MA 351 Fall 2007 Exam #1 Review Solutions 1

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1 MA 35 Fall 27 Exam # Review Solutions THERE MAY BE TYPOS in these solutions. Please let me know if you find any.. Consider the two surfaces ρ 3 csc θ in spherical coordinates and r 3 in cylindrical coordinates. Are they the same surface, or different? Explain your answer. Several ways to do this: First curve: x ρ cos θ sin φ y ρ sin θ sin φ z ρ cos φ Second curve: x r cos θ y r sin θ z z So if we let φ for the first curve, x, y and z ρ 3 csc θ. For the second curve, when x, that would mean cos θ since r 3. Then on the second curve, y 3 when x. Thus these cannot be the same surface. 2. Let a (x + yi + 2j + yk and b 3i + (4x + y + j + 4k. Find the relationship between x and y such that a and b are orthogonal. a and b are orthogonal a b a b 3(x + y + 2(4x + y + + 4y 3x + 3y + 8x + 2y y x + 9y 2 x + 9y 2 (or some equivalent equation for this line 3. Find a vector function that represents the curve of intersection of the paraboloid z 3x 2 + y 2 and the parabolic cylinder y x 2. C : z 3x 2 + y 2, C 2 : y x 2 By plugging in the equation for C 2 into C, we get z 3y + y 2 So if we let x t for t R, then y t 2 and z 3t 2 + t 4 to get r(t ti + t 2 j + (3t 2 + t 4 k 4. Show that the line given by x t, y 3t 2, z t intersects the plane x + y + z and find the point of intersection. Plug in the parametric equations for the line into the equation of the plane to get: x + y + z t + 3t 2 t 3t 2 t So the line intersects the plane when t, or more specifically, at the point x, y and z Point of intersection: (,,

2 MA 35 Fall 27 Exam # Review Solutions 2 5. Sketch the portion of the plane x + y + z that lies in the first octant. To sketch, try and find intercepts. x-int: when y z x (,, y-int: when x z y (,, z-int: when x y z (,, Since 3 points determine a plane, can use these points to draw the sketch (sketch not shown 6. Suppose that a, b and c are three distinct unit vectors in R 3 which satify the following two conditions: b c and a (b c. Show that a is perpendicular to both b and c. (Hint: use the definition of the cross product. b c b c b c sin θ means that sin θ and we also have b c since they are unit vectors. Thus b c sin θ. We are given that a (b c. But a (b c a b c sin θ 2 and since a, we get a (b c sin θ sin θ 2 The only way for this to happen is for sin θ 2, in other words, θ 2, the angle between a and b c is or π. So a is parallel to b c. Thus a is orthogonal to b and c since b c is orthogonal to both b and c. 7. Find the angle θ between the planes x + y + z and x 2y + 3z. Look at the normal vectors n and n 2. n <,, > n 2 <, 2, 3 > n n 2 n n 2 cos θ cos θ n n 2 n n ( 2 θ cos (this is the exact value radians or Find the symmetric equation for the line of intersection between the two planes x + y + z and x 2y + 3z. Try setting z into both: x + y x 2y 3y y x So a point on the line is (,, Now we need the direction vector for the line. The normal vectors n and n 2 for both planes are perpendicular to the line; i.e., the line is orthogonal to both normal vectors. Thus, the line is in the same direction as n n 2. n n 2 <,, > <, 2, 3 > i j k 2 3 i(3 ( 2 j(3 + k( 2 5i 2j 3k

3 MA 35 Fall 27 Exam # Review Solutions 3 x 5 y 2 z 3 9. Let r(t sin(2ti + 3tj + cos(2tk where π t π. Find r (t, r (t, T( and r (t r (t. r (t 2 cos(2ti + 3j 2 sin(2tk r (t 4 sin(2ti + j 4 cos(2tk T( r ( r (, r ( < 2, 3, >, r ( T(,, 3 3 i j k r (t r (t 2 cos(2t 3 2 sin(2t 4 sin(2t 4 cos(2t i( 2 cos(2t j( 8 cos 2 (2t 8 sin 2 (2t + k(2 sin(2t 2 cos(2ti + 8j + 2 sin(2tk r (t 2 cos(2ti + 3j 2 sin(2tk r (t 4 sin(2ti + j 4 cos(2tk 2 3 T(,, r (t r (t 2 cos(2ti + 8j + 2 sin(2tk 3 3. Let y x 2 be a parabola in the xy-plane parametrized by r(t < t, t 2, >. Find the unit tangent vector at the origin. r (t <, 2t, > r ( <,, > r ( T( <,, >. Determine whether the following curves are smooth. (a r(t < t 3, t 4, t 5 > r (t < 3t 2, 4t 3, 5t 4 > is continuous for any t when t r(t is not smooth (b r(t < t 3 + t, t 4, t 5 > r (t < 3t 2 +, 4t 3, 5t 4 > is continuous for any t for any t r(t is smooth

