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nomadreid

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On http://mathworld.wolfram.com/HeartCurve.html, the nicest heart-shaped curve is given by

x=16sin

y = 13 cos t-5 cos(2t) -2 cos (3t) - cos(4t)

There are evidently two values of t, one in the first, one in the second quadrant, where you can get a horizontal tangent to this shape. Asking Wolfram alpha to solve dy/dt= -13 sin t+10 sin(2t) +6 sin(3t) +4 sin(4t)=0 (because I want dy/dx = (dy/dt)/(dx/dt)=0)), I get

http://www.wolframalpha.com/input/?i=-13*sin+x++10*sin(2x)++6*sin+(3x)+++4*sin(4x)+=0+

and weeding out other inflection points, there appears a graph with a reasonable solution of t= 0.908 for the first quadrant, but

(a) the symmetrical solution of 2π - 0.908 is missing on the graph

(b) in the listing of the numerical solutions, except for t=0, only complex non-real solutions are given. I have two objections to this: one, there are obviously real solutions, since there are the two tangents. Two, usually Wolfram alpha, when there is a nonzero imaginary part to a solution, puts the imaginary part in a different color on the graph, something they did not do here.

Then, worse, when I want the vertical tangents by setting dx/dt=0, I get only the points at n*π/2, which is obviously not right (as a glance of the curve will show).

What am I doing wrong here? thanks in advance

PS Believe it or not, this is not a homework question. (What heartless teacher would give a heart to differentiate?) But if the mentors think that it is too textbooky style, I have no objections to moving it.

x=16sin

^{3}ty = 13 cos t-5 cos(2t) -2 cos (3t) - cos(4t)

There are evidently two values of t, one in the first, one in the second quadrant, where you can get a horizontal tangent to this shape. Asking Wolfram alpha to solve dy/dt= -13 sin t+10 sin(2t) +6 sin(3t) +4 sin(4t)=0 (because I want dy/dx = (dy/dt)/(dx/dt)=0)), I get

http://www.wolframalpha.com/input/?i=-13*sin+x++10*sin(2x)++6*sin+(3x)+++4*sin(4x)+=0+

and weeding out other inflection points, there appears a graph with a reasonable solution of t= 0.908 for the first quadrant, but

(a) the symmetrical solution of 2π - 0.908 is missing on the graph

(b) in the listing of the numerical solutions, except for t=0, only complex non-real solutions are given. I have two objections to this: one, there are obviously real solutions, since there are the two tangents. Two, usually Wolfram alpha, when there is a nonzero imaginary part to a solution, puts the imaginary part in a different color on the graph, something they did not do here.

Then, worse, when I want the vertical tangents by setting dx/dt=0, I get only the points at n*π/2, which is obviously not right (as a glance of the curve will show).

What am I doing wrong here? thanks in advance

PS Believe it or not, this is not a homework question. (What heartless teacher would give a heart to differentiate?) But if the mentors think that it is too textbooky style, I have no objections to moving it.

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