Worked examples · Sheet 1

Regular curves and tangent vector

Theory used: Regular curve Tangent vector and tangent line Piecewise regular curve

Exercise 1

Determine whether the curve

γ (t) = (3 t + 1, t^{2}), t \in [0, 2],

is regular and compute the tangent vector at the point $γ (1)$ .

γ^{'} (t) = (3, 2 t)

γ^{'} (t) \neq = (0, 0) for every t \in [0, 2]

γ (1) = (4, 1), γ^{'} (1) = (3, 2)

The trace of

γ

and the tangent vector

γ^{'} (1) = (3, 2)

applied at the point

γ (1) = (4, 1)

The components $3 t + 1$ and $t^{2}$ are polynomials, hence the curve is of class $C^{1}$ and the derivative vector is computed component by component. The first component of $γ^{'} (t)$ equals $3$ for every $t$ , therefore the vector $γ^{'} (t)$ is never zero and the curve is regular on the whole of $[0, 2]$ . The tangent vector at the point $γ (1) = (4, 1)$ is $γ^{'} (1) = (3, 2)$ .

Exercise 2

Verify that the helix

γ (t) = (cos t, sin t, t), t \in [0, 2 π],

is regular and compute the unit tangent vector at the point $γ (π)$ .

γ^{'} (t) = (- sin t, cos t, 1)

∥ γ^{'} (t)∥ = sin^{2} t + cos^{2} t + 1 = 2

γ^{'} (π) = (0, - 1, 1), T (π) = \frac{γ ^{'} ( π )}{2} = (0, - \frac{1}{2}, \frac{1}{2})

The components are functions of class $C^{1}$ and the derivative vector has third component constantly equal to $1$ , hence $γ^{'} (t)$ never vanishes and the curve is regular. The norm of the tangent vector equals $2$ for every $t$ , since $sin^{2} t + cos^{2} t = 1$ . At the point $γ (π) = (- 1, 0, π)$ the tangent vector is $γ^{'} (π) = (0, - 1, 1)$ ; dividing by the norm $2$ one obtains the unit tangent vector.

Exercise 3

Determine whether the curve

γ (t) = (t^{2}, t^{3}), t \in [- 1, 1],

is regular or piecewise regular.

γ^{'} (t) = (2 t, 3 t^{2})

γ^{'} (t) = (0, 0) ⟺ {2 t = 0 3 t^{2} = 0 ⟺ t = 0

γ regular on [- 1, 0] and on [0, 1] ⟹ γ piecewise regular on [- 1, 1]

The curve is of class $C^{1}$ because the components are polynomials. The derivative vector vanishes when both components are zero, and the system has the unique solution $t = 0$ , which is an interior point of the interval $[- 1, 1]$ ; hence the curve is not regular on $[- 1, 1]$ . On the two subintervals $[- 1, 0]$ and $[0, 1]$ the point $t = 0$ is an endpoint, therefore $γ^{'} (t)$ does not vanish at any interior point of the two pieces and each restriction is regular. The curve is therefore piecewise regular with the partition $- 1 < 0 < 1$ . At the point $γ (0) = (0, 0)$ the trace has a cusp: at the instant $t = 0$ the velocity vanishes and the first component $x^{'} (t) = 2 t$ changes sign, so the horizontal direction of motion is reversed.