{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

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
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

{[ snackBarMessage ]}

### Page1 / 23

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online