How do you use cos(t) and sin(t), with positive coefficients, to parametrize the intersection of the surfaces `x^2+y^2=25` and `z=3x^2`?

Use the following parametrization for the curve `s` generated by the intersection:

`s(t)=(x(t), y(t), z(t)), t in [0, 2pi)`
`x = 5cos(t)`
`y = 5sin(t)`
`z=75cos^2(t)`

Note that `s(t): RR -> RR^3` is a vector valued function of a real variable.

To reach this result, consider the curves that these equations define on certain planes.

The equation `x^2+y^2=25` defines a circle of radius `5` centered on the `z`-axis on the planes `z=c_1`, where `c_1 in RR` is any constant.
The equation `z=3x^2` defines a parabola on any plane `y=c_2`, where `c_2 in RR` is another constant.
The surfaces are, therefore, those obtained by translating the circle along the `z`-axis and the parabola along the `y`-axis.

To obtain a parametrization for the intersection curve `s`, we must find equations for `x`, `y` and `z` as functions of `t` that obey both equations given in the problem.

Consider the standard parametrization for a circle `C` of radius `r` (it's easy to see that this parametrization fulfils the condition `x^2+y^2=r^2`):

`C(t)=(rcos(t), rsin(t)), t in [0,2pi)`

Checking the first equation, we get `r=5` and

`x = 5cos(t)`
`y = 5sin(t)`

Now, we already have an expression for `x(t)`. So, in order to obey the second condition, we make:

`z=3x^3=3(5cos(t))^2=75cos^2(t)`

And we have the parametrization `s(t)=(x(t), y(t), z(t)), t in [0, 2pi)` for `s`.