Lecture Notes

User Tools

Site Tools

You are here: Course Notes » Math 121B: Mathematics for Physical Sciences » Exercises on Tangent Vectors and Metric Tensor
math121b:ex3

Differences

This shows you the differences between two versions of the page.

Link to this comparison view

Both sides previous revision Previous revision
Next revision 		Previous revision
math121b:ex3 [2020/02/10 13:24]
pzhou 		math121b:ex3 [2020/02/20 15:02] (current)
pzhou
Line 12: Line 12:
  
 2. Area 2. Area
-  * Consider the vector space R2\R^2. Let v⃗1=(1,1),v⃗2=(0,2)\vec v_1 = (1,1), \vec v_2 = (0,2). Let P(v⃗1,v⃗2)P(\vec v_1, \vec v_2) denote the area of the parallelogram (skewed rectangle) generated by v⃗1,v⃗2\vec v_1, \vec v_2.    P(\vec v_1, \vec v_2) = ? +  * Consider the vector space R2\R^2. Let v⃗1=(1,1),v⃗2=(0,2)\vec v_1 = (1,1), \vec v_2 = (0,2). Let P(v⃗1,v⃗2)P(\vec v_1, \vec v_2) denote the **signed** area ((Signed area means $P(v,w) = - P(w,v)$)) of the parallelogram (skewed rectangle) generated by v⃗1,v⃗2\vec v_1, \vec v_2.    P(\vec v_1, \vec v_2) = ?
   * From the above computation, can you deduce P(v⃗1+3v⃗2,v⃗2)=? P(\vec v_1 + 3 \vec v_2, \vec v_2) = ? which formula did you use?    * From the above computation, can you deduce P(v⃗1+3v⃗2,v⃗2)=? P(\vec v_1 + 3 \vec v_2, \vec v_2) = ? which formula did you use? 
   * How about P(av⃗1+bv⃗2,cv⃗1+dv⃗2)=? P(a \vec v_1 + b \vec v_2, c \vec v_1 + d \vec v_2)=?    * How about P(av⃗1+bv⃗2,cv⃗1+dv⃗2)=? P(a \vec v_1 + b \vec v_2, c \vec v_1 + d \vec v_2)=?
Line 43: Line 43:
  
 ===== Solution ===== ===== Solution =====
 +==== Metric Tensor. Length, Area and Volume element ====
 1. True of False 1. True of False
   * False. Although, you can equip any finite dimensional vector space with an inner product, there is no one holier than the other.    * False. Although, you can equip any finite dimensional vector space with an inner product, there is no one holier than the other. 
Line 60: Line 61:
   * ∥e1+e2∥2=g(e1,e1)+g(e2,e2)+2g(e1,e2)=3+2+2=7 \| e_1 + e_2 \|^2 = g(e_1, e_1) + g(e_2, e_2) + 2 g(e_1, e_2) = 3 + 2 + 2 = 7 , hence  ∥e1+e2∥=7\| e_1 + e_2 \| = \sqrt{7} . Another way to get ∥e1+e2∥2\| e_1 + e_2 \|^2 is by    * ∥e1+e2∥2=g(e1,e1)+g(e2,e2)+2g(e1,e2)=3+2+2=7 \| e_1 + e_2 \|^2 = g(e_1, e_1) + g(e_2, e_2) + 2 g(e_1, e_2) = 3 + 2 + 2 = 7 , hence  ∥e1+e2∥=7\| e_1 + e_2 \| = \sqrt{7} . Another way to get ∥e1+e2∥2\| e_1 + e_2 \|^2 is by 
  (11 )(3112)(11)=7  \begin{pmatrix} 1 & 1  \end{pmatrix} \begin{pmatrix} 3 & 1 \cr 1 & 2 \end{pmatrix} \begin{pmatrix} 1 \cr 1 \end{pmatrix} = 7   (11 )(3112)(11)=7  \begin{pmatrix} 1 & 1  \end{pmatrix} \begin{pmatrix} 3 & 1 \cr 1 & 2 \end{pmatrix} \begin{pmatrix} 1 \cr 1 \end{pmatrix} = 7
-  * ∥e1−2e2∥= (1−2 )(3112)(1−2)=7 \| e_1 - 2 e_2 \| =  \begin{pmatrix} 1 & -2  \end{pmatrix} \begin{pmatrix} 3 & 1 \cr 1 & 2 \end{pmatrix} \begin{pmatrix} 1 \cr -2 \end{pmatrix} = 7  +  * $ \| e_1 - 2 e_2 \|^2 =  \begin{pmatrix} 1 & -2  \end{pmatrix} \begin{pmatrix} 3 & 1 \cr 1 & 2 \end{pmatrix} \begin{pmatrix} 1 \cr -2 \end{pmatrix} = 7 $ 
-  * $P(e_1, e_2) = \sqrt{\det(g)} = \sqrt{4} = 2$?+  * $P(e_1, e_2) = \sqrt{\det(g)} = \sqrt{5}$
   * Can you find two vectors v1,v2∈R2v_1, v_2 \in \R^2, such that v1,v2v_1, v_2 has the same properties as e1,e2e_1, e_2? v1=(3,0),v2=(2cos⁡θ,2sin⁡θ)v_1 = (\sqrt{3}, 0), v_2 = (\sqrt{2} \cos\theta, \sqrt{2} \sin \theta) where cos⁡θ=123\cos\theta = \frac{1}{\sqrt{2} \sqrt{3}}. We are using the formula    * Can you find two vectors v1,v2∈R2v_1, v_2 \in \R^2, such that v1,v2v_1, v_2 has the same properties as e1,e2e_1, e_2? v1=(3,0),v2=(2cos⁡θ,2sin⁡θ)v_1 = (\sqrt{3}, 0), v_2 = (\sqrt{2} \cos\theta, \sqrt{2} \sin \theta) where cos⁡θ=123\cos\theta = \frac{1}{\sqrt{2} \sqrt{3}}. We are using the formula 
 v⋅w=∥v∥∥w∥cos⁡θ v \cdot w = \| v \| \|w \| \cos \theta  v⋅w=∥v∥∥w∥cos⁡θ v \cdot w = \| v \| \|w \| \cos \theta 
math121b/ex3.1581369859.txt.gz · Last modified: 2020/02/10 13:24 by pzhou

Page Tools

Except where otherwise noted, content on this wiki is licensed under the following license: CC Attribution-Share Alike 4.0 International
CC Attribution-Share Alike 4.0 International Donate Powered by PHP Valid HTML5 Valid CSS Driven by DokuWiki