13.1 Vector Functions and Space Curves #

Vector Functions #

We can write a vector function as $$\vec r(t) = \langle f(t), g(t), h(t) \rangle = f(t) \hat i + g(t) \hat j + h(t) \hat k$$
The domain of .$\vec r$ consists of all values of .$t$ for which each of the terms are defined
The limit of a vector function is $$\lim_{t\to a} \vec r(t) = \big\langle \lim_{t\to a} f(t), \lim_{t\to a} g(t), \lim_{t\to a} h(t) \big\rangle$$
.$\vec r$ is continuous at time .$a$ if .$\lim_{t\to a} \vec r(t) = \vec r(a)$

The set .$C$ of all points defined by a vector function .$\vec r$ over interval .$I$ is called a space curve.
Think of .$C$ as being traced out by a moving particle whose position follows .$\vec r$
Space curves are parametrized by a vector function but isn’t necessarily that vector function!
- E.x. .$\vec r(t) \langle \cos t, \sin t, t \rangle \neq \vec q(t) = \langle \cos 3t, \sin 3t, 3t \rangle$
- Space curves exist on the same point, but don’t grow at the same rate

$$\frac{d\vec r}{dt} = \vec r’(t) = \lim_{h \to 0} \frac{\vec r(t+h) - r(t)}{h}$$ $$… = \big\langle f’(t), g’(t), h’(t) \big\rangle$$

Note that the last equation only works if each component is differentiable
The direction is the tangent line and the magnitude is the rate at which the particle is moving at that point
If we just want the tangent line, we can write the unit tangent vector: $$\vec T(t) = \frac{\vec r’(t)}{\Vert \vec r ’ (t) \Vert}$$
Differentiation rules (notice .$f(t)$ is a scalar function) $$ \frac{d}{dt} \big[f(t) \vec u(t) \big] = f’(t) \vec u(t) + f(t) \vec u’(t)$$ $$ \frac{d}{dt}\big[\vec u(t) \cdot \vec v(t)\big] = \vec u’(t) \cdot \vec v(t) + \vec u(t) \cdot \vec v’(t) $$ $$ \frac{d}{dt} \big[\vec u(t) \times \vec v(t)\big] = \vec u’ (t) \times \vec v(t) + \vec u(t) + \vec v’(t)$$ $$ \frac{d}{dt} \big[ \vec u(f(t)) \big] = f’(t) \vec u’ (f(t))$$

Let .$\vec r (t) = \langle f(t), g(t), h(t) \rangle$, then $$\int_a^b \vec r(t)\ dt = \bigg\langle \int_a^b f(t)\ dt, \int_a^b g(t) \ dt, \int_a^b h(t)\ dt \bigg\rangle$$
Fundamental theorem: If .$\vec R(t)$ is an anti-derivative of .$\vec r(t)$, (i.e. .$\vec R’(t) = \vec r(t)$), then $$ \int_a^b \vec r(t) \ dt = \Big[\vec R(t)\Big]_a^b = \vec R(b) - \vec R(a)$$