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: Chapter Review Exercises 585 SOLUTION (a) Differentiating the vector e θ = t  sin θ , cos θ t with respect to θ gives: d e θ d θ = d d θ t  sin θ , cos θ t = t  cos θ , sin θ t =t cos θ , sin θ t = e r (b) Eq. (2) is the following equality: r t t ( t ) = k t r t 3 r ( t ) = G M t r t 3 r ( t ) Writing r = t r t e r and r t t ( t ) = d dt r t ( t ) = d v dt we get: d v dt = G M t r t 3 t r t e r = G M t r t 2 e r = G M r 2 e r That is, d v dt = G M r 2 e r (1) By the Chain Rule, d v dt = d v d θ · d θ dt and by part (a) d e θ d θ = e r . Substituting in (1), we get: d v d θ d θ dt = G M r 2 d e θ d θ (c) In Exercise 24 we showed that 2 d A dt = t r × r t t = t J t ⇒ d A dt = 1 2 t J t and d A dt = 1 2 r 2 d θ dt Combining the two equalities we get: 1 2 r 2 d θ dt = 1 2 t J t ⇒ d θ dt = t J t r 2 Substituting in the equality obtained in part (b) we obtain: d v d θ t J t r 2 = G M r 2 d e θ d θ Denoting C = G M t J t we obtain: d v d θ = G M t J t d e θ d θ = C d e θ d θ (d) Integrating the two sides of d v d θ = C d e θ d θ we have v ( θ ) = y d v d θ d θ = C y d e θ d θ d θ = C e θ + u where u is a constant vector. Notice that t v ( θ ) u t = t C e θ t =  C  , which is the equation of a circle of radius C (recall, C = G M / t J t = k / t J t ) centered at the terminal point of u . CHAPTER REVIEW EXERCISES 1. Determine the domains of the vectorvalued functions. (a) r 1 ( t ) = y t 1 , ( t + 1 ) 1 , sin 1 t y (b) r 2 ( t ) = y y 8 t 3 , ln t , e √ t y SOLUTION 586 C H A P T E R 14 CALCULUS OF VECTORVALUED FUNCTIONS (ET CHAPTER 13) (a) We find the domain of r 1 ( t ) = y t 1 , ( t + 1 ) 1 , sin 1 t y . The function t 1 is defined for t = 0. ( t + 1 ) 1 is defined for t = 1 and sin 1 t is defined for 1 ≤ t ≤ 1. Hence, the domain of r 1 ( t ) is defined by the following inequalities: t = t = 1 ⇒ 1 < t < 1 ≤ t ≤ 1 or < t ≤ 1 (b) We find the domain of r 2 ( t ) = y y 8 t 3 , ln t , e √ t y . The domain of y 8 t 3 is 8 t 3 ≥ 0. The domain of ln t is t > 0 and e √ t is defined for t ≥ 0. Hence, the domain of r 2 ( t ) is defined by the following inequalities: 8 t 3 ≥ t > t ≥ ⇒ t 3 ≤ 8 t > ⇒ < t ≤ 2 2. Sketch the paths r 1 ( θ ) = t θ , cos θ t and r 2 ( θ ) = t cos θ , θ t in the x yplane. SOLUTION The parametric equations of r 1 ( θ ) = ( θ , cos θ ) are x = θ , y = cos θ . Therefore, y = cos x . The parametric equations of r 2 ( θ ) = ( cos θ , θ ) are x = cos θ , y = θ . Therefore, x = cos y . We can sketch the graphs of r 1 ( θ ) and r 2 ( θ ) in the x yplane, using the explicit relations between y and x for the two parametric representations. We obtain: r 1 ( ) = ( , cos ), y = cos x 2 6 4 2 4 6 2 1 2 1 x 1 3 2 1 2 3 4 2 4 2 x y r 2 ( ) = (cos , ), x = cos y y As seen in the graph, although r 2 ( θ ) = ( cos θ , θ ) is a function of θ , y is not a function of x ....
View
Full
Document
This note was uploaded on 07/15/2009 for the course MATH 210 taught by Professor Hubscher during the Spring '08 term at University of Illinois at Urbana–Champaign.
 Spring '08
 Hubscher

Click to edit the document details