Chpt 13 Review Solns-sm.dvi - Chapter Review Exercises 585...

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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 vector-valued 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 VECTOR-VALUED 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 y-plane. 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 y-plane, 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 ....
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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.

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Chpt 13 Review Solns-sm.dvi - Chapter Review Exercises 585...

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