How is the parametrization of the second path found.
http://tutorial.math.lamar.edu/Classes/CalcIII/LineIntegralsPtI_files/image002.gif
It's 1- x^3.
I want an explanation.
http://tutorial.math.lamar.edu/Classes/CalcIII/LineIntegralsPtI_files/image002.gif
It's 1- x^3.
I want an explanation.
-
He explains it on the first page of that section. If you have a polynomial function of the form
y = f(x)
Then all you have to to obtain the parametrization is to let x = t. Then the parameters are:
x = t
y = f(t)
So in that example, we let x = t. Then y = 1 - t^3. That's it.
Done!
P.S. Obviously it's more complicated for circles and things like that, but for polynomials, it's really just that simple.
y = f(x)
Then all you have to to obtain the parametrization is to let x = t. Then the parameters are:
x = t
y = f(t)
So in that example, we let x = t. Then y = 1 - t^3. That's it.
Done!
P.S. Obviously it's more complicated for circles and things like that, but for polynomials, it's really just that simple.