**Unformatted text preview: **SOLUTION TO QUIZ 1
Question 1. Find a parametrization of the intersection of x2 + y 2 = 4
and the surface z = xy.
Solution: Any points on the cylinder x2 + y 2 = 4 can be written as (2 cos t, 2 sin t, z). If the point is also on the surface z = xy,
then we must have z = (2 cos t)(2 sin t) = 4 sin t cos t. Therefore, the
parametrization is given by (2 cos t, 2 sin t, 4 sin t cos t).
Question 2. At what point do the curves r1 (t) = (t, 1 − t, 3 + t2 ) and
r2 (s) = (3 − s, s − 2, s2 ) intersect? Find their angle of intersection .
Solution: At the intersection, we have r1 (t) = r2 (s), so that we have
3 equations: t = 3 − s
1−t=s−2 3 + t2 = s2
Solving the equations, we get t = 1 and s = 2. Therefore, the intersection is r1 (1) = r2 (2) = (1, 0, 4).
The angle of intersection is the angle between the tangent vectors.
Since r01 (t) = (1, −1, 2t) and r02 (s) = (−1, 1, 2s), the tangent vector to
the curve r1 (t) at (1, 0, 4) is r01 (1) = (1, −1, 2) and the tangent vector
to the curve r2 (s) at (1, 0, 4) is r02 (2) = (−1, 1, 4). If the angle between
them is θ, then
r0 · r0
cos θ = 10 0 2
|r1 |r2 |
(1, −1, 2) · (−1, 1, 4)
=
|(1, −1, 2)||(−1, 1, 4)|
1
=√ .
3
−1 √1
Therefore, θ = cos
.
3
Question 3. Suppose you start at the point (0, 0, 3) and move 5 units
along the curve r(t) = (3 sin t, 4t, 3 cos t) in the positive direction, where
are you now?
Solution. Note that r(0) = (0, 0, 3). As we are moving 5 units in the
positive direction along the curve, the destination would be r(T ) with T
1 2 SOLUTION TO QUIZ 1 RT
satisfying 5 = 0 |r0 (t)|dt. Since r0 (t) = (3 cos t, 4, −3 sin t), |r0 (t)| = 5.
RT
Therefore, 5 = 0 5dt = 5T and hence T = 1. The destination is
r(1) = (3 sin 1, 4, 3 cos 1). ...

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- Fall '14
- Cos, Manifold, parametrization, tangent vector