MATH 255 Applied Honors Calculus III Winter Midterm 1 Review Solutions

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1 MATH 55 Applied Honors Calculus III Winter 11 Midterm 1 Review Solutions 11.1: #19 Particle starts at point ( 1,, traces out a semicircle in the counterclockwise direction, ending at the point (1,. 11.1: #1 Particle starts at point (, 3, traces out an ellipse in the clockwise direction, ending at the same point it started. 11.1: #35 (a x = a cos t; y = b sin t. (b See figure below. (c Graph stretches/contracts in y-direction as b increases/decreases. 11.: #31 a A = = ab a π a ydx = a (1 cos(θdθ = abπ. b sin θ( a sin θdθ = ab π sin θdθ 11.: #53 We use e = a b. According to arc length formula, we have c π π L = a cos θ + b sin θ dθ = a cos θ + b a sin θ dθ π π ( = a 1 sin θ + b a sin θ dθ = a 1 1 b sin θ dθ a π = a 1 e sin θ dθ. By the symmetry sin θ = sin(π θ, we can replace the integral from to π with twice the integral from to π/, giving us the formula π/ L = 4a 1 e sin θ dθ. 1

2 11.3: #1 x = 3, just substitute x = r cos θ, obtaining r cos θ = : #7 (a θ = π/6 (b x = : #9 This is a circle which is traced out entirely when θ varies from to π. The area of the circle is A = 1 π r dθ = 1 π 9 cos θ dθ = 9 4 π (1 + cos(θdθ = 9π : #14 This is a curve which is traced out entirely when θ varies from to π. The area is given by A = 1 π = 4π + 1 r dθ = 1 π π (4 + 4 cos(θ + cos (θ dθ (1 + cos(4θdθ = 4π + π = 9π. Ch. 11 Review: #1 x = t + 4t, y = t, 4 t 1. Graph is shown below.

3 To write the Cartesian equation, solve the second equation for t, t = y, and substitute into the first equation, so that x = ( y + 4( y, which simplifies to x = y = 8y + 1. Ch. 11 Review: #15 x + y =. Substitute the usual polar expressions for x and y, which gives r(cos θ + sin θ = #19 (a We have to check that the point ( x 1 +x, y 1+y, z 1+z is on the line segment between P1 and P, and that is it equidistant from P 1 and P. The line segment from P 1 to P is parametrized by the equation < x, y, z >=< tx 1 + (1 tx, ty 1 + (1 ty, tz 1 + (1 tz >, and thus the point ( x 1 +x, y 1+y, z 1+z corresponds to the point t = 1 and is indeed on the line segment. To check that it is equidistant from P 1 and P, we simply use the distance formula, obtaining ( ( x1 + x dist P 1,, y 1 + y, z 1 + z (x1 x = + (y 1 y + (z 1 + z (( x1 + x = dist, y 1 + y, z 1 + z, P. Let me remark that we also could have done a geometric proof using similar triangles. (b ( dist A, B + C ( ( 4 = dist (1,, 3,, 1 +, = + ( 3 + = 5. ( ( dist B, A + C ( = dist (,, 5, (9 ( 3 = + ( dist C, A + B = dist ( (4, 1, 5, + 1 = ( + 1 (9 = =, 1 +, , +, #33 Traces for fixed y are closed discs of radius 3. Thus the graph is a closed disc of radius 3 in the xz-plane and its translates parallel to the y-axis. 13. #9 We have and F 1 =< 1 cos(135, 1 sin(135 >, F =< 1 cos(3, 1 sin(3 > F = F 1 +F =< 1 cos( cos(3, 1 sin( sin(3 >=< , 1 +6 >.

4 Thus the magnitude of F is given by ( F = The angle is given by and thus θ 76. tan θ = ( lb #31 Consider nort is the positive y-direction and east is the positive x-direction. The the woman s velocity vector is given by v =<, > + < 3, >=< 3, > and her speed is the magnitude of the velocity, which is v = ( 3 + = 493. mph. Consider her direction relative to north. The cosine of the angle between her direction and north is given by <, > < 3, > cos θ = = Taking inverse cosine gives θ 7.77, so her direction is 7.77 west of north #1 (a scalar vector is meaningless (b scalar times scalar has meaning (c scalar times scalar has meaning (d vector vector has meaning (e scalar + vector is meaningless (f scalar vector is meaningless unless the product is interpreted as a scalar product not as a dot product #14 u v = u v cos 6 = 1 u w = u w cos 1 = #6 b should solve 6b + b 3 + b =, or equivalently b 3 = 4b, thus we can have b =,, #39 comp a b = < 1,, 1 > < 1, 1, > = = 1. proj a b = (comp a b a a = 1 a = a =< 1,, 1 > #45 The displacement vector is D <, 6, 15 > thus the work done is W = F D =< 1, 18, 6 > <, 6, 15 >= = 38 J.

