Pre-Calc Homework Solutions 427

6 the polar solutions are 0 a given k the line curve

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Unformatted text preview: , a given k, the line curve at 0, k 2 for k 0, 1, 2, 3, 4, and for k appears to be tangent to the 2 [ 3, 3] by [ 2, 2] A trace of the graph suggests three tangent lines, one with positive slope for 6 k . This can be confirmed analytically by 2 dy noting that the slope of the curve, , equals the slope of dx k the line, tan . So the tangent lines are 0 [y 0] and 2 3 [x 0]. , and 2 are duplicate 2 2 , a vertical one for 5 . 6 2 , and one solutions. with negative slope for Confirm analytically: dy d dx d 6 sin 3 sin 6 sin 3 cos 6 2 cos 3 cos 2 cos 3 sin . 5 6 and 0, dy dx dy dx dy dx dy dx , 0, 2 , and 0, are all solutions. dy/d , and so dx/d 6(1)(1/2) 2(0)( 3/2) 6(1)( 3/2) 2(0)(1/2) /6 /2 1 3 ; 6( 1)(1) 6( 1)(0) 2(0)(0) , which is undefined; and 2(0)(1) 1 3 5 /6 6(1)(1/2) 2(0)( 3/2) 6(1)( 3/2) 2(0)(1/2) 6 1 3 . The tangent 2 lines have equations and 5 6 y 1 3 x , [x 0], y x ....
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

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