Inquiry-Based Learning
- (T/F) If I get a problem wrong on one of the weekly mastery-based quizzes, I will have a chance to demonstrate mastery in future weeks.
Answer: True - Expand to see more.
True. However, you may have to wait 2 weeks to reassess on some objectives.
- The first time we have a quiz on new material, only the new material will appear on the quiz. This week, the quiz will contain only problems related to the objectives from the second unit.
- In off weeks (when we are in the middle of a new unit), the quiz will contain problems from all previous units. During these weeks, you'll be able to reassess any objective you missed in a previous week.
- (T/F) To pass the class, I need to present in class.
Answer: True - Expand to see more.
True. You get one participation point each time you present in class. I will prioritize picking presenters who have the fewest presentation points.
- You earn participation points when you present things in class.
- To pass the class, your presentation score needs to be at least 80% of the median score. If the median score is 5, you need a score of 4. If the median score is 10, you need at least 8.
- Regularly come to class with things ready to share, and you'll definitely pass this portion of the class.
- There is no requirement that you get everything correct, nor finish every problem. Try everything.
- (T/F) To get higher than a C, I need to complete SDL (self-directed learning) projects.
Answer: True - Expand to see more.
True. Each SDL project increases your grade by a step. Completing one project for each unit (so 6 projects) will get you an A.
- The time commitment is 5-8 hours.
- There is no firm deadline. They can be completed when you have time. I put a due date in I-Learn, spacing things out every 2 weeks.
- Completing one project every other week will get you to an A with a weekly 2-4 hour time commitment.
- You may only propose and complete 1 project a week. So if you wait too long to start, then you may not have time to complete them all.
- If you submit a proposal and it gets rejected, resubmit the proposal after revising based on the feedback you get.
Brain Gains
1. Compute the work done by the force $\vec F = \left<1,2,-2\right>$ acting through the displacement $\vec d = \left<0,3,4\right>$.
Answer:
Work is given by $W = \vec F \cdot \vec d$, so $W = \left<1,2,-2\right> \cdot \left<0,3,4\right> = 2\cdot3-2\cdot4 = -2$.
2. Compute the work done by the force $\vec F = \left<2t,t-1,t^2\right>$ (assuming $t$ is a constant) acting through the displacement $d \vec r = \left<dx,dy,dz\right>$.
Answer:
Work is given by $W = \vec F \cdot d\vec r$, so $$W = \left<2t,t-1,t^2\right> \cdot \left<dx,dy,dz\right> = 2tdx+(t-1)dy+t^2dz.$$
3. Compute the integral $\int \sqrt{x}\,dx$.
Answer:
The solution is $$\ds \int \sqrt{x}\, dx=\int x^{1/2}\,dx = \frac{x^{3/2}}{3/2}=\frac{2x^{3/2}}{3}.$$ Of course there is a plus $C$ that should be added. I'll leave it off, just as software does.
4. Compute the integral $\int t\sqrt{t^2+4}\,dt$.
Answer:
Let $u=t^2+4$. Then $du = 2t\,dt$, or $dt = \frac{du}{2t}$. Substituting these into the integral yields $$ \begin{align} \int t \sqrt{t^2+4} \, dt &= \int t \sqrt{u} \frac{du}{2t}\\ &= \frac{1}{2}\int \sqrt{u} \, du \\ &= \frac{1}{2} \frac{2u^{3/2}}{3} \\ &= \frac{(t^2+4)^{3/2}}{3}. \end{align}$$ Of course there is a plus $C$ that should be added. I'll leave it off, just as software does. Another option is $$ \begin{align} \int t \sqrt{t^2+4} \, dt &= \frac{1}{2}\int 2t \sqrt{t^2+4} \, dt\\ &= \frac{1}{2}\int \sqrt{t^2+4} \, 2t \, dt \\ &= \frac{1}{2} \int \sqrt{u} \, du \\ &= \frac{1}{2} \frac{2u^{3/2}}{3} \\ &= \frac{(t^2+4)^{3/2}}{3}. \end{align}$$
5. Recall the arc length formula is $$\int_C ds = \int_a^b\left|\dfrac{d\vec r}{dt}\right|dt=\int_a^b\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt.$$ Set up an integral to find the arc length of the curve $\vec r(t) = (t^2, t^3)$ for $t\in [-1,3] $.
Answer:
The derivative (velocity) is $\vec v(t) =(2t,3t^2)$. The magnitude of the derivative (speed) is $v = |\vec v| = \sqrt{(2t)^2+(3t^2)^2}$. A little bit of distance is equal to the speed multiplied by a little bit of time, or $$ds = \sqrt{(2t)^2+(3t^2)^2}dt.$$ Summing this over the time values for $t\in [-1,3] $ gives the total distance as $$s = \int_C ds = \int_{-1}^{3}\sqrt{(2t)^2+(3t^2)^2}dt.$$
Group problems
Pass the chalk (PTC) is short for letting someone else take a turn writing on your group Jamboard.
- Compute the integral $\ds \int x \sin (x^2)dx$. Pass the chalk (PTC).
- Consider the curve $C$ parametrized by $\vec r(t) = (3-2t^2,4t+5)$ for $-1\leq t\leq 3$.
- Give a vector equation of the tangent line to the curve at $t=2$. PTC
- Recall the arc length formula is $$\int_C ds = \int_a^b\left|\dfrac{d\vec r}{dt}\right|dt=\int_a^b\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt.$$ Set up a formula to compute the length of this curve. Just set it up. PTC
- Consider the curve $C$ parametrized by $\vec r(t) = (t^2, t^3)$ for $0\leq t\leq 2$.
- Give a vector equation of the tangent line to the curve at $t=1$. PTC
- Find the length of this curve. Actually compute the integral. PTC
- A wire lies along the curve $C$. The density (mass per length) of the wire at a point $(x,y)$ on the curve is given by $\delta(x,y) = y+2$. This means a little bits of mass along a small length $ds$ is given by $$dm = \delta ds = (y+2)ds = (y(t)+2)\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt.$$ Set up an integral formula that gives the total mass of the wire (put everything in terms of $t$. PTC
- The wire contains charged particles. The charge density (charge per length) at a point $(x,y)$ on the curve is given by the product $q(x,y)=xy$. This means the charge along a little length $ds$ is given by $dQ = (q) ds = (xy)ds$. Set up an integral formula that gives the total charge on the wire.
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