5 13.3 #49 The direction of the line is given by the displacement vector as we move from one point to another on the line. The displacement vector from the x-intercept (, c/b and the y-intercept ( c/a, is d 1 =< c/a, c/b >. From this, it is clear that the vector < a, b > is perpendicular to the line. Pick a point on the line, say the x-intercept, and consider the displacement vector from this point to the point P 1. If we use the x-intercept, this vector is d =< x 1, y + c/b >. The distance from P 1 to the line is the absolute value of the component of the vector d in the direction perpendicular to the line, that is in the direction of the vector< a, b >. In symbols, this is dist(p 1, line = comp <a,b> < x 1, y 1 +c/b > = < a, b > < x 1, y 1 + c/b > = ax 1 + by 1 + c. a + b a + b Thus, the distance from the point (, 3 to the line 3x 4y + 5 = is 3( + 3( = #53 The displacement vector from the centriod to the vertex (1, 1, 1 is d 1 =< 1/, 1/, 1/ >. The displacement vector from the centriod to the vertex (, 1, is d =< 1/, 1/, 1/ >. Thus the cosine of the angles between these vectors is cos θ = < 1/, 1/, 1/ > < 1/, 1/, 1/ > ( 1/ + ( 1/ + ( 1/ (1/ + ( 1/ + (1/ = 1/4 3/4 = 1 3. Taking inverse cosine gives θ #9 (a vector dot vector makes sense (b vector cross scalar makes no sense (c vector cross vector makes sense (d scalar cross vector makes no sense (e scalar cross scalar makes no sense (f vector dot vector makes sense 13.4 #19 Let a =< a 1, a, a 3 > and b =< b 1, b, b 3 >. We have a b =< a b 3 a 3 b, a 3 b 1 a 1 b 3, a 1 b a b 1 >= < b a 3 b 3 a, b 3 a 1 b 1 a 3, b 1 a b a 1 > = b a #33 The vectors are coplanar if the scalar triple product a (b c =. We have a (b c = a < ( 1( ((3, ((7 (1(, (1(3 ( 1(7 >=<, 3, 1 > <,, 1 > = ( + 3( + 1 = #4 (a The distance from the point P to the plane is the absolute value of the component of the vector c in the direction of a vector perpendicular to the plane, for which we can use the vector a b. This is exactly the formula that is given. (b Using these points, we have a =< 1,, >, b =< 1,, 3 >, and c =< 1, 1, 4 >. Thus we see that a b =< (3, ( 1 ( 1(3, ( 1 ( 1 >=< 6, 3, >.

6 Thus, applying the above formula gives d = < 6, 3, > < 1, 1, 4 > = = #45 (a No. As a counterexample, take a =< 1,, >, b =<, 1, >, c =<,, 1 >. (b No. As a counterexample, take a =< 1,, >, b =<,, >, c =< 3,, >. (c Yes. If a b = a c, then a, b, and c are coplanar (otherwise the normal vectors a b and a c wouldn t point in the same direction. Furthermore, according to the right hand rule, these vectors are oriented in the same way with respect to the vector a. We are left to show that the vectors b and c have the same magnitude, and make the same angle with the vector a. Let θ 1 [, π] be the angle between a and b, and θ [, π] be the angle between a and c. According to the sine and cosine formulas for the cross and dot products, we have b cos θ 1 = c cos θ b sin θ 1 = c sin θ. Squaring both equations and adding them together gives b (cos θ 1 + sin θ 1 = c (cos θ + sin θ, which proves that b = c. This implies that cos θ 1 = cos θ and sin θ 1 = sin θ, which imply that θ 1 = θ #1 (a True, they must have the same direction. (b False, there are many directions perpendicular to a given line. (c True, they must have the same normal. (d True, they must have the same normal. (e False, consider a translation of any two lines in the plane. (f True, their direction is determined by the normal to the plane. (g False, the planes are determined by a normal to the given line, of which there are many. (h True, the direction of the line is the normal of the plane. (i True. (j False, they may be skew. (k True #1 L 1 has direction < 1,, 3 >. L has direction < 4, 3, >. These vectors are not multiples of each other, thus the lines are not parallel. To see if they intersect, we first write the parametric equations L 1 : x = t 1, y = t 1 + 1, z = 3t 1 + L : x = 4t + 3, y = 3t + 3, z = t + 1 and look for t 1 and t such that all coordinates coincide. That is, we look for a solution to the system of equations t 1 = 4t + 3, t = 3t + 3, 3t 1 + = t + 1. The first equation is already solved for t 1. Plugging this expression into the second equation gives ( 4t = 3t + 3, which has solution t = 4/5, which implies t 1 = 1/5. Plugging these into the third equation gives 3/5 + = 8/5 + 1, which is not a true statement. It follows that the lines do not intersect, and are therefore skew.

