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Bryan Li's Homepage
About Me
Hi, I'm a freshman intending to major in mathematics and CS.
Notes
You can find my course notes typeset in $\LaTeX$ here
Homework
Week 1
- The gravitational force is equal to $10m$ downwards, and so the normal and frictional forces must sum to the opposite the gravitational force $(0, -10m)$. This gives us the equation $N \cdot (1/2, \sqrt{3}/2) + \mu \cdot (-\sqrt{3}/2, 1/2) = (0, 10m)$, where $N$ and $\mu$ are the magnitude of the normal and frictional forces respectively. Solving this system of linear equations gives us $\mu = 5m$ and $N = 5\sqrt{3}m$, implying that the normal and frictional forces are $F_N = (5\sqrt{3}m/2, 15m/2)$ and $F_\mu = (-5\sqrt{3}m/2, 5m/2)$ respectively.
- Assume the river flows horizontally, then the horizontal component of the velocity vector of the ship has magnitude $3$, which means the vertical component must have magnitude $4$ by the Pythagorean theorem. To travel $1.2$ miles in a round trip with a velocity of $4$ mph would take $0.3$ hours.
- We proceed by induction on the number of breaks in the line segment. For 0 breaks, this is trivially true. Now, suppose we have a broken segment $A_1\cdots A_{n + 1}$. Then, by the induction hypothesis we have $\overrightarrow{A_1A_2} + \cdots + \overrightarrow{A_{n - 1}A_n} + \overrightarrow{A_nA_1} = 0$, thus $\overrightarrow{A_1A_2} + \cdots + \overrightarrow{A_nA_{n + 1}} + \overrightarrow{A_{n + 1}A_1} = -\overrightarrow{A_nA_1} + \overrightarrow{A_nA_{n + 1}} + \overrightarrow{A_{n + 1}A_1} = \overrightarrow{A_1A_n} + \overrightarrow{A_nA_{n + 1}} + \overrightarrow{A_{n + 1}A_1} = \overrightarrow{A_1A_{n + 1}} + \overrightarrow{A_{n + 1}A_1} = 0$.
- The medians passing through $A$ can be given by $\frac{1}{2}(\overrightarrow{AB} + \overrightarrow{AC})$, with analogous formulas for the other vertices. The barycenterr lies on these medians so $\overrightarrow{AM} = \frac{1}{2}a(\overrightarrow{AB} + \overrightarrow{AC})$, with similar identities holding for the other vertices.
We then get
$$3\overrightarrow{OM} = \frac{1}{2}a(\overrightarrow{AB} + \overrightarrow{AC}) + \frac{1}{2}b(\overrightarrow{BA} + \overrightarrow{BC}) + \frac{1}{2}c(\overrightarrow{CB} + \overrightarrow{CA}) + \overrightarrow{OA}+ \overrightarrow{OB}+ \overrightarrow{OC}$$
