Welcome back, everyone. In our last lecture, we learned how to describe curves in space using vector functions, where the position vector $\mathbf{r}(t)$ traces out the path of a particle over time. Now that we have functions that describe motion, the natural next step is to ask: how do we apply the tools of calculus to this motion?
The great news is that everything we learned about derivatives and integrals in single-variable calculus extends beautifully to vector functions. The core idea is simple: to differentiate or integrate a vector, we simply differentiate or integrate each of its components. This will allow us to formally define and calculate crucial physical concepts like velocity and acceleration for objects moving along space curves. 🚀
Just as in Calculus I, the derivative of a vector function represents an instantaneous rate of change. Geometrically, it gives us a vector that is tangent to the curve, pointing in the direction of motion.
The derivative of a vector function $\mathbf{r}(t)$ is defined by the limit:
$$ \mathbf{r}'(t) = \lim_{h \to 0} \frac{\mathbf{r}(t+h) - \mathbf{r}(t)}{h} $$This new vector $\mathbf{r}'(t)$ is called the tangent vector to the curve at the point corresponding to $t$. If $\mathbf{r}(t)$ is a position function, then $\mathbf{r}'(t)$ is the velocity vector, $\mathbf{v}(t)$.
While the limit definition is the foundation, in practice we almost always calculate the derivative by differentiating each component function separately. If $\mathbf{r}(t) = \langle f(t), g(t), h(t) \rangle$, then:
$$ \mathbf{r}'(t) = \langle f'(t), g'(t), h'(t) \rangle $$Find the derivative of the vector function $\mathbf{r}(t) = \langle t^3, \cos(t), te^t \rangle$.
We simply take the derivative of each component:
So, the derivative is $\mathbf{r}'(t) = \langle 3t^2, -\sin(t), e^t(1+t) \rangle$.
The velocity vector $\mathbf{r}'(t)$ tells us two things: the direction of motion and the speed of the motion (which is its magnitude, $|\mathbf{r}'(t)|$). Sometimes, we only care about the direction. For this, we use the unit tangent vector.
The unit tangent vector, denoted by $\mathbf{T}(t)$, is found by dividing the derivative by its magnitude:
$$ \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|} $$The unit tangent vector is useful because it isolates the direction of motion from the speed. It always has a magnitude of 1, so it purely answers the question, "Which way is the particle heading at this instant?"
The familiar differentiation rules from single-variable calculus have direct analogs for vector functions. Note that the dot and cross product rules are analogous to the Product Rule from single-variable calculus.
Suppose $\mathbf{u}$ and $\mathbf{v}$ are differentiable vector functions, $c$ is a scalar, and $f$ is a real-valued function. Then:
Let $\mathbf{u}(t) = \langle \cos t, \sin t, t \rangle$ and let $f(t) = t^2$. Find the derivative of the composite function $\mathbf{r}(t) = \mathbf{u}(f(t))$.
We want to find the derivative of $\mathbf{r}(t) = \mathbf{u}(t^2)$. The Chain Rule states that $\mathbf{r}'(t) = f'(t)\mathbf{u}'(f(t))$.
First, we find the necessary derivatives:
Now, we evaluate $\mathbf{u}'$ at $f(t)=t^2$:
$$ \mathbf{u}'(f(t)) = \mathbf{u}'(t^2) = \langle -\sin(t^2), \cos(t^2), 1 \rangle $$
Finally, we multiply by $f'(t)$:
$$ \mathbf{r}'(t) = 2t \langle -\sin(t^2), \cos(t^2), 1 \rangle = \langle -2t\sin(t^2), 2t\cos(t^2), 2t \rangle $$
Prove that if a particle moves on the surface of a sphere, its velocity vector is always orthogonal to its position vector.
If the particle is on a sphere centered at the origin, its distance from the origin is constant. Let's say this constant distance (the radius) is $c$. This means the magnitude of the position vector is constant: $|\mathbf{r}(t)| = c$.
We can also write this as $|\mathbf{r}(t)|^2 = c^2$, which is equivalent to $\mathbf{r}(t) \cdot \mathbf{r}(t) = c^2$.
Now, we differentiate both sides with respect to $t$:
$$ \frac{d}{dt}[\mathbf{r}(t) \cdot \mathbf{r}(t)] = \frac{d}{dt}[c^2] $$Using the Dot Product Rule on the left side and noting that the derivative of a constant is 0 on the right side, we get:
$$ \mathbf{r}'(t) \cdot \mathbf{r}(t) + \mathbf{r}(t) \cdot \mathbf{r}'(t) = 0 $$ $$ 2[\mathbf{r}(t) \cdot \mathbf{r}'(t)] = 0 $$ $$ \mathbf{r}(t) \cdot \mathbf{r}'(t) = 0 $$Since the dot product of the position vector $\mathbf{r}(t)$ and the velocity vector $\mathbf{r}'(t)$ is zero, they must be orthogonal at all times.
