This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Solutions for Homework 1 1. 2.1.1 Fixed points are points where sin x = 0, that is, integer multiples of . 2.1.2 The greatest velocity to the right occurs where sin x is the greatest. That is, where x is / 2 plus an integer multiple of 2 . 2.1.3 a) x = sin x implies x = ( cos x ) x by the chain rule, so x = cos x sin x = sin ( 2 x ) / 2. Did you remember that cos x sin x = sin ( 2 x ) / 2? If not, here is an easy way to memorize it: Among the trigonometric formulas, the only ones that I memorize are sin ( a + b ) = sin a cos b + cos a sin b and cos ( a + b )= cos a cos b sin a sin b . All others follow from these very quickly. For example, the first with a = b = x gives you what you need for this problem. b) The acceleration to the right is greatest where sin ( 2 x ) is greatest, that is where x = / 4 plus an integer multiple of . 2. 2.2.1. Sketch of the vector field:42 2 42010 10 20 30 40 x 4 x 216 x = 2 is a stable fixed point, and x = 2 is unstable, as is apparent from the vector field. Here is what the solutions look like: 0.1 0.2 0.3 0.4321 1 2 3 t x (Solutions with x > 2 blow up in finite time.) Solution in closed form: If x = 2, then x ( t ) = 2 for all t , since 2 and 2 are fixed points. Lets assume that x negationslash = 2 now. Then x ( t ) negationslash = 2 for all t (this is true by the uniqueness part of 1 the existence and uniqueness theorem). This will be used in the following manipulations. x = 4 x 2 16 x x 2 4 = 4 4 x ( x 2 )( x + 2 ) = 16 x x 2 x x + 2 = 16 d dt ln (  x 2  ) d dt ln (  x + 2  ) = 16 d dt ln parenleftbiggvextendsingle vextendsingle vextendsingle vextendsingle x 2 x + 2 vextendsingle vextendsingle vextendsingle vextendsingle parenrightbigg = 16 ln parenleftbiggvextendsingle vextendsingle vextendsingle vextendsingle x 2 x + 2 vextendsingle vextendsingle vextendsingle vextendsingle parenrightbigg = 16 t + C for some constant C , which is found by plugging in t = 0. Using the notation x ( ) = x , we find C = ln parenleftbiggvextendsingle vextendsingle vextendsingle vextendsingle x 2 x + 2 vextendsingle vextendsingle...
View
Full
Document
This note was uploaded on 02/28/2011 for the course MATH 150 taught by Professor Dr.borgers during the Spring '09 term at Tufts.
 Spring '09
 Dr.Borgers

Click to edit the document details