102y06mt1

# 102y06mt1 - he line of the intersection of two planes is...

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Unformatted text preview: he line of the intersection of two planes is perpendicular to the planes' normal vectors n1 and n2 , and therefore parallel to n1 n2 . In other words, the line of intersection is a nonzero scalar multiple of n1 n2 . In our case, i j k 3 -6 -2 2 1 -2 n1 n2 = = 14i + 2j + 15k Now, we need to find a point on the line to write the equation of the line. To find a point on the line, we should take a point common to the two planes. So we substitute z = 0 in the plane equations and solve the equations for x and y simultaneously. 3x - 6y = 15 2x + y = 5 x = 3 y = -1 So we found the point (3, -1, 0). Hence the line is x(t) = 3 + 14t, y(t) = -1 + 2t, z(t) = 15t Note that (3, -1, 0) is not the only intersection point. You may find another intersection point and write the equation of the line according to this point. 3. If r(t) is a differentiable vector-valued function of t of constant length, then show that r(t)r (t). Solution: First consider that But we know that So d [ r(t) 2 ] = 0 dt dr = 0. In other words r(t)r (t). dt d dr dr dr r(t) r(t) = r(t) + r(t) = 2r(t) dt dt dt dt d d r(t) r(t) = [ r(t) 2 ] where r(t) dt dt 2 is a scalar. Hence r(t) 4. The vector-valued function r(t) is given by r(t) =< cos t, sin t, t >. (i) Find the arc length from t = 0 to t = 2. Solution: r(t) = < cos t, sin t, t > r (t) = < - sin t, cos t, 1 > r (t) = then and 2 (- sin t)2 + (cos t)2 + 1 = Let L be the arc length of r(t) from t = 0 to t = 2 2 L= 0 dr dt = dt 2 0 2dt = 2 2 0 dt = t=2 2t t=0 = 2 2 (ii) Find the principle unit tangent vector T . Solution: The principle unit tangent vector to the graph of r(t) at t is T (t) = Since r (t) =< - sin t, cos t, 1 > and r (t) = T (t) = 2 we obtain r (t) r (t) . - sin t cos t 1 , , 2 2 2 (iii) Find the principle unit normal vector N . Solution: The principle unit normal vector to the graph of r(t) at t is N (t) = T (t) T (t) 1 2 . Since T (t) = - cos t - sin t , ,0 2 2 and T (t) = cos2 t sin2 t + = 2 2 we obtain N (t) =< - cos t, - sin t, 0 >= - cos ti - sin tj B U Department of Mathematics Math 102 Calculus II Fall 2004 First Midterm Calculus archive is a property of Boazii University Mathematics Department. The purpose of this archive is to organise and centralise the distribution of the exam questions and their solutions. g c This archive is a non-profit service and it must remain so. Do not let anyone sell and do not buy this archive, or any portion of it. Reproduction or distribution of this archive, or any portion of it, without non-profit purpose may result in severe civil and criminal penalties. 1.) a) Sketch the region R that lies inside the curve r = 2 and outside the curve r = 2 - 2 cos . b) Find the area of the region R. Solution: a) The curve r = 2 is a circle of radius 2 with its center located at the origin. To draw the curve r = 2 - 2 cos we determine some points that this curve passes through such as: ( = 0, r = 0), ( = /2, r = 2), ( = , r = 4), ( = 3/2, r = 2), ( = 2, r = 0). These two curve intersect when 2 = 2 - cos , i.e, at = /2 and = 3/2. So we obtain, theta=pi/2 2 -4 -2 o -2 b) The area of the region R can be obtained by the integral /2 /2 -/2 A = -/2 1 2 [2 - (2 - 2 cos )2 ]d = 2 2 [2 cos - ( /2 = 2 -/2 1 + cos 2 )]d = 8 - 2 Note that since our figure is symmetric with respect to x-axis, this area can also be /2 calculated from A=2 0 1 [22 - (2 - 2 cos )2 ]d. 2 2.) Find an equation of the plane that passes through the points P1 (0, -3, 2) and P2 (1, 2, 3) and parallel to the line of intersection of the planes 2x + y - z = 1 and x - 2y + z = 7. Solution: The vector N1 =< 2, 1, -1 > is normal to the first plane and N2 =< 1, -2, 1 > is normal to the second