- 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)
1. For $\vec r(t) = (\cos t, \sin t, t)$, compute $\vec T = \dfrac{d\vec r}{ds}$, $\vec N = \dfrac{d\vec T/dt}{|d\vec T/dt|}$, and $\vec B = \vec T\times \vec N$.
Solution and animation
We have $$\ds\frac{d\vec r}{dt} = (-\sin t, \cos t, 1)$$ which means $\ds\left|\frac{d\vec r}{dt}\right| = \sqrt{2}$. This gives $$\ds\vec T = \frac{ (-\sin t, \cos t, 1) }{\sqrt{2}}.$$ We then compute $$\ds\frac{d\vec T}{dt} = \frac{ (-\cos t, -\sin t, 0) }{\sqrt{2}}$$ which means $\ds\left|\frac{d\vec T}{dt}\right| = \frac{1}{\sqrt{2}}$. We then have $$\vec N = (-\cos t, -\sin t, 0).$$ The cross product gives $$\vec B = \frac{ (\sin t,-\cos t, 1) }{\sqrt{2}}.$$
Here's an animation of these topics.
2. For the circle $\vec r(t) = (3\cos t, 3\sin t)$, compute the curvature $\kappa = \left|\dfrac{d\vec T}{ds}\right|$ and radius of curvature $\dfrac{1}{\kappa}$.
Solution
We need $\kappa = |\ds\frac{d\vec T}{ds}| = \frac{|d\vec T/dt|}{ds/dt} = \frac{|d\vec T/dt|}{|dr/dt|}$.
We have $$\ds\frac{d\vec r}{dt} = (-3\sin t, 3\cos t)$$ which means $\ds\left|\frac{d\vec r}{dt}\right| = 3$. This gives $$\ds\vec T = \frac{ (-3\sin t, 3\cos t) }{3} = (-\sin t, \cos t).$$ We then compute $$\ds\frac{d\vec T}{dt} = (-\cos t, -\sin t)$$ which means $\ds\left|\frac{d\vec T}{dt}\right| = 1$. Putting this together gives $$\kappa = \frac{|d\vec T/dt|}{|dr/dt|} = \frac{1}{3}.$$
For a circle of radius 3, the curvature is 1/3, and the radius of curvature is 3.
3. Find a nonzero vector that is orthogonal to the two vectors $(-1,2,5)$ and $(3,0,4)$.
Solution
The cross product of the two vectors, in either order, will provide an answer. So one answer is $$(-1,2,5)\cross (3,0,4) = (8, 19, -6).$$
Group Problems
1. For $\vec r(t) = (4\cos t, 4\sin t, 3t)$, compute $\vec T = \dfrac{d\vec r}{ds}$, $\vec N = \dfrac{d\vec T/dt}{|d\vec T/dt|}$, $\vec B = \vec T\times \vec N$, and $\kappa = \left|\dfrac{d\vec T}{ds}\right|$.
2. A plane passes through the points $P=(2,0,1)$, $Q=(1,-2,0)$, and $R=(3,3,2)$. Give a nonzero vector $\vec n$ that is orthogonal to both $\vec {PQ}$ and $\vec{PR}$, which we call a normal vector to the plane (it will be orthogonal to every vector in the plane). Then let $S = (x,y,z)$ be another point in the plane. Use that fact that $\vec PS$ is a vector in the plane, and $\vec n$ is a normal vector to the plane, to give an equation of the plane.
3. For $\vec r(t) = (t^2, 0,t)$, at $t=1$ please compute $\vec T = \dfrac{d\vec r}{ds}$, $\vec N = \dfrac{d\vec T/dt}{|d\vec T/dt|}$, $\vec B = \vec T\times \vec N$, and $\kappa = \left|\dfrac{d\vec T}{ds}\right|$. The computations, by hand, will get quite ugly very quickly. Once you've computed $d\vec T/dt$, you don't need any more derivatives, which means at that point you can plug in $t=1$ and then simplify before continuing on to get $\vec N$, $\vec B$, and $\kappa$.
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