Unformatted text preview: Section 10.6
5. 7. 427 [ 3.8, 3.8] by [ 2.5, 2.5] [ 1.5, 1.5] by [ 1, 1] The graph passes through the pole when r which occurs when interval 0 only consider
2 3 cos
3 . Since the 2 0, and when The polar solutions are 0, a given k, the line curve at 0, k 5 for k 0, 1, 2, 3, 4, and for 1 produces the entire graph, we need
2 k appears to be tangent to the 5 . At this point, there appears to be a
2 vertical tangent line with equation Confirm analytically: x y
dy d (or x 0). (3 cos ) cos (3 cos ) sin ( 3 sin )sin
dx d 3 cos2 (3 cos )cos 3(cos2 sin2 ) and 6 cos ( sin ). k . This can be confirmed analytically by 5 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], 5 2 2 y tan x , y tan x , 5 5 5 5 3 3 4 4 y tan x , and y tan x . 5 5 5 5 8. dx , 0, and At 0, /2 2 d dy dx 3(02 12) 3. So at 0, , 0 /2 2 d d dy dy 0, so is undefined and the tangent line is and dx d [ 3, 3] by [ 2, 2] vertical. 6. The polar solutions are 0...
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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.
 Spring '08
 GERMAN

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