To find : the normal line which is parallel to 2x+y=0 .
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Answer to Problem 57E
The normal equation to curve at (−1,−1) is given by y=−2x−3 .
The normal equation to curve at (3,−3) is given by y=−2x+3 .
Explanation of Solution
Given information :
The curve is xy+2x−y=0 and line parallel to normal line is 2x+y=0 .
Formula needed :
Chain rule of derivative: ddx(f(v))=f′(v)⋅dvdx .
Power rule of derivative: ddx(xn)=nxn−1
The normal equation to curve at (a,b) is given by y−b=−1m(x−a) .
Compare y−b=−1m(x−a) with 2x+y=0 .
−1m=−2m=12
For slope, differentiate xy+2x−y=0 with respect to x .
ddx(xy+2x−y=0)=0ddx(xy)+ddx(2x)−ddx(y)=0xddx(y)+yddx(x)+2−dydx=0xdydx+y+2−dydx=0
Solving further,
(x−1)dydx+(y+2)=0(x−1)dydx=−(y+2)dydx=−(y+2)(x−1)
Take both slopes equally.
−(y+2)(x−1)=12−2(y+2)=(x−1)−2y−4=x−1y=1−2(x+3)
Plug above equation in the curve.
x(1−2(x+3))+2x−(1−2(x+3))=0−x22−3x2+2x+x2+32=0−x22+x+32=0−x2+2x+3=0x=−1, 3
The coordinates are (−1,−1) and (3,−3) .
The normal equation to curve at (−1,−1) is given by y+1=−2(x+1) .
The normal equation to curve at (3,−3) is given by y+3=−2(x−3) .
Chapter 3 Solutions
CALCULUS-W/XL ACCESS
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