13.2.b-sm.dvi - 464 C H A P T E R 14 CALCULUS OF...

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Unformatted text preview: 464 C H A P T E R 14 CALCULUS OF VECTOR-VALUED FUNCTIONS (ET CHAPTER 13) (A) R 2 (B) Q 1 Q 2 F 2 F 1 P P K R 1 FIGURE 15 SOLUTION To show that C is an ellipse, we show that every point P on C satisfies: F 1 P + F 2 P = K We denote the points of intersection of the vertical line through P with the equators of the two spheres by R 1 and R 2 (see figure). R 2 F 2 F 1 P K R 1 We denote by O 1 and O 2 the centers of the spheres. F 1 O 1 P r Since F 1 is the tangency point, the radius O 1 F 1 is perpendicular to the plane of the curve C , and therefore it is orthogonal to the segment P F 1 on this plane. Hence, O 1 F 1 P is a right triangle and by Pythagoras Theorem we have: O 1 F 1 2 + P F 1 2 = O 1 P 2 r 2 + P F 1 2 = O 1 P 2 P F 1 = t O 1 P 2- r 2 (1) R 1 O 1 P r O 1 R 1 P is also a right triangle, hence by Pythagoras Theorem we have: O 1 R 1 2 + R 1 P 2 = O 1 P 2 S E C T I O N 14.1 Vector-Valued Functions (ET Section 13.1) 465 r 2 + R 1 P 2 = O 1 P 2 P R 1 = t O 1 P 2- r 2 (2) Combining (1) and (2) we get: P F 1 = P R 1 (3) Similarly we have: P F 2 = P R 2 (4) We now combine (3), (4) and the equality P R 1 + P R 2 = K to obtain: F 1 P + F 2 P = P R 1 + P R 2 = K Thus, the sum of the distances of the points P on C to the two fixed points F 1 and F 2 is a constant K > 0, hence C is an ellipse. 41. Now reprove the result of Exercise 40 using vector geometry. Assume that the cylinder has equation x 2 + y 2 = r 2 and the plane has equation z = ax + by . (a) Show that the upper and lower spheres in Figure 15 have centers C 1 = t , , r t a 2 + b 2 + 1 t C 2 = t , ,- r t a 2 + b 2 + 1 t (b) Show that the points where the plane is tangent to the sphere are F 1 = r t a 2 + b 2 + 1 t a , b , a 2 + b 2 t F 2 =- r t a 2 + b 2 + 1 t a , b , a 2 + b 2 t Hint: Show that C 1 F 1 and C 2 F 2 have length r and are orthogonal to the plane. (c) Verify, with the aid of a computer algebra system, that Eq. (2) holds with K = 2 r t a 2 + b 2 + 1. To simplify the algebra, observe that since a and b are arbitrary, it suffices to verify Eq. (2) for the point P = ( r , , ar ) . SOLUTION (a) and (b) Since F 1 is the tangency point of the sphere and the plane, the radius to F 1 is orthogonal to the plane. Therefore to show that the center of the sphere is at C 1 and the tangency point is the given point we must show that:-- C 1 F 1 = r (1)-- C 1 F 1 is orthogonal to the plane. (2) We compute the vector-- C 1 F 1 :-- C 1 F 1 = t r a t a 2 + b 2 + 1 , r b t a 2 + b 2 + 1 , r ( a 2 + b 2 ) t a 2 + b 2 + 1- r t a 2 + b 2 + 1 = r t a 2 + b 2 + 1 a , b ,- 1 Hence,-- C 1 F 1 = r t a 2 + b 2 + 1 a , b ,- 1 = r t a 2 + b 2 + 1 t a 2 + b 2 + (- 1 ) 2 = r We, thus, proved that (1) is satisfied. To show (2) we must show that-- C 1 F 1 is parallel to the normal vector a , b ,- 1 to the plane z = ax + by (i.e., ax + by- z = 0). The two vectors are parallel since by (1)-- C 1 F 1 is a constant multiple of a , b ,- 1 . In a similar manner one can show (1) and (2) for the vector....
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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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13.2.b-sm.dvi - 464 C H A P T E R 14 CALCULUS OF...

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