7 13.5 #3 The vector equation is The corresponding scalar equation is 13.5 #5 The vector equation is The corresponding scalar equation is <, 1, 5 > < x 6, y 3, z >=. (x 6 + (y 3 + 5(z =. < 1, 1, 1 > < x 1, y + 1, z 1 >=. (x 1 + (y + 3 (z 1 = #7 The vector equation is <, 1, 3 > < x, y, z >=. The corresponding scalar equation is x y + 3z = #31 Two vectors on the plane are < 1, 1, > and < 1,, 1 >, thus we can take the normal to be n =< 1, 1, > < 1,, 1 >=< ( 1( 1, 1( 1, 1( 1 >=< 1, 1, 1 >. The vector equation is then The corresponding scalar equation is or < 1, 1, 1 > < x 1, y 1, z >=. x 1 + y 1 + z = x + y + z = #35 The direction of the line is <, 5, 4 >, which is a vector on the plane. To get another vector on the plane, we take the displacement vector from any point on the line, say (4, 3, 7 to the point (6,,. This vector is <, 3, 9 >. Thus we can take as our normal vector n =<, 5, 4 > <, 3, 9 >=< 5( 9 4( 3, 4( ( ( 9, ( ( 3 5( >=< 33, 6, 4 >. The vector equation is then The corresponding scalar equation is < 33, 1, 4 > < x 6, y, z + >=. 33(x 6 1y 4(z + = #39 Substituting the parametric expressions for x, y, and z into the equation for the plane gives 3 t ( + t + (5t = 9. Solving this equation gives t = 1, and we find that the intersection occurs at (, 3, 5.

8 13.5 #55 From distance formula, we have Expanding, we get (x 1 + (y 1 + z = x + (y 1 + (z 1. x x y y z = x + y y z z + 1 which simplifies to Notice this is a plane. x = z #59 The normal vector to the plane is n =< 1, 1, 1 >. The direction of our line is orthogonal to both n and to the line given in the problem, which has direction < 1, 1, >. Thus the direction of our line is < 1, 1, 1 > < 1, 1, >=< 1( 1( 1, 1(1 1(, 1( 1 1(1 >=< 3, 1, >. The parametric equations for this line could therefore be x = 3t, y = t + 1, z = t #9 θ should be in the 4th quadrant, thus Thus, coordinates are ( (, 7π/4, 4. r = x + y = θ = arctan(y/x = arctan( 1 = π/ #1 We have ρ = x + y + z = =. This point is below the xy-plane, thus we should have φ (π/.π. This is given by the equation z = ρ cos φ, which is 1 = cos φ, thus φ = arccos( 1/ = 3π/4. θ should be on the negative y-axis in the xy-plane, thus we should take θ = π/. The spherical coordinates are therefore (, π/, 3π/ #3 We have r + z = ρ, which, in this case gives = ρ, or ρ =. To find φ, we can use z = ρ cos φ to get cos φ = 3/. Thus φ = arccos( 3/ = π/6. The spherical coordinates are thus (, π/6, π/ #49 (a z = x + y z = r

