Implicitly defined parametrizations

  • Thread starter miglo
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  • #1
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Homework Statement


x^3+2t^2=9, 2y^3-3t^2=4, t=2
find the slope of the curve at the given value of t

Homework Equations





The Attempt at a Solution


i know dy/dx=(dy/dt)/(dx/dt) for parametrized curves
but how do i use implicit differentiation on parametrized curves?
i tried d/dt(2y^3-3t^2)=d/dt(4)
=6y^2*dy/dt-6t=0
=6y^2*dy/dt=6t
=dy/dt=t/y^2 ? but shouldnt dy/dt be defined in t's only? im confused
 

Answers and Replies

  • #2
SammyS
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Homework Statement


x^3+2t^2=9, 2y^3-3t^2=4, t=2
find the slope of the curve at the given value of t

Homework Equations



The Attempt at a Solution


i know dy/dx=(dy/dt)/(dx/dt) for parametrized curves
but how do i use implicit differentiation on parametrized curves?
i tried d/dt(2y^3-3t^2)=d/dt(4)
=6y^2*dy/dt-6t=0
=6y^2*dy/dt=6t
=dy/dt=t/y^2 ? but shouldn't dy/dt be defined in t's only? im confused
Find dx/dt in a similar fashion. Also find x & y, when t = 2.

Then combine the results: [itex]\displaystyle \frac{dy/dt}{dx/dt}[/itex] will have t cancel out, in this case.
 
  • #3
97
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i thought about trying that out before asking for help
but i kept thinking there had to be an easier way
well thanks for the help
 

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