Let's visualize this with the classic circular helix, given by the vector equation $\mathbf{r}(t) = \langle \cos(t), \sin(t), t \rangle$. The velocity vector is $\mathbf{v}(t) = \langle -\sin(t), \cos(t), 1 \rangle$. Notice the vertical component is always 1, indicating a constant upward speed. The acceleration vector is $\mathbf{a}(t) = \langle -\cos(t), -\sin(t), 0 \rangle$. This vector is always horizontal, has a constant magnitude of 1, and points directly from the particle's position on the helix towards the z-axis. This is a perfect example of centripetal acceleration; it is entirely responsible for changing the direction of the velocity vector to keep the particle moving in a circle, without changing its speed.
Just as with derivatives, we find the integral of a vector function by integrating each of its components separately.
This is a very important point of clarification. In single-variable calculus, $\int_a^b f(x) dx$ represents the area under a curve. However, the definite integral of a vector function, $\int_a^b \mathbf{r}(t) dt$, results in a vector, not a scalar area.
If $\mathbf{v}(t)$ represents a velocity vector, then its definite integral represents the displacement vector, $\Delta\mathbf{r}$. This is the straight-line vector that points from the particle's starting position at $t=a$ to its ending position at $t=b$.
Just as in Calculus I, we can use antiderivatives to evaluate definite integrals. If $\mathbf{R}(t)$ is an antiderivative of $\mathbf{r}(t)$ (meaning $\mathbf{R}'(t) = \mathbf{r}(t)$), then:
$$ \int_a^b \mathbf{r}(t) \,dt = \mathbf{R}(b) - \mathbf{R}(a) $$Find the position vector $\mathbf{r}(t)$ if the velocity vector is $\mathbf{v}(t) = \mathbf{r}'(t) = \langle 2t, 3t^2, \sqrt{t} \rangle$ and the initial position is $\mathbf{r}(1) = \langle 1, 0, 2 \rangle$.
First, we find the indefinite integral (the antiderivative) of $\mathbf{v}(t)$ component by component:
$$ \mathbf{r}(t) = \int \langle 2t, 3t^2, t^{1/2} \rangle \,dt = \left\langle t^2, t^3, \frac{2}{3}t^{3/2} \right\rangle + \mathbf{C} $$
Now, we use the initial condition at $t=1$ to solve for the constant vector $\mathbf{C}$:
$$ \mathbf{r}(1) = \left\langle 1^2, 1^3, \frac{2}{3}(1)^{3/2} \right\rangle + \mathbf{C} = \langle 1, 0, 2 \rangle $$
$$ \langle 1, 1, 2/3 \rangle + \mathbf{C} = \langle 1, 0, 2 \rangle $$
$$ \mathbf{C} = \langle 1, 0, 2 \rangle - \langle 1, 1, 2/3 \rangle = \langle 0, -1, 4/3 \rangle $$
Substituting $\mathbf{C}$ back in gives our final position vector:
$$ \mathbf{r}(t) = \left\langle t^2, t^3 - 1, \frac{2}{3}t^{3/2} + \frac{4}{3} \right\rangle $$
For the vector function $\mathbf{r}(t) = \langle t, 2\cos t, \sin t \rangle$, find the derivative $\mathbf{r}'(t)$ and the unit tangent vector $\mathbf{T}(t)$ at the point $t=0$.
Let $\mathbf{u}(t) = \langle t, t^2, t^3 \rangle$ and $\mathbf{v}(t) = \langle \sin t, \cos t, t \rangle$. Find $\frac{d}{dt}[\mathbf{u}(t) \cdot \mathbf{v}(t)]$ at $t=0$.
An object has an acceleration vector $\mathbf{a}(t) = \langle 0, 2, 6t \rangle$. Find its velocity and position vectors given an initial velocity $\mathbf{v}(0) = \langle 1, 0, 0 \rangle$ and initial position $\mathbf{r}(0) = \langle 0, 1, -1 \rangle$.
A particle's velocity is given by $\mathbf{v}(t) = \langle 3t^2, 2t, \frac{1}{t+1} \rangle$. Find the total displacement of the particle from $t=0$ to $t=1$.
Today, we've seen that the calculus of vector functions is a natural extension of single-variable calculus. We simply apply our known rules for derivatives and integrals to each component of the vector. This allows us to move from just describing a particle's position with $\mathbf{r}(t)$ to analyzing its motion through its velocity vector $\mathbf{r}'(t)$ and its acceleration vector $\mathbf{r}''(t)$. In our next lecture, we'll use these tools to measure geometric properties of the curve itself, such as its length and how much it bends, which are known as arc length and curvature.