9 (b z = x + y ρ cos φ = ρ sin φ cos θ + ρ sin φ sin θ cos φ = ρ sin φ(cos θ + sin θ cos φ = ρ sin φ cos φ = ρ sin φ ρ = cos φ sin φ ρ = cot φ csc φ. Ch 13 Review #11 (a Two vectors in the plane are < 1,, 1 >, and <, 4, 3 >. Their cross product is perpendicular to the plane. < 1,, 1 > <, 4, 3 >=< ( 14, ( 1 1(3, 1(4 >=< 4, 3, 4 >. (b The area of the triangle ABC is half of the area of the parallelogram spanned by the vectors < 1, 3, 1 >, and <, 1, 3 >. This area is the length of the cross product computed above. Thus, the area of the triangle is ( = =. Ch 13 Review #13 Let F 1 be the force pulling up (in the picture, F be the force pulling down and F = F 1 + F be their resultant. Also, let a = F 1 and b = F. Since the vertical components of the forces should be of equal magnitude but in opposite directions, we should have a sin( = b sin(3, b or a =. Furthermore, since the sum of their horizontal components should be 55, we sin( should have a cos( + b cos(3 = 55. Substituting the first equation into the second gives b cot( + b = 55, or b Thus a sin( Ch 13 Review #41 This is a line in polar coordinates, a plane in cylindrical coordinates, and a half-plane in spherical coordinates. Ch 13 Review #43 Let s convert to Cartesian coordinates. ρ = 3 sec φ ρ = 3 cos φ ρ cos φ = 3 z = 3. We thus see that this is a plane parallel to the xy-plane. Additional problems #1 Let the vertices of the quadrilateral in R, taken in counterclockwise order, be V 1 = (x 1, y 1, V = (x, y, V 3 = (x 3, y 3, V 4 = (x 4, y 4. The quadrilateral formed by joining the midpoints of the segments of the quadrilateral V 1 V V 3 V 4 has vertices W 1 = ( x 1 +x, y 1+y,W = ( x +x 3, y +y 3, W3 = ( x 3 +x 4, y 3+y 4, W4 = ( x 4 +x 1, y 4+y 1. Consider the vectors W 1 W =<

10 x 3 x 1, y 3 y 1 > and W 3 W 4 =< x 1 x 3, y 1 y 4 >. Clearly these vectors point in opposite directions, and thus the line segments W 1 W and W 3 W 4 are parallel to each other. Similarly, the line segments W W 3 and W 4 W 1 are parallel to each other, thus the quadrilateral W 1 W W 3 W 4 is a parallelogram. Additional problems # The vectors <, a 3, a > and < a, a 1, > are both orthogonal to < a 1, a, a 3 > and do not point in the same direction. The vectors < x, y, z > which are orthogonal to the vector < a 1, a, a 3 > give a plane which passes through the origin. Additional problems #3 To find the distance from the origin to the plane, I fix a point on the plane, say (,, 3, and consider the displacement vector from the origin to this point, which is <,, 3 >. Then the distance from the origin to the plane is the absolute value of the scalar projection of this vector onto the normal vector, which we can read off of the equation for the plane as n =<, 1, 1 >. Thus the distance is d = <,, 3 > <, 1, 1 > = 3 6 = 3. To find the distance from the origin to the line, we need to identify the plane that contains the line and the origin. This plane contains the direction vector of the line, < 1, 1, >, and the displacement vector from the origin to any point on the line, say (1,, 1, so the displacement vector is < 1,, 1 >. The cross product of these two vectors gives the normal vector to the plane: n 1 =< 1, 1, > < 1,, 1 >=< ( 1( 1 ((, (1 1( 1, 1( ( 1(1 >=< 3, 3, 3 >. Consider the plane which also contains the line but is orthogonal to this plane. Its normal vector is orthogonal to both n 1 and to the line. Thus, its normal vector can be obtained as the cross product of n 1 and the directions vector of the line: n =< 3, 3, 3 > < 1, 1, >=< 3( 3( 1, 3(1 ( 3, ( 3( 1 3(1 >=< 9, 9, >. The plane with normal vector n passing through the given line is the plane containing the line but perpendicular to the plane which contains both the line and the origin. Thus, the distance from the origin to this plane is the same as the distance from the origin to the line. This distance is, of course, the absolute value of the scalar projection of the displacement vector from the origin to an arbitrary point in the plane, say (1,, 1, in the direction of the normal vector n. This is < 1,, 1 > < 9, 9, > d = = = 3. Additional problems #4 (1 z = x y z = r cos θ r sin θ z = r (cos θ sin θ z = r cos(θ

11 ( z = x y ρ cos φ = ρ sin φ cos θ ρ sin φ sin θ cos φ = ρ sin φ(cos θ sin θ cos φ = ρ sin φ cos(θ ρ = cos φ sin φ cos(θ Additional problems #5 (a False, the normal to the plane, < 1,, 1 > is also normal to the line, which has directions < 1, 1, 1 >. (b True, this means that the volume of the parallelpiped spanned by these vectors is zero, which can happen only if the vector are coplanar. (c False. If it were true then the equation sin t + cos t + t = 1 would be valid for all t, but it is only valid for t =. (d False. Consider as a counterexample v 1 =< 1,, >, v =<, 1, >, v 3 =<,, 1 >. (e True. They have the same direction and both pass through the point (1,,.