How do you use cos(t) and sin(t), with positive coefficients, to parametrize the intersection of the surfaces x^2+y^2=25x2+y2=25 and z=3x^2z=3x2?

1 Answer

Use the following parametrization for the curve ss generated by the intersection:

s(t)=(x(t), y(t), z(t)), t in [0, 2pi)s(t)=(x(t),y(t),z(t)),t[0,2π)
x = 5cos(t)x=5cos(t)
y = 5sin(t)y=5sin(t)
z=75cos^2(t)z=75cos2(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.