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Pre-Calc Homework Solutions 110

# Pre-Calc Homework Solutions 110 - 110 49 Section 3.7 xy...

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49. xy 1 2 x 2 y 5 0 } d d x } ( xy ) 1 } d d x } (2 x ) 2 } d d x } ( y ) 5 0 x } d d y x } 1 ( y )(1) 1 2 2 } d d y x } 5 0 ( x 2 1) } d d y x } 5 2 2 2 y } d d y x } 5 } 2 x 2 2 2 1 y } 5 } 2 1 1 2 y x } Since the slope of the line 2 x 1 y 5 0 is 2 2, we wish to find points where the normal has slope 2 2, that is, where the tangent has slope } 1 2 } . Thus, we have } 2 1 1 2 y x } 5 } 1 2 } 2(2 1 y ) 5 1 2 x 4 1 2 y 5 1 2 x x 5 2 2 y 2 3 Substituting 2 2 y 2 3 in the original equation, we have: xy 1 2 x 2 y 5 0 ( 2 2 y 2 3) y 1 2( 2 2 y 2 3) 2 y 5 0 2 2 y 2 2 8 y 2 6 5 0 2 2( y 1 1)( y 1 3) 5 0 y 5 2 1 or y 5 2 3 At y 5 2 1, x 5 2 2 y 2 3 5 2 2 3 5 2 1. At y 5 2 3: x 5 2 2 y 2 3 5 6 2 3 5 3. The desired points are ( 2 1, 2 1) and (3, 2 3). Finally, we find the desired normals to the curve, which are the lines of slope 2 2 passing through each of these points. At ( 2 1, 2 1), the normal line is y 5 2 2( x 1 1) 2 1 or y 5 2 2 x 2 3. At (3, 2 3), the normal line is y 5 2 2( x 2 3) 2 3 or y 5 2 2 x 1 3. 50. x 5 y 2 } d d x } ( x ) 5 } d d x } ( y 2 ) 1 5 2 y } d d y x } } d d y x } 5 } 2 1 y } The normal line at ( x , y ) has slope 2 2 y . Thus, the normal
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