How Do You Solve for t in a Cubic Parametric Equation?

  • Thread starter Thread starter jjj888
  • Start date Start date
  • Tags Tags
    Parametric
AI Thread Summary
To solve for t in the cubic parametric equation x = at^3 + bt^2 + ct + d, one must simplify the equation by eliminating t. The attempt at a solution leads to the expression (x-d)/t = t^2(a + (b/t) + (c/t^2)), but the solver feels stuck. It is noted that expressing the roots of a cubic equation is more complex than using the quadratic formula. The coefficients a, b, c, and d are confirmed to be single real number values. Understanding cubic functions is essential for tackling this problem.
jjj888
Messages
23
Reaction score
0

Homework Statement



Here's the equation:

slove for t

x=at^3+bt^2+ct+d

Homework Equations



Parametic Equations, removing the t value and such, to "simplify" the equation.

The Attempt at a Solution



This is as far as I got:

(x-d)/t = t^2(a+(b/t)+(c/t^2))

I'm stumped. Is this even possible?

Thanks
 
Physics news on Phys.org
There is a general method of expressing the roots of a cubic equation in terms of its coefficients, but it is far more complicated than the quadratic formula. You can find it here: http://en.wikipedia.org/wiki/Cubic_function

Are there any constraints on a, b, c, and d?
 
Thanks, yea that looks pretty crazy, but interesting. I had a feeling it was headed that way, I wasn't aware of the cubic function. a,b,c,and d are basically a single real number value, if that's what you mean.

Thanks!
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top