Explanation Given: x 2 y − 2 = y 2 − 3 x − 5 at ( 1 , 3 ) Concept used: Slope of the normal = − 1 Slope of the tangent Equation of the line passing through the point ( a , b ) with slope m is, y − b = m ( x − a ) Calculation: Slopeof the tangent is the derivative of the curve at the given point. So differentiating given curve with respect to x, we get ⇒ d ( x 2 ) d x ⋅ y − 2 + d ( y − 2 ) d x ⋅ x 2 = 2 y ⋅ d y d x − 3 ( 1 ) − 0 ⇒ 2 x ⋅ y − 2 + 1 2 ⋅ y − 2 ⋅ d y d x ⋅ x 2 = 2 y ⋅ d y d x − 3 ⇒ 1 2 ⋅ y − 2 ⋅ d y d x ⋅ x 2 − 2 y ⋅ d y d x = − 3 − 2 x ⋅ y − 2 ⇒ d y d x ( x 2 2 y − 2 − 2 y ) = − 3 − 2 x ⋅ y − 2 ⇒ d y d x ( x 2 − 4 y y − 2 2 y − 2 ) = − 3 − 2 x ⋅ y − 2 ⇒ d y d x = ( 2 y − 2 ) ( − 3

7th Edition

ISBN: 9781524916817

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 3.6, Problem 36PS

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To find the equation of the normal line to the given curve at the given point.

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Chapter 3.6, Problem 35PS

Chapter 3.6, Problem 37PS

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Chapter 3 Solutions

Calculus

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To find the equation of the normal line to the given curve at the given point.

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Chapter 3 Solutions