Math 214 - Multivariable Calculus - Schedule

Prep

We're in Unit 1 - Motion. Your homework assignment each day is to spend 1-2 hours working on the next 4 tasks from the current unit's prep.

For the curve $\vec r(t) = (t,\cos t, \sin t)$, compute $\ds \int_0^a \left|\frac{d\vec r}{dt}\right|dt$.
For the same curve $\vec r(t) = (t,\cos t, \sin t)$, compute the arc length parameter $\ds s(t) = \int_0^t \left|\frac{d\vec r}{d\tau}\right|d\tau$.
Compute $\ds \frac{d}{dt}\left(\int_0^t x^3 dx\right)$.
Compute $\ds \frac{d}{dt}\left(\int_0^t \sqrt{1+\tau^2+\cos(\tau)} d\tau\right)$.
Compute $\ds \frac{ds}{dt}$, so compute $\ds \frac{d}{dt}\left(\int_0^t \left|\frac{d\vec r}{d\tau}\right|d\tau\right)$.
For the curve $\vec r(t) = (t,t^2)$, compute both $\dfrac{d\vec r}{dt}$ and $\dfrac{d\vec r}{ds}$, giving both in terms of $t$.

Solutions

For the curve $r(t) = (t,\cos t, \sin t)$, we have $r'(t) = (1,-\sin t, \cos t)$ and $|\vec r'(t)|=\sqrt{2}$. This gives $$\ds \int_0^a \left|\frac{d\vec r}{dt}\right|dt = \int_0^a\sqrt{2}dt = a\sqrt{2}.$$
The arc length parameter is $$\ds s(t) = \int_0^t \left|\frac{d\vec r}{d\tau}\right|d\tau= \int_0^t\sqrt{2}d\tau = t\sqrt{2}.$$ Why the $\tau$? We want length $s$ to be a function of $t$, which means the "dummy" variable of integration we need to change away from $t$ to something else (we could have used $w$ or $p$, but for historically people use $\tau$).
The fundamental theorem of calculus gives $\ds \frac{d}{dt}\left(\int_0^t x^3 dx\right) = t^3$.
The fundamental theorem of calculus gives $\ds \frac{d}{dt}\left(\int_0^t \sqrt{1+\tau^2+\cos(\tau)} d\tau\right)=\sqrt{1+t^2+\cos(t)}$. Yes, it is that simple.
Again, the fundamental theorem of calculus gives $\ds \frac{d}{dt}\left(\int_0^t \left|\frac{d\vec r}{d\tau}\right|d\tau\right) = \left|\frac{d\vec r}{d t}\right|$, so $ds/dt$ is just the speed.
For the curve $\vec r(t) = (t,t^2)$, we have $\dfrac{d\vec r}{dt} = (1,2t)$ and $\dfrac{d\vec r}{ds} = \frac{d\vec r/dt}{ds/dt} = \frac{ (1,2t) }{ \sqrt{1^2+(2t)^2} }$. Remember that $ds/dt$ is just the speed.

Compute the work done by $\vec F(x,y) = (-3y,3x)$ along $\vec r(t)=(5\cos t, 5\sin t)$ for $t\in[0,2\pi] $. The work integral is written using any of $$\ds\int_C Mdx+Ndy=\ds\int_C \vec F\cdot d\vec r=\ds\int_{a}^{b}\vec F(\vec r(t))\cdot \frac{d\vec r}{dt}dt.$$
Let $r(t) = (2\cos t, 3t, 2\sin t)$.
1. Compute (and simplify) the arc length parameter $\ds s(t) = \int_0^t \left|\frac{d\vec r}{d\tau}\right|d\tau$.
2. Compute $ds/dt$, $d\vec r/dt$, $|d\vec r/dt|$, and $d\vec r/ds$.
3. State the unit tangent vector $\vec T$ at time $t$, and identify how to obtain it from the parts above.
4. Compute $d\vec T/dt$, and show that this vector is orthogonal to $\vec T$.
For the curve $r(t) = (t^2, 3t)$, set up an integral that gives the arc length parameter $s(t)$.
1. Compute $ds/dt$, $d\vec r/dt$, $|d\vec r/dt|$, and $d\vec r/ds$.
2. State the unit tangent vector $\vec T$ at time $t$, and identify how to obtain it from the parts above.
3. Compute $d\vec T/dt$, and show that this vector is orthogonal to $\vec T$.