4 MA 35 Fall 27 Exam # Review Solutions 4 2. Evaluate the integrals. ( 4 (a + t 2 j + 2t + t 2 k dt ( 4 + t 2 j + 2t ( ( + t 2 k 4 dt + t 2 dt 2t j + + t 2 dt k ( 4 tan t ( j + ln( + t 2 k πj + (ln 2k (b ((cos πti + (sin πtj + tk dt ( ((cos πti + (sin πtj + tk dt ( cos πt dt i + π sin πt, π cos πt, 2 t2 ( sin πt dt j + t dt k (c 4 ( ti + te t j + t 2 k dt 4 ( ti + te t j + ( 4 t 2 k dt ( 2 ( 4 t /2 dt i + ( 4 te t dt j + t 2 dt k 3 t3/2 4 i + ( te t 4 ( 4 + e t dt j + t 4 k ( i + ( 4e 4 + e (e t 4 3 j + ( 4 + k 4 3 i + ( 5e 4 + 2e j k 3. A particle moves along a curve defined by r(t < 2 cos(2t, 2 sin(2t, 3t >. (a Reparametrize the curve in terms of arc length s. r (t < 4 sin(2t, 4 cos(2t, 3 > r (t 6 sin 2 (2t + 6 cos 2 (2t s t r (u du s 5t t s 5 r(t(s r(s 2 cos t 5 du 5t ( 2s, 2 sin 5 ( 2s, 3s 5 5

5 MA 35 Fall 27 Exam # Review Solutions 5 (b Find a unit tangent vector to the curve in terms of s. r (s 4 ( 2s 5 sin 5 6 ( 2s r (s 25 sin T(s r (s r (s (c Show that the curve has constant curvature., 4 ( 2s 5 cos cos2, 3 5 ( 2s ( 2s 5 sin, 4 ( 2s 5 5 cos, κ T (s (using the defn, not our usual formula T (s 8 25 cos(2s/5, 8 25 sin(2s/5, 64 κ T (s 625 cos2 (2s/ sin2 (2s/ no matter what s is so curvature is constant 25 TYPO IN STATEMENT OF FOLLOWING PROBLEM: SHOULD BE 4. For the curve given by r(t 3 t3, 2 t2, t, find (a T(t, (b N(t, r (t t 2, t,, r (t t 4 + t 2 + T(t r (t r (t t 2, t, t4 + t 2 + N(t T T and using Rule 3 on page 859: [ T (t 4 + t 2 + /2] < t 2, t, > +(t 4 + t 2 + /2 < 2t,, > 2 (t4 + t 2 + 3/2 (4t 3 + 2t < t 2, t, > +(t 4 + t 2 + /2 < 2t,, > 2t 3 t (t 4 + t 2 + < 3/2 t2, t, > + t4 + t 2 + < 2t,, > (t 4 + t 2 + 3/2 (t 4 + t 2 + < 3/2 t3 + 2t, t 4 +, 2t 3 t > T t6 + 4t (t 4 + t t 2 + t 8 2t t 6 + 4t 4 + t 2 3/2 t8 + 5t (t 4 + t t 4 + 5t 2 + 3/2 N(t t8 + 5t 6 + 6t 4 + 5t 2 + < t3 + 2t, t 4 +, 2t 3 t >

