Worked examples · Sheet 1

Normal vector and tangent plane to parametric surfaces

Theory used: Regular surface Normal vector Tangent plane to a surface

Exercise 1

Find the normal vector and the tangent plane to the surface

r (u, v) = (u + v, u - v, uv)

at the point corresponding to $(u_{0}, v_{0}) = (1, 1)$ .

P_{0} = r (1, 1) = (2, 0, 1)

r_{u} = (1, 1, v), r_{v} = (1, - 1, u)

r_{u} (1, 1) = (1, 1, 1), r_{v} (1, 1) = (1, - 1, 1)

N = r_{u} \land r_{v} = i 11 j 1 - 1 k 11 = (2, 0, - 2)

2 (x - 2) - 2 (z - 1) = 0 ⟹ x - z = 1

The point of the surface is obtained by substituting $(u_{0}, v_{0}) = (1, 1)$ into the parametrization, so $P_{0} = (2, 0, 1)$ . The derivative vectors $r_{u}$ and $r_{v}$ are computed by differentiating the three components with respect to $u$ and with respect to $v$ , and are then evaluated at the point. The normal vector is their cross product, which is expanded with the symbolic determinant: the first component is $1 \cdot 1 - 1 \cdot (- 1) = 2$ , the second is $- (1 \cdot 1 - 1 \cdot 1) = 0$ , the third is $1 \cdot (- 1) - 1 \cdot 1 = - 2$ . Since $N = (2, 0, - 2) \neq = 0$ , the surface is regular at the point and the tangent plane exists. The tangent plane passes through $P_{0}$ and is orthogonal to $N$ , therefore its equation is $2 (x - 2) + 0 \cdot (y - 0) - 2 (z - 1) = 0$ , which simplified becomes $x - z = 1$ .

Exercise 2

Find the normal vector and the tangent plane to the helicoid

r (u, v) = (u cos v, u sin v, 2 v)

at the point corresponding to $(u_{0}, v_{0}) = (1, \frac{π}{2})$ .

P_{0} = r (1, \frac{π}{2}) = (0, 1, π)

r_{u} = (cos v, sin v, 0), r_{v} = (- u sin v, u cos v, 2)

r_{u} (1, \frac{π}{2}) = (0, 1, 0), r_{v} (1, \frac{π}{2}) = (- 1, 0, 2)

N = r_{u} \land r_{v} = i 0 - 1 j 10 k 02 = (2, 0, 1)

2 (x - 0) + (z - π) = 0 ⟹ 2 x + z = π

The point is obtained by substitution: since $cos \frac{π}{2} = 0$ and $sin \frac{π}{2} = 1$ , we get $P_{0} = (0, 1, π)$ . The vector $r_{u}$ is obtained by differentiating with respect to $u$ with $v$ held constant, the vector $r_{v}$ by differentiating with respect to $v$ with $u$ held constant. The cross product of the two vectors evaluated at the point gives $N = (1 \cdot 2 - 0 \cdot 0, - (0 \cdot 2 - 0 \cdot (- 1)), 0 \cdot 0 - 1 \cdot (- 1)) = (2, 0, 1)$ . The normal vector is not zero, so the tangent plane is the plane through $P_{0}$ orthogonal to $N$ , that is, $2 x + z = π$ .

Exercise 3

Given the sphere of radius $2$ parametrized by

r (φ, θ) = (2 sin φ cos θ, 2 sin φ sin θ, 2 cos φ),

compute the normal vector at the point corresponding to $(φ_{0}, θ_{0}) = (\frac{π}{3}, 0)$ , verify that it is parallel to the position vector and write the tangent plane.

P_{0} = r (\frac{π}{3}, 0) = (3, 0, 1)

r_{φ} = (2 cos φ cos θ, 2 cos φ sin θ, - 2 sin φ), r_{θ} = (- 2 sin φ sin θ, 2 sin φ cos θ, 0)

r_{φ} (\frac{π}{3}, 0) = (1, 0, - 3), r_{θ} (\frac{π}{3}, 0) = (0, 3, 0)

N = r_{φ} \land r_{θ} = i 10 j 03 k - 3 0 = (3, 0, 3) = 3 (3, 0, 1)

3 (x - 3) + (z - 1) = 0 ⟹ 3 x + z = 4

Cross-section with the plane

y = 0

: the normal vector at

P_{0}

has the direction of the radius and the trace of the tangent plane

3 x + z = 4

is tangent to the circle

At the given point we have $sin \frac{π}{3} = \frac{3}{2}$ and $cos \frac{π}{3} = \frac{1}{2}$ , so $P_{0} = (3, 0, 1)$ . The two derivative vectors, evaluated at the point, are $(1, 0, - 3)$ and $(0, 3, 0)$ , and their cross product is $N = (0 \cdot 0 - (- 3) \cdot 3, - (1 \cdot 0 - (- 3) \cdot 0), 1 \cdot 3 - 0) = (3, 0, 3)$ . Factoring out $3$ we obtain $N = 3 (3, 0, 1)$ , that is, a positive multiple of the position vector $P_{0}$ ; therefore the normal to the sphere is radial and, with this parametrization, points outward. The tangent plane is the plane through $P_{0}$ orthogonal to the direction $(3, 0, 1)$ , that is, $3 x + z = 4$ . The constant term $4$ coincides with the square of the radius, in agreement with the general formula $x x_{0} + y y_{0} + z z_{0} = R^{2}$ for the tangent plane to a sphere centered at the origin.