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math54-f22:s:bryanli [2022/08/26 05:50]
alphyte [Homework] 		math54-f22:s:bryanli [2026/02/21 14:41] (current)
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         <li> 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.         <li> 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.
         <li> 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 $\overrightarrow{A_1A_2} + \cdots + \overrightarrow{A_nA_{n + 1}} + \overrightarrow{A_{n + 1}A_1} = -\overrightarrow{A_nA_1} + \overrightarrow{A_nA_{n + 1}} + \overrightarrow{A_{n + 1}A_1} = \overrightarrow{A_1A_n} + \overrightarrow{A_nA_{n + 1}} + \overrightarrow{A_{n + 1}A_1} = \overrightarrow{A_1A_{n + 1}} + \overrightarrow{A_{n + 1}A_1} = 0$.         <li> 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 $\overrightarrow{A_1A_2} + \cdots + \overrightarrow{A_nA_{n + 1}} + \overrightarrow{A_{n + 1}A_1} = -\overrightarrow{A_nA_1} + \overrightarrow{A_nA_{n + 1}} + \overrightarrow{A_{n + 1}A_1} = \overrightarrow{A_1A_n} + \overrightarrow{A_nA_{n + 1}} + \overrightarrow{A_{n + 1}A_1} = \overrightarrow{A_1A_{n + 1}} + \overrightarrow{A_{n + 1}A_1} = 0$.
-        <li>+        <li> $3\overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{AM} + \overrightarrow{OB} + \overrightarrow{BM} + \overrightarrow{OC} + \overrightarrow{CM} = \overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC}$ by Problem 5. 
 +        <li> $\overrightarrow{AM} = \frac{2}{3}\overrightarrow{AA'}$ with similar formulas for the other vertices by geometry. By Problem 7, we can then write $\overrightarrow{MA} + \overrightarrow{MB} + \overrightarrow{MC}$ as $-\frac{1}{3}(\overrightarrow{AB} + \overrightarrow{AC}) + \frac{1}{3}(\overrightarrow{BA} + \overrightarrow{BC}) + \frac{1}{3}(\overrightarrow{CA} + \overrightarrow{CB})$ which clearly sums to $0$. 
 +        <li> If $z_1, z_2, z_3$ are the centers of the vertices' circles, then $v_i = z_i + c_ie^{i(\omega t + \phi_1)} = z_i + k_ie^{i\omega t}$, then the barycenter follows the motion $M = \frac{1}{3}(v_1 + v_2 + v_3 + (k_1 + k_2 + k_3)e^{i\omega t}$ 
 +        <li> This is by definition of a median. The vector $\overrightarrow{AB} + \overrightarrow{AC}$ passes between $B$ and $C$, but is twice the length, so we scale it by $1/2$.
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math54-f22/s/bryanli.1661493010.txt.gz · Last modified: 2026/02/21 14:44 (external edit)

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