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# 12.7-sm.dvi - S E C T I O N 13.7 SOLUTION Cylindrical and...

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S E C T I O N 13.7 Cylindrical and Spherical Coordinates (ET Section 12.7) 411 SOLUTION A point P on the parabola C has the form P = x 0 , ax 2 0 , c , hence the parametric equations of the line through the origin and P are x = tx 0 , y = tax 2 0 , z = tc To find a direct relation between xy and z we notice that yz = tax 2 0 ct = ac ( tx 0 ) 2 = acx 2 Now, defining new variables z = u - v and y = u + v . This equation becomes ( u + v)( u - v) = acx 2 u 2 - v 2 = acx 2 u 2 = acx 2 + v 2 This is the equation of an elliptic cone in the variables x , v , u . We, thus, showed that the cone on the parabola C is transformed to an elliptic cone by the transformation (change of variables) y = u + v , z = u - v , x = x . 13.7 Cylindrical and Spherical Coordinates (ET Section 12.7) Preliminary Questions 1. Describe the surfaces r = R in cylindrical coordinates and ρ = R in spherical coordinates. SOLUTION The surface r = R consists of all points located at a distance R from the z -axis. This surface is the cylinder of radius R whose axis is the z -axis. The surface ρ = R consists of all points located at a distance R from the origin. This is the sphere of radius R centered at the origin. 2. Which statement about the cylindrical coordinates is correct? (a) If θ = 0, then P lies on the z -axis. (b) If θ = 0, then P lies in the xz -plane. SOLUTION The equation θ = 0 defines the half-plane of all points that project onto the ray θ = 0, that is, onto the nonnegative x -axis. This half plane is part of the ( x , z ) -plane, therefore if θ = 0, then P lies in the ( x , z ) -plane. z y x The half-plane q = 0 For instance, the point P = ( 1 , 0 , 1 ) satisfies θ = 0, but it does not lie on the z -axis. We conclude that statement (b) is correct and statement (a) is false. 3. Which statement about spherical coordinates is correct? (a) If φ = 0, then P lies on the z -axis. (b) If φ = 0, then P lies in the xy -plane. SOLUTION The equation φ = 0 describes the nonnegative z -axis. Therefore, if φ = 0, P lies on the z -axis as stated in (a). Statement (b) is false, since the point ( 0 , 0 , 1 ) satisfies φ = 0, but it does not lie in the ( x , y ) -plane. 4. The level surface φ = φ 0 in spherical coordinates, usually a cone, reduces to a half-line for two values of φ 0 . Which two values? SOLUTION For φ 0 = 0, the level surface φ = 0 is the upper part of the z -axis. For φ 0 = π , the level surface φ = π is the lower part of the z -axis. These are the two values of φ 0 where the level surface φ = φ 0 reduces to a half-line. 5. For which value of φ 0 is φ = φ 0 a plane? Which plane? SOLUTION For φ 0 = π 2 , the level surface φ = π 2 is the xy -plane.

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412 C H A P T E R 13 VECTOR GEOMETRY (ET CHAPTER 12) z y P P x π 2 π 2 Exercises In Exercises 1–4, convert from cylindrical to rectangular coordinates. 1. ( 4 , π , 4 ) SOLUTION By the given data r = 4, θ = π and z = 4. Hence, x = r cos θ = 4 cos π = 4 · ( - 1 ) = - 4 y = r sin θ = 4 sin π = 4 · 0 z = 4 ( x , y , z ) = ( - 4 , 0 , 4 ) 2. 2 , π 3 , - 8 SOLUTION We are given that ( r , θ , z ) = ( 2 , π 3 , - 8 ) . Hence, x = r cos θ = 2 cos π 3 = 2 · 1 2 = 1 y = r sin θ = 2 sin π 3 = 2 3 2 = 3 z = - 8 ( x , y , z ) = 1 , 3 , - 8 3. 0 , π 5 , 1 2 SOLUTION We have r = 0, θ = π 5 , z = 1 2 . Thus, x = r cos θ = 0 · cos π 5 = 0 y = r sin θ = 0 · sin π 5 = 0 z = 1 2 ( x , y , z ) = 0 , 0 , 1 2 4. 1 , π 2 , - 2 SOLUTION
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12.7-sm.dvi - S E C T I O N 13.7 SOLUTION Cylindrical and...

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