Rapid Recall
1. Compute the work done by the vector field $\vec F(x,y,z) = \left<1,2,-2\right>$ 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 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 $+C$.
3. 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 $$ \begin{align} \int t \sqrt{t^2+4} \, dt &= \frac{1}{2}\int 2t \sqrt{t^2+4} \, dt\\ &= \frac{1}{2}\int \sqrt{t^2+4} \, 2t \, dt \\ &= \frac{1}{2} \int \sqrt{u} \, du \\ &= \frac{1}{2} \frac{2u^{3/2}}{3} \\ &= \frac{(t^2+4)^{3/2}}{3}. \end{align}$$
4. 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 $$s = \int_C ds = \int_{-1}^{3}\sqrt{(2t)^2+(3t^2)^2}dt.$$
5. (T/F) If I miss a due date on a quiz, I can submit it late without penalty (for a time).
Answer:
True. However, earlier completion of a quiz process will facilitate focus on current materials
Group problems
- 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 $\delta(x,y) = y+2$. Set up an integral formula that gives the total mass of the wire. 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 $q(x,y)=xy$. Set up an integral formula that gives the total charge on the wire.
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