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

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Chapter Review Exercises 585 SOLUTION (a) Differentiating the vector e θ = - sin θ , cos θ with respect to θ gives: d e θ d θ = d d θ - sin θ , cos θ = - cos θ , - sin θ = - cos θ , sin θ = - e r (b) Eq. (2) is the following equality: r ( t ) = - k r 3 r ( t ) = - GM r 3 r ( t ) Writing r = r e r and r ( t ) = d dt r ( t ) = d v dt we get: d v dt = - GM r 3 r e r = - GM r 2 e r = - GM r 2 e r That is, d v dt = - GM 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 = GM r 2 d e θ d θ (c) In Exercise 24 we showed that 2 d A dt = r × r = J d A dt = 1 2 J 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 J d θ dt = J r 2 Substituting in the equality obtained in part (b) we obtain: d v d θ J r 2 = GM r 2 d e θ d θ Denoting C = GM J we obtain: d v d θ = GM J 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 ) = d v d θ d θ = C d e θ d θ d θ = C e θ + u where u is a constant vector. Notice that v ( θ ) - u = C e θ = | C | , which is the equation of a circle of radius C (recall, C = GM / J = k / J ) centered at the terminal point of u . CHAPTER REVIEW EXERCISES 1. Determine the domains of the vector-valued functions. (a) r 1 ( t ) = t - 1 , ( t + 1 ) - 1 , sin - 1 t (b) r 2 ( t ) = 8 - t 3 , ln t , e t SOLUTION
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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 ) = t - 1 , ( t + 1 ) - 1 , sin - 1 t . 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 = 0 t = - 1 - 1 < t < 0 - 1 t 1 or 0 < t 1 (b) We find the domain of r 2 ( t ) = 8 - t 3 , ln t , e t . The domain of 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 0 t > 0 t 0 t 3 8 t > 0 0 < t 2 2. Sketch the paths r 1 ( θ ) = θ , cos θ and r 2 ( θ ) = cos θ , θ in the xy -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 xy -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 . 3. Find a vector parametrization of the intersection of the surfaces x 2 + y 4 + 2 z 3 = 6 and x = y 2 in R 3 . SOLUTION We need to find a vector parametrization r ( t ) = x ( t ), y ( t ), z ( t ) for the intersection curve. Using t = y as a parameter, we have x = t 2 and y = t . We substitute in the equation of the surface x 2 + y 4 + 2 z 3 = 6 and solve for z in terms of t . This gives: t 4 + t 4 + 2 z 3 = 6 2 t 4 + 2 z 3 = 6 z 3 = 3 - t 4 z = 3 3 - t 4 We obtain the following parametrization of the intersection curve: r ( t ) = t 2 , t , 3 3 - t 4 .
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