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math54-f22:s:bryanli

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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\LaTeX here

Homework

Week 1
  1. The gravitational force is equal to 10m10m downwards, and so the normal and frictional forces must sum to the opposite the gravitational force (0,−10m)(0, -10m). This gives us the equation N⋅(1/2,3/2)+μ⋅(−3/2,1/2)=(0,10m)N \cdot (1/2, \sqrt{3}/2) + \mu \cdot (-\sqrt{3}/2, 1/2) = (0, 10m), where NN and μ\mu are the magnitude of the normal and frictional forces respectively. Solving this system of linear equations gives us μ=5m\mu = 5m and N=53mN = 5\sqrt{3}m, implying that the normal and frictional forces are FN=(53m/2,15m/2)F_N = (5\sqrt{3}m/2, 15m/2) and Fμ=(−53m/2,5m/2)F_\mu = (-5\sqrt{3}m/2, 5m/2) respectively.
  2. Assume the river flows horizontally, then the horizontal component of the velocity vector of the ship has magnitude 33, which means the vertical component must have magnitude 44 by the Pythagorean theorem. To travel 1.21.2 miles in a round trip with a velocity of 44 mph would take 0.30.3 hours.
  3. 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 A1⋯An+1A_1\cdots A_{n + 1}. Then, by the induction hypothesis we have A1A2→+⋯+An−1An→+AnA1→=0\overrightarrow{A_1A_2} + \cdots + \overrightarrow{A_{n - 1}A_n} + \overrightarrow{A_nA_1} = 0, thus A1A2→+⋯+AnAn+1→+An+1A1→=−AnA1→+AnAn+1→+An+1A1→=A1An→+AnAn+1→+An+1A1→=A1An+1→+An+1A1→=0\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.
  4. 3OM→=OA→+AM→+OB→+BM→+OC→+CM→=OA→+OB→+OC→3\overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{AM} + \overrightarrow{OB} + \overrightarrow{BM} + \overrightarrow{OC} + \overrightarrow{CM} = \overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} by Problem 5.
  5. AM→=23AA′→\overrightarrow{AM} = \frac{2}{3}\overrightarrow{AA'} with similar formulas for the other vertices by geometry. By Problem 7, we can then write MA→+MB→+MC→\overrightarrow{MA} + \overrightarrow{MB} + \overrightarrow{MC} as −13(AB→+AC→)+13(BA→+BC→)+13(CA→+CB→)-\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 00.
  6. If z1,z2,z3z_1, z_2, z_3 are the centers of the vertices' circles, then vi=zi+ciei(ωt+ϕ1)=zi+kieiωtv_i = z_i + c_ie^{i(\omega t + \phi_1)} = z_i + k_ie^{i\omega t}, then the barycenter follows the motion M=13(v1+v2+v3+(k1+k2+k3)eiωtM = \frac{1}{3}(v_1 + v_2 + v_3 + (k_1 + k_2 + k_3)e^{i\omega t}
  7. This is by definition of a median. The vector AB→+AC→\overrightarrow{AB} + \overrightarrow{AC} passes between BB and CC, but is twice the length, so we scale it by 1/21/2.

math54-f22/s/bryanli.txt · Last modified: 2022/08/28 12:33 by alphyte

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