- I-Learn, Class Pictures, Learning Targets, Text Book Practice
- Prep Tasks: Unit 1 - Motion, Unit 2 - Derivatives, Unit 3 - Integration, Unit 4 - Vector Calculus
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.
Brain Gains (Rapid Recall, Jivin' Generation)
1. For the curve $\vec r(t) = (t^2+2, -3t+4)$, note that $\frac{d\vec r}{dt} =(2t,-3)$. Give a vector equation of the tangent line to $\vec r(t)$ at $t=2$. In other words, give a vector equation of a line that passes through $\vec r(2) = (6,-2)$ and is parallel to $\frac{d\vec r}{dt}(2) = (4,-3)$.
Solution
Passing through $(6,-2)$ and parallel to $(4,-3)$ means an equation is $$(x,y) = (4,-3)t+(6,-2)\quad \text{or}\quad \vec r(t) = (4,-3)t+(6,-2).$$ This is the same as $$(x,y) = (4t+6,-3t-2).$$ The Mathematica Code below defined both $\vec r(t)$ and the tangent line (as $\vec r2$), and then plots them on the same axes.
r[t_] := {t^2 + 2, -3 t + 4}
r'[t]
r'[2]
r2[t_] := r'[2] t + r[2]
r2[t]
Show[
ParametricPlot[r[t], {t, 0, 4}],
ParametricPlot[r2[t], {t, -1, 1}, PlotStyle -> Red]
]
2. The curve above represents the path of an object. Find the object's velocity and speed at any time $t$, and then at $t=1$.
Solution
The derivative $\frac{d\vec r}{dt} =(2t,-3)$ gives us the velocity at any time $t$. So at $t=1$ we have the velocity as $$\vec v = (2,-3).$$ The speed is the magnitude of the velocity, which gives the speed as $$v=|\vec v|=||\vec v|| = \sqrt{4t^2+9}.$$ At $t=1$ we have $v(1) = \sqrt{13}$. Note that we may use the same letter, without a vector symbol, to represent the magnitude of the corresponding vector, and you'll find that some people prefer double bars $||\vec v||$ instead of single bards $|\vec v|$.
We can quickly perform all these computations in Mathematica. The ComplexExpand[] command will expand an expression, assuming all variables involve are real.
r'[t] Norm[r'[t]] Norm[r'[t]] // ComplexExpand r'[2] Norm[r'[2]]
3. The (vector) projection of $\vec P$ onto $\vec Q$ is given by the formula $$\ds \text{proj}_\vec Q\vec P = \frac{\vec P\cdot \vec Q}{\vec Q\cdot \vec Q}\vec Q.$$ Compute the projection of $\vec P$ onto $\vec Q$ using $P=(3,4)$ and $Q=(-2,5)$.
Solution
We compute $$\ds \text{proj}_\vec Q\vec P = \frac{\vec P\cdot \vec Q}{\vec Q\cdot \vec Q}\vec Q = \frac{(3)(-2)+(4)(5)}{(-2)^2+5^2}(-2,5) = \frac{14}{29}(-2,5) .$$ Note that this is a little less than half $\vec Q$.
In Mathematica, the Projection[] command will do this for us.
Projection[{3, 4}, {-2, 5}]
4. Use the law of cosines in dot product form, so $\vec u\cdot \vec v = |\vec u||\vec v|\cos\theta$, to find the angle between $(-2,1)$ and $(4,3)$.
Solution
We have $$\cos\theta = \frac{\vec u\cdot \vec v}{|\vec u||\vec v|} = \frac{(-2)(4)+(1)(3)}{\sqrt{(-2)^2+1^2}\sqrt{4^2+3^2}} = \frac{-5}{\sqrt{5}5} = \frac{-1}{\sqrt{5}}.$$ This gives the angle as $\arccos(\frac{-1}{\sqrt{5}}) \approx 116.6^\circ$.
Using Norm[] to compute magnitude, and a period (.) for the dot product, Mathematica gives the following. The N[] command will numerically approximate an expression. We can use % for the previous output, and % // N is shorthand for N[%] (so numerically approximate the last output).
u = {-2, 1}
v = {4, 3}
ArcCos[u . v/(Norm[u] Norm[v])]
% // N (*radians by default*)
%*180/Pi (*degrees*)
Group Problems
Remember to pass the chalk after each problem (or part of a problem).
- Give a vector equation of the line that passes through the point $(1,2,3)$ and $(-2,4,9)$.
- An object starts at $P=(1,2,3)$ and each unit of time its displacement is $\vec v=(-4,5,1)$. Give an equation for the position $(x,y,z)$ at any time $t$.
- What is the speed of an object that follows the path described above?
- Draw the parametric curve $\vec r(t) = (1+t^2,3t-2)$ for $-2\leq t\leq 3$.
- Give $\frac{d\vec r}{dt}$. Then state $\frac{dx}{dt}$, $\frac{dy}{dt}$, and $\frac{dy}{dx}$.
- Give a vector equation of the tangent line to the curve at $t=2$.
- Let $P=(3,4)$ and $Q=(2,0)$.
- Using the formula $$\ds \text{proj}_\vec Q\vec P = \frac{\vec P\cdot \vec Q}{\vec Q\cdot \vec Q}\vec Q,$$ compute $\text{proj}_\vec Q\vec P$, the project of $\vec P$ onto $\vec Q$. We may also write this as $\vec P _{\parallel \vec Q}$, which we read as, "The vector component of $\vec P$ that is parallel to $\vec Q$."
- Draw $\vec P$, $\vec Q$ and $\text{proj}_\vec Q\vec P $ on the same grid with their base at the origin.
- Add to your picture the vector difference $\vec P_{\perp \vec Q} = \vec P - \text{proj}_{\vec Q}\vec P$, which we call "The vector component of $\vec P$ that is orthogonal to $\vec Q$." How is $\vec P_{\perp \vec Q}$ related to the other vectors?
- Now compute the projection of $\vec Q$ onto $\vec P$ (so swap which vector is projected onto the other).
- Draw $\vec P$, $\vec Q$ and $\text{proj}_\vec P\vec Q $ on the same grid with their base at the origin, and add to your picture the vector difference $\vec Q_{\perp \vec P} = \vec Q - \text{proj}_{\vec P}\vec Q $.
- Repeat the previous problem with $P=(3,4)$ and $Q=(-1,1)$.
|
Sun |
Mon |
Tue |
Wed |
Thu |
Fri |
Sat |