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#.