NEED a TRUE CalCulus Genius?

Use a graphing calculator to find the point at which curve C defined by parametric equations x=sin t , y=sin(t+sint), crosses itself for two different values of t.





a)Find and state the two values of t.


b). Show that C has two tangent lines at the origin and find their equations.

NEED a TRUE CalCulus Genius?
Hi There. Do you have a graphing calculator?? We're looking at the graph and it looks like a figure 8 in the 1st and 3rd quadraths. It crosses itself at (0,0) and again at (0,0) so all you need to do is set x and y to be zero and solve for t





for example





0=sint for example t=0 or pi (i'm not really sure how far to go with the answer because t could also equal 2*pi because of the behaviour of the sine function





when y =0


0=sin(t+sint)





means t could also equal zero here too





This is just a guess because I haven't studied this kind of mathematics for years - but hopefully I give you a start??





C does have two tangent lines at the origin - one with a positive gradient that looks almost like y=x (but it isn't) and another that looks like y=0 (but it isn't but close to it)





Sorry I'm not much more help than that!





Cheers, Laura
Reply:I'm using a TI-83 calculator :





Set it up for parametric functions and radians


input the two functions in terms of t





In trace mode, enter x = 0 you get y = 0


enter x = π you again get y = 0





This suggests that these two points are on the graph, and that t = 0 and t = π give the same point.





To get the equations of the tangents, remember that the tangent is the limit of a chord with one point fixed and the other getting ever closer to the fixed point.





To do this choose a very small value of t (close to t = 0), say t = 0.01 look at the x and y, t = 0.001 look at the x and y. You will notice that the y is approaching twice the value of x, so, since (0, 0) is the other point, the gradient is approaching 2. y = 2x is the equation of the tangent.





Likewise, choose a value very close to π, say π + 0.01 and adopt the same method, getting ever closer to π. Here the y-coordinate approaches zero, whilst the x-coordinate merely gets smaller. Conclusion ? The gradient approaches zero, so there IS a second tangent, and it has equation y = 0.





Hope the method is useful in future - it is really only doing gradients from first principles.

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