6 MA 35 Fall 27 Exam # Review Solutions 6 (c B(t B T N t 2, t, t4 + t 2 + t8 + 5t 6 + 6t 4 + 5t 2 + < t3 + 2t, t 4 +, 2t 3 t > using Rule 2 on page 88: t4 + t 2 + t ( < t 2, t, > < t 3 + 2t, t 4 +, 2t 3 t > 8 + 5t 6 + 6t 4 + 5t 2 + t 2 + 6t + 2t 8 + 6t 6 + 2t 4 + 6t 2 + i j k t 2 t t 3 + 2t t 4 + 2t 3 t i( 2t4 t 2 + t 4 j( 2t 5 t 3 t 3 2t + k( t 6 + t 2 t 4 2t 2 t 2 + 6t + 2t 8 + 6t 6 + 2t 4 + 6t 2 + ( t4 t 2 i + (2t 5 + 2t 3 + 2tj + ( t 6 t 4 t 2 k t 2 + 6t + 2t 8 + 6t 6 + 2t 4 + 6t Let r(t sin(2ti + 3tj + cos(2tk where π t π be the position vector of a particle at time t. (a Show that the velocity and acceleration vectors are always perpendicular. v(t r (t < 2 cos(2t, 3, 2 sin(2t > a(t v (t < 4 sin(2t,, 4 cos(2t > a(t v(t 8 sin(2t cos(2t + 8 cos(2t sin(2t Since the dot product of the velocity and acceleration vectors is always, they are always perpendicular. (b Is there a time t when the position and velocity vectors are perpendicular? If so, find all such time(s t and the corresponding position(s of the particle. r(t v(t 2 sin(2t cos(2t + 9t 2 cos(2t sin(2t 9t t r( < sin(,, cos( ><,, > They are perpendicular only when t. The position of the particle at this time is <,, >.

7 MA 35 Fall 27 Exam # Review Solutions 7 6. Let y x 2 be a parabola in the xy-plane parametrized by r(t < t, t 2, >. Find the unit tangent vector, the unit normal vector and the binormal vector at the origin. r (t <, 2t, > r (t + 4t 2, T(t, + 4t 2 2t, + 4t 2 <,, > t, so T( <,, > T (t 2 ( + 4t2 3/2 2( + 4t 2 /2 2t 2 (8t, ( + 4t2 /2 (8t + 4t 2, 4t ( + 4t 2, 2( + 4t2 /2 2 8t (+4t 2 /2 4t 3/2 + 4t 2, ( + 4t 2, 2( + 4t 2 8t 2, 3/2 ( + 4t 2 3/2 4t ( + 4t 2, 2 3/2 ( + 4t 2, T ( <, 2, >, T ( /2 N( T ( T N(t <,, > ( i j k B( T( N( i j + k k B( <,, > 7. Let C be the circle of radius 3 centered at the origin in the xy-plane. (a Find a parametrization for C in the form x f(t, y g(t for t in some interval and use it to obtain the position vector r(t for C. x 2 + y x 3 cos t, y 3 sin t, t [, 2π] r(t < 3 cos t, 3 sin t > (b Give the arc length parametrization r(s of this curve, starting at the point (3,. (3, t r (t < 3 sin t, 3 cos t > r (t 9 sin 2 t + 9 cos 2 t 3 s t t s/3 r (u du t 3 du 3t r(s < 3 cos(s/3, 3 sin(s/3 >

8 MA 35 Fall 27 Exam # Review Solutions 8 (c Find the curvature κ of C. κ T (t r (t T(t < 3 sin t, 3 cos t > 3 < sin t, cos t > T (t < cos t, sin t > T (t cos 2 t + sin 2 t, r (t 3 κ 3 OR κ T (s r (s < sin(s/3, cos(s/3 > r (s sin 2 (s/3 + cos 2 (s/3 T(s < sin(s/3, cos(s/3 > T (s 3 cos(s/3, 3 sin(s/3 κ 9 cos2 (s/3 + 9 sin2 (s/ A moving particle starts at an initial position r( (4, 2, with initial velocity v( 2i + 3j k. Its acceleration is a(t e 2t j 2k. (a Find the velocity vector v(t for the particle. v(t a(t dt + C, C < c, c 2, c 3 >, 2 e2t, 2t + < c, c 2, c 3 > v( < 2, 3, > c, 2 + c 2, c 3 c 2 c 2 5 2, c 3 v(t 2, e 2t , 2t (b Find the position vector r(t for the particle. r(t v(t dt + D, D < d, d 2, d 3 > e 2t r(t 2t + d, t + d 2, t 2 t + d 3 r( < 4, 2, > d, 4 + d 2, d 3 d 4 d e 2t r(t 2t + 4, t + 7 4, t2 t d 3