Asked by Daniel
Find an equation of the tangent(s) to the curve at a given point. Then graph the curve and the tangent(s).
x = 6sin(t), y = t^2 + t; (0,0)
How do I do this without the value of t? I don't see a way to eliminate the parameter.
x = 6sin(t), y = t^2 + t; (0,0)
How do I do this without the value of t? I don't see a way to eliminate the parameter.
Answers
Answered by
Daniel
Oh also if you can, how do you find d2y/d2x for parametrics?
Answered by
Steve
The nice thing about parametric equations is that you don't have to eliminate the parameter.
dy/dx = dy/dt / dx/dt
dy/dx = (2t+1)/6cos(t)
This is zero at t = -1/2 and has a vertical asymptote whenever t = an odd multiple of pi/2.
The graph is a kind of double wave sine surve. It wanders back and forth in the x direction, but the escillations get longer in the y direction, since y uses t^2, instead of just t.
Mosey on over to wolfram dot com to see the big picture. Type in a command like
graph x=6*sin(t), y=t^2 + t, t=-5..5
try
graph x=6*sin(t), y=t^2 + t, t=-2..1.5 to see the horizontal tangent at t=-1/2
dy/dx = dy/dt / dx/dt
dy/dx = (2t+1)/6cos(t)
This is zero at t = -1/2 and has a vertical asymptote whenever t = an odd multiple of pi/2.
The graph is a kind of double wave sine surve. It wanders back and forth in the x direction, but the escillations get longer in the y direction, since y uses t^2, instead of just t.
Mosey on over to wolfram dot com to see the big picture. Type in a command like
graph x=6*sin(t), y=t^2 + t, t=-5..5
try
graph x=6*sin(t), y=t^2 + t, t=-2..1.5 to see the horizontal tangent at t=-1/2
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