hws15

# hws15 - Homework 14 Solutions 1 Comment It is my hope that...

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Homework 14 Solutions 1 Comment It is my hope that the graphs accompanying some of the implicit diFerentiation exercises made it easier to see if your solutions made sense. You should be using the point-slope formula for lines, since you have the point and the slope: ( y - b ) = m ( x - a ). There is no need (and it is a waste of time) to reduce the equation to the slope intercept form. 2 Implicit Diferentiation Exercises Page 125, Number 17: ±ind an equation of the tangent line to the ellipse x 2 + xy + y 2 = 3 at (1 , 1). We have (using the power rule, product rule and chain rule) 2 x + ( y + xy 0 ) + 2 yy 0 = 0 2 + (1 + y 0 ) + 2 y 0 = 0 3 + 3 y 0 = 0 so y 0 = - 1. Using the point-slope formula we have y - 1 = - ( x - 1) as an equation of the tangent line to the ellipse at (1 , 1). Page 125, Number 18: ±ind an equation of the tangent line to the hyperbola x 2 + 2 xy - y 2 + x = 2 at (1 , 2). We have (using the power rule, product rule and chain rule): 2 x + 2( y + xy 0 ) - 2 yy 0 + 1 = 0 2 + 2(2 + y 0 ) - 4 y 0 + 1 = 0 7 - 2 y 0 = 0 so y 0 = 7 / 2. Using the point-slope formula we have ( y - 2) = 7( x - 1) / 2 or, if you hate fractions, 2( y - 2) = 7( x - 1) as equations of the tangent line to the hyperbola at (1 , 2). Page 126, Number 19: ±ind an equation of the tangent line to the cardioid x 2 + y 2 = (2 x 2 + 2 y 2 - x ) 2 at (0 , 1 / 2). Judging from the graph this line should have a positive slope that is slightly more than 1. We have (using the power rule, product rule and chain rule): 2 x + 2 yy 0 = 2(2 x 2 + 2 y 2 - x )(4 x + 4 yy 0 - 1) 0 + y 0 = 2(0 + (1 / 2) - 0)(0 + 2 y 0 - 1) y 0 = 2 y 0 - 1 so y 0 = 1. Using the point-slope formula we have ( y - (1 / 2)) = x or, if you hate fractions, 2 y - 1 = 2 x . 1

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Page 126, Number 20: Find an equation of the tangent line to the astroid x 2 / 3 + y 2 / 3 = 4 at ( - 3 3 , 1). Note that 3 3 = 3 3 / 2 . Judging from the graph this line should have a positive slope that is slightly less than 1. We have (using the power rule, product rule and chain rule):
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## This note was uploaded on 12/22/2011 for the course MATH 1337 taught by Professor Xu during the Fall '08 term at SMU.

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hws15 - Homework 14 Solutions 1 Comment It is my hope that...

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