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Unformatted text preview: 1 Tutorial 1 1.1 Mistakes in class On qn5, L has a unit : the same unit of the time t , since ˙ a = d a/ d t where d t has the unit of time. So that both sides of the equation L 2 (˙ a ) 2 = a 2- 2 a 2 + 1 . ( L &gt; 0) have no unit. On qn4, solve d.e. k = tan φ = r d θ d r . k = tan φ = r d θ d r , k d r r = d θ, k ln r = θ + C ( r &gt; 0) , e k ln r = e θ C ( n ot ke ln r ) , r k = e θ C r = e θ k C 1 . 1.2 Graphical demo on qn4 Open the file by graphmatica * Click on the curves, you will see the corresponding equation on the top bar. Adjust tan φ to observe different outcome. (I use 10,- 10 and 10 30 .) 1.3 Solving the d.e. in qn5 L 2 (˙ a ) 2 = a 2- 2 a 2 + 1 . ( L &gt; 0) (1) with initial condition a (0) = 1. The above is a first order separable d.e.: d a d t = q a 2- 2 a 2 + 1 L d a q a 2- 2 a 2 + 1 = d t L * Download the tiny software from http://www.graphmatica.com . It can be successfully installed even on a computer’s guest account (change the installation folder to desktop). 1 a d a √ a 4 + a 2- 2 = d t L Remark. When finding the square root of ˙ a 2 , we actually assume that ˙ a &gt; 0. However, there is no word indicating that ˙ a can’t be negative. Thus, we also need to consider thecan’t be negative....
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This note was uploaded on 12/09/2011 for the course ELECTRICAL 101 taught by Professor Huanhoang during the Fall '11 term at National University of Singapore.
- Fall '11