Is Curve x=3t^2, y=e^2t+1 Concave Up/Down at (3,e^3)?

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SUMMARY

The curve defined by the parametric equations x = 3t^2 and y = e^(2t) + 1 requires analysis to determine its concavity at the point (3, e^3). To assess concavity, one must compute the second derivative of y with respect to x. The first derivative dy/dx is derived from the first derivatives of x and y with respect to t, leading to dy/dx = (2e^(2t))/(6t). The second derivative, necessary for concavity determination, is calculated as d²y/dx² = (4e^(2t))/(6t) - (2e^(2t))/(36t^2), evaluated at the appropriate t-value corresponding to x = 3.

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Homework Statement



Determine whether the curve x = 3t^2, y = e^2t+1 is concave up or concave down at the point (3, e^3)

thanks for any help with this one.


Homework Equations





The Attempt at a Solution



know i need to take the double derivative. Also that would be 6
and 4e^2t+1

what point do i put back into the double derivative. Also, do I use 3 and e^3? they represent x, y, so do they plug into the equation appropriately, or back into t? please help.
 
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If you know how to determine concavity or concavity when given a function y(x), then I suggest you try to "eliminate" that nasty t and try to write y as a function of x!
 

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