In this recitation, we covered motion and paths in space as well as arclength.

Motions and Paths

One key application of everything we’ve been talking about is analyzing motion in space. We did this kind of analysis in Calc 1 by thinking about \(s(t)\) as a function which gives the position of a particle at time \(t\). Remember that

• the velocity is given by \(v(t) := s'(t)\),
• the acceleration is given by \(a(t) := v'(t)\),
• the speed is given by \(S(t) := \mid v(t) \mid\),
• and the displacement over the interval \([a,b]\) is given by

\[\int_a^b s(t)dt.\]

It turns out that this is exactly the same for vector valued functions. Let’s use this to study projectile motion. Suppose someone tosses a baseball. Let’s assume that we’re viewing the person in a two dimensional plane, and (with respect to some frame of reference) we know that the baseball has initial position \(\vec{s}(0) = \langle x(0), y(0) \rangle\). Let’s also assume that the baseball leaves the persons hand with initial velocity given by \(\vec{v}(0) = v_0 \langle \cos(\theta), \sin(\theta) \rangle\), where \(\theta\) is referred to as the angle of elevation. Once the baseball leaves the persons hand, the only force acting on it is gravity, which (in our picture) acts only on the \(y\)-coordinate. Translating this, the only thing we know is that \(\vec{a}(t) = \langle 0, -g \rangle\) (here, \(g\) is the gravitational constant). Let’s use this to try to determine the velocity. We have

\[\vec{v}(t) = \int \vec{a}(t) dt = \int \langle 0, -g \rangle dt = \langle 0, -gt \rangle + \vec{C},\]

where \(\vec{C} = \langle C_1, C_2 \rangle\) is a constant vector. We know that

\[\vec{v}(0) = \langle v_0 \cos(\theta), v_0 \sin(\theta) \rangle,\]

so we combine the two expressions:

\[\langle v_0 \cos(\theta), v_0 \sin(\theta) \rangle = \langle C_1, C_2 - g \cdot 0 \rangle = \langle C_1, C_2 \rangle.\]

We see that \(C_1 = v_0 \cos(\theta)\) and \(C_2 = v_0 \sin(\theta)\), so we have

\[\vec{v}(t) = \langle v_0 \cos(\theta), v_0 \sin(\theta) - gt \rangle.\]

Now let’s find the position. Integrate again to get

\[\vec{s}(t) = \int \vec{v}(t) dt = \langle t v_0 \cos(\theta), t v_0 \sin(\theta) - gt^2/2 \rangle + \vec{C},\]

where again \(\vec{C} = \langle C_1, C_2 \rangle\) is some constant vector. Plug in \(t = 0\) to get

\[\langle x(0), y(0) \rangle = \langle C_1, C_2 \rangle,\]

so we put it all together and we get

\[\vec{s}(t) = \langle x(0) + t v_0 \cos(\theta), y(0) + tv_0 \sin(\theta) - gt^2/2 \rangle.\]

In other words, the \(x\)-position can be modeled by

\[x(t) = x(0) + t v_0 \cos(\theta),\]

and the \(y\)-position can be modeled by

\[y(t) = y(0) + tv_0 \sin(\theta) - gt^2/2.\]

We can also derive the equation for the speed at time \(t\) and the displacement of the ball, but this is just an easy calculus exercise at this point.

Arclength

Let’s go back to geometry for a moment. Remember that we the length of a curve \(y = f(x)\) on an interval \([a,b]\) is given by

\[\int_a^b \sqrt{1 + f'(x)^2}dx.\]

The way we “derived” this was actually by pretending it was a parameterized curve already. If we consider the curve \(\vec{r}(t) = (x(t), y(t))\), then our analysis showed us that the length of the curve on the interval \([a,b]\) should be something like

\[\int_a^b \sqrt{x'(t)^2 + y'(t)^2}dt.\]

From our work prior, we know that we can write this as just

\[\int_a^b \mid \vec{r}'(t) \mid dt.\]

In fact, this formula works in general. Going back to physical applications, if we know that the position of a particle can be modeled by the vector-valued function \(\vec{s}(t)\), then the distance traveled over the interval \([a,b]\) is given by

\[\int_a^b \mid \vec{s}'(t) \mid dt = \int_a^b \mid \vec{v}(t) \mid dt.\]

In other words, just integrate the speed from \(t=a\) to \(t=b\).