Question

At what points does the curve r(t) = ti (4t - t²)k intersect the paraboloid z = x² y²?

Answer 1

The curve r(t) = ti + (4t - t²)k intersects the paraboloid z = x²y² at the points when t equals 0 or 4, because at these values the z-component of the curve is equal to zero, which is the result of x²y² when y(t) = 0.

Step-by-step explanation:

To determine at what points the curve given by r(t) = ti + (4t - t²)k intersects the paraboloid defined by z = x²y², we need to set the z-component of the curve equal to the equation of the paraboloid.

First, we express r(t) in its component form:

• x(t) = t
• y(t) = 0 (since the j-component is missing)
• z(t) = 4t - t²

Next, we substitute these into the equation of the paraboloid:

1. z = x²y² becomes 4t - t² = t²(0)
2. Since y(t) = 0, x²y² is always 0.
3. Thus, z(t) must also be 0 for an intersection to occur.
4. Therefore, 4t - t² = 0.
5. Solve for t: t(4 - t) = 0, gives t = 0 or t = 4.

There are two times when this intersection occurs, at t = 0 and t = 4. Thus, the curve intersects the paraboloid at the points where t is either 0 or 4.