Rapid Recall
- Write the following antiderivatives:
- $\int x^3-2x^2+x+5\,dx$
- $\displaystyle \int e^{3x}\,dx$
- $\displaystyle \int x\sin x^2\,dx$
Solution
The solutions are
- $\displaystyle \frac{x^4}{4}-\frac{2x^3}{3}+\frac{x^2}{2}+5x+C$,
- $\displaystyle \frac{e^{3x}}{3}+C$, and
- $-\frac{1}{2}\cos(x^2)+C$.
For the last problem, did you remember substitution? If we let $u=x^2$, then $du =2xdx$, or $\frac{du}{2} = xdx$. This means $$ \begin{align} \displaystyle \int x\sin x^2\,dx &=\displaystyle \int \sin u (x\,dx)\\ &=\displaystyle \int \sin u(\frac{1}{2}\,du)\\ &=\displaystyle \frac{1}{2}(-\cos u)+C\\ &=\displaystyle -\frac{1}{2}\cos(x^2)+C. \end{align} $$ Anyone forget the $+C$?
- Give a vector equation of a line that passes through $(1,2)$ and is parallel to $(3,4)$.
Solution
$\vec r(t)=\left<3,4\right>\,t + \left<1,2\right>$
- Give a vector of length 4 that is parallel to the vector $\vec v = (-1,2,-2)$.
Solution
A quick answer is $$\frac{4}{3}(-1,2,-2)=\left(-\frac{4}{3}, \frac{8}{3}, -\frac{8}{3} \right).$$ The length of $\vec v$ is $|\vec v| = \sqrt{(-1)^2+(2)^2+(-2)^2} = \sqrt{9} =3$. A unit vector is then $$\hat v = \frac{\vec v}{|\vec v|} = \frac{(-1,2,-2)}{3}=\left(-\frac{1}{3}, \frac{2}{3}, -\frac{2}{3} \right).$$ The requested vector is then $$\vec w = 4\hat u = \frac{4}{3}(-1,2,-2)=\left(-\frac{4}{3}, \frac{8}{3}, -\frac{8}{3} \right).$$
- Use the law of cosines ($c^2=a^2+b^2-2ab\cos\theta$) to find the angle between the vectors $\vec u = (-1,5)$ and $\vec v = (2,4)$.
Solution
The vectors have lengths $a=|\vec u| = \sqrt{(-1)^2+(5)^2} = \sqrt{26}$ and $b=|\vec v| = \sqrt{(2)^2+(4)^2} = \sqrt{20}$. The difference is $\vec v-\vec u = (3, -1)$, and has length $c = |\vec v - \vec u| = \sqrt{10}$. From the law of cosines, we get $10 = 26+20-2\sqrt{26}\sqrt{10}\cos \theta$. Solving for $\cos\theta$ gives $\cos \theta = \frac{10-26-20}{-2\sqrt{26}\sqrt{10}}$, which means $$\theta = \arccos\left(\frac{10-26-20}{-2\sqrt{26}\sqrt{10}}\right).$$
Group problems
(Don't forget to PTC after each problem)
- Give a vector equation of the line that passes through the point $(1,2,3)$ and $(-2,4,9)$ (all distances are in meters, and times in minutes).
- Modify your vector equation from the previous part so that the speed of an object that is tracked with this equation is 3 meters per unit time.
- An object starts at $P=(1,2,3)$ and each unit of time its displacement is 2 units in the direction of $\vec v=(-4,5,1)$. Give an equation for the position $(x,y,z)$ at any time $t$.
- Use the law of cosines ($c^2=a^2+b^2-2ab\cos\theta$ or $\vec u\cdot \vec v = |\vec u||\vec v|\cos\theta$) to find the angle between each pair of vectors below.
- $(-2,1)$ and $(1,3)$.
- $(2,3)$ and $(-1,4)$
- $(\pi,e)$ and $(\sqrt{17},c)$
- $(1,2,3)$ and $(-7,2,1)$
- $(1,2,3)$ and $(x,y,z)$.
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