- I-Learn, Class Pictures, Learning Targets, Text Book Practice
- Prep Tasks: Unit 1 - Motion, Unit 2 - Derivatives, Unit 3 - Integration, Unit 4 - Vector Calculus
Brain Gains (Rapid Recall, Jivin' Generation)
- 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 $t$.
- For the vectors $\vec u = (2,-1,5)$ and $\vec v = (-3, 4,6)$, compute the cross product $\vec u\times \vec v$.
- Compute $\vec v\times \vec u$, and verify that $\vec v\times \vec u = - \vec u\times \vec v$.
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.
- The cross product of $\vec u = (2,-1,5)$ and $\vec v = (-3, 4,6)$ is $((-1)(6)-(4)(5), (-3)(5) - (2)(6), (2)(4)-(-3)(-1) ) = (-26,-27,5)$. We can compute that in Mathematica using the following code.
Cross[{2, -1, 5}, {-3, 4, 6}] - We have $\vec v\times \vec u = (26,27,-5)$, which shows $\vec v\times \vec u = - \vec u\times \vec v$.
Group Problems
- For the curve $r(t) = (2\cos t, 3t, 2\sin t)$, compute the arc length parameter $s(t)$.
- Compute $ds/dt$, $d\vec r/dt$, $|d\vec r/dt|$, and $d\vec r/ds$.
- State the unit tangent vector $\vec T$ at time $t$, and identify how to obtain it from the parts above.
- 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 $s(t)$.
- Compute $ds/dt$, $d\vec r/dt$, $|d\vec r/dt|$, and $d\vec r/ds$.
- State the unit tangent vector $\vec T$ at time $t$, and identify how to obtain it from the parts above.
- Compute $d\vec T/dt$, and show that this vector is orthogonal to $\vec T$.
- For the vectors $\vec u = (1,2,3)$ and $\vec v = (4, 5,6)$, compute the cross product $\vec u\times \vec v$ and $\vec v\times \vec u$. Then change the vectors to something else, and practice doing this multiple times.
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