²) 12 dx On simplification, 2x² + 2xyy' + 2xy + 2y²y¹ − x² − x²y² − y² - y²y¹ = 0 y' = (−x² + y² − 2xy)/ (−x² + y² + 2xy) - (x²-y²+2xy) x²-y²-2xy = x² - y² + y2xy + 2y²x² - y² + 2xy = 0 - 1²). y (−x² + y² + 2xy) = −x² + y² − 2xy y' = This is the nx"-1
Transcribed Image Text:PLOT THE GRAPH FOR THE GIVEN CURVES AND CONDITIONS BELOW Need to find out the differential equation that represents the circles with center on the line y=x and passes through the origin. Standard equation of the circle is, (x-h)² + (y-k)² = r² As it passes through the origin, Put x=0, y=0 (0h)² + (0-k)² = r² h² +k² = r² Put h=k, as x=y Now, r² = 2h² (x-h)² + (y-h)² = 2h² x² + h² − 2xh + y² + h² − 2yh = 2h² (Since, (a - b)² = a² + b² - 2ab) x² + y² - h (2x + 2y) = 0 (x²+x²) = 2h x+y Now differentiate it with respect to x, (x+y)(2x+2yy)-(x²+y²)(1+y) = 0 (x+y)² Quotient rule is used: (#)' = -u', (x²) 12 dx On simplification, 2x² + 2xyy' + 2xy + 2y²y¹ − x² − x²y² − y² - y²y¹ = 0 y' = (−x² + y² − 2xy)/ (−x² + y² + 2xy) - (x²-y²+2xy) x²-y²-2xy = x² - y² + y2xy + 2y²x² - y² + 2xy = 0 - 1²). y (−x² + y² + 2xy) = −x² + y² − 2xy y' = This is the nx"-1
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