Rapid Recall
- Compute the first derivative of the following functions:
- $f(x) = x^2 \ln 3x$
- $g(x) = \sqrt{x^2+9}$
- $h(x) = \sin x^3$
Solution
- $\displaystyle f'(x) = 2x \ln 3x + x^2\frac{1}{3x}(3) = 2x\ln 3x + x$
- $\displaystyle g'(x) = \frac{1}{2}\frac{1}{\sqrt{x^2+9}}(2x)=\frac{x}{\sqrt{x^2+9}}$
- $h'(x) = 3x^2 \cos x^3$
- 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 $$\vec w = 4\hat u = \frac{4}{3}(-1,2,-2)=\left(-\frac{4}{3}, \frac{8}{3}, -\frac{8}{3} \right).$$
- Let $\vec u = (2,3,-1)$ and $\vec v = (-4,1,7)$. Compute the dot product $\vec u\cdot \vec v$.
- Using the same vectors as above, compute $\vec u\cdot \vec u$.
- Using the same vectors as above, compute the magnitude $\left|\vec u\right|$.
Group problems
(Don't forget to PTC after each problem) Use the law of cosines (the dot product version $\vec u\cdot \vec v = |\vec u||\vec v|\cos\theta$), to find the angle between each pair of vectors below. Take turns doing each problem, one at a time. Remember to pass the chalk between each problem. Use a calculator to compute the inverse cosine, as needed, and make a picture to verify that your answer is reasonable.
- $(-2,1)$ and $(1,3)$
- $(2,3)$ and $(-1,4)$
- $(-5,1)$ and $(2,10)$
- $(a,b)$ and $(c,d)$ (a general formula)
- $(1,2,3)$ and $(-7,2,1)$
- $(1,2,3)$ and $(x,y,z)$.
- Give values for $x,y,z$ above so that the angle is 90 degrees.
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