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Calculus/parametric equation and gradient

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Expert: - 7/20/2010


Question
a curve is defined parametrically by the equation
X=t(cubed) -6t+4
y= t-3+2/t find dy/dx

Answer
It can be seen that dy/dt and dx/dt can be found.
It is also known that these behave just like fractions.
That is, (dy/dt)/(dx/dt) can be treated just like a fraction, where the dt's cancel.
After doing this, we have dy/dx.

Differentials are just like fractions.
For example, (5/2)/(7/2) = 5/7, since the 2's cancel.
In the same way, the dt's cancel.

Looking back at the problem, dy/dt = 1 - 2/t² = (t²-2)/t² and dx/dt = 3t² - 6.
Using both of these, dy/dx  = ((t²-2)/t²)/(3t²-6) = (t²-2)/[t²(3t²-6)].

Now as far as putting dy/dx in terms of x, the equation x = t³-6t+4 would have to be solve fot t.
Once this was done, it could be put back in what dy/dx is equal to.  Solving cubics, however, is not such a simple matter.

To solve that cubic, I would look up "cubic solving" on the internet.
I have tried to do so, but currently my computer won't open another internet window,
even though I was away for around an hour.  I was and it probably will again, but not now.

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