Vector Functions — Question 9

Question 9

A curve is known to lie on the sphere $x^2+y^2+z^2=25$ and in the plane $z=4$. Construct two vector parametrizations that traverse the entire curve exactly once in opposite directions. Choose both to start at $(3,0,4)$.

Tasks

1. Identify the intersection curve and its dimensions.

2. Construct the two oppositely oriented parametrizations.

3. Verify the surface constraints, starting point, and one-pass intervals.

Question 9 – Solution

Strategy. Substituting the fixed height into the sphere reveals a circle. The sign of the sine component controls orientation.

Step 1: Identify the circle With $z=4$, the sphere equation becomes $$x^2+y^2+16=25\quad\Longrightarrow\quad x^2+y^2=9.$$ The intersection is a radius-$3$ circle centered at $(0,0,4)$.

Step 2: First orientation A standard counterclockwise parametrization as viewed from the positive $z$-axis is $$\boxed{\mathbf r_+(t)=\left\langle 3\cos t,\ 3\sin t,\ 4\right\rangle}, \qquad 0\le t<2\pi.$$ It begins at $\mathbf r_+(0)=\left\langle 3,0,4\right\rangle$ and initially enters the half-space $y>0$.

Step 3: Opposite orientation Reverse the sign of the sine component: $$\boxed{\mathbf r_-(t)=\left\langle 3\cos t,\ -3\sin t,\ 4\right\rangle}, \qquad 0\le t<2\pi.$$ It has the same starting point but initially enters $y<0$, so its orientation is opposite.

Verification. For either formula, $$x^2+y^2+z^2=9(\cos^2t+\sin^2t)+16=25.$$ The half-open interval covers every angular position once without duplicating the start at $2\pi$.