plane. Therefore the line of intersection of these two planes is parallel to the vector R 2 theta=0 [2 cos - cos2 ]d V = N1 N2 = -i -3j -5 k. The vector P1 P2 =< 1, 5, 1 > lies on the required plane. Now, also the vector V can be carried onto this plane since this plane is parallel to the line of intersection. Thus, N3 = P1 P2 V = -22 i + 4 j + 2 k is a normal vector for the required plane. To write down the equation of this plane both of the points P1 and P2 can be used. Using P1 we get -22x + 4y + 2z + 8 = 0 3.) (a) Find an equation of the plane parallel to the plane 3x - y + 2z + 3 = 0 if the point (2,2,-1) is equidistant from both planes. Solution: Since the required plane is parallel to the plane 3x - y + 2z + 3 = 0 , < 3, -1, 2 > is a normal vector for both of them. Therefore the required plane has an equation of the form 3x - y + 2z + D = 0 where D is a real number. Now since the point (2,2,-1) is equidistant from both planes we have |3.2 - 1.2 + 2. - 1 + 3| 32 + (-1)2 + 22 = |3.2 - 1.2 + 2. - 1 + D| 32 + (-1)2 + 22 . This equation implies |D+2| = 5 which has two solutions D = 3 and D = -7. The D = 3 solution corresponds to the first plane, therefore the required plane has the equation: 3x - y + 2z - 7 = 0 (b) Consider the straight line through the point (3,2,3) and perpendicular to the plane given by 2x - y + 3z + 1 = 0. Find the coordinates of the point of the intersection of that line and that plane. Solution: The plane has a normal vector < 2, -1, 3 >. Since the required line is perpendicular to the plane, this vector is parallel to the line. Hence, the parametric equation of the line is: x=3+ 2t , y=2-t , z=3+3t , - < t < . To find the value of t at which the line and the plane intersect, we insert the above result into the plane equation: 2(3+2t) - (2-t) + 3(3+3t) +1 =0, which gives t = -1. The intersection point is obtained by putting t = -1 in the parametric equation of the line which gives its coordinates as (1,3,0). (c) Can a vector (whose initial point is at the origin) have direction angles 1 = /4, 2 = 3/4, 3 = /3 where 1 , 2 , 3 are the angles between the vector and x,y,z coordinates respectively? Solution: For a vector V whose direction angles are 1 , 2 , 3 we have V .i= cos 1 |V | , V .j= cos 2 |V | , V .k= cos 3 |V | , which leads to, V |V | = cos 1 i + cos 2 j + cos 3 k Since this is a unit vector, direction cosines should satisfy cos2 1 + cos2 2 + cos2 3 =1 In our problem this is not satisfied since, cos2 (/4) = cos2 (/4) = 1/2 and cos2 (/3) = 1/4. Therefore, such a vector does not exist. (d) Find equations in rectangular and cylindrical coordinates for the surface = 2 sin given in spherical coordinates. Solution: We first multiply the equation of this surface by which gives 2 = 2 sin . To convert this into rectangular coordinates we use 2 = x2 + y 2 + z 2 , sin = x2 + y 2 and obtain x2 + y 2 + z 2 = 2 x2 + y 2 (rectangular). Now we can convert the final expression into the cylindrical coordinates by applying x2 + y 2 = r2 which gives r2 + z 2 = 2r , r 0 (cylindrical). 4.) Let r(t)= sin 3ti +cos 3tj + 2tk be a vector valued function. (a) Find the unit tangent vector T(t) and the principal unit normal vector N(t). Solution: Let ' (prime) denote the derivative with respect to t. Then, 3 cos 3ti - 3 sin 3tj + 2k r(t) = |r(t) | 13 T (t) |T (t) | = - sin 3ti - cos 3tj T (t) = N (t) = (b) Find the curvature (t). Solution: |T (t) | 9 = |r(t) | 13 (t) = (c) Find the arc length parametrization of this curve by taking (0,1,0) as the reference point. Solution: The point (0,1,0) corresponds to t = 0. Taking this point as the reference we can find the arc lengt parametrization in the increasing t direction from the integral t s= 0 dr(u) du = 13t du This gives t = s/ 13. Therefore, r(s) = sin 3s 13 i + cos 3s 13 2s j+ k 13 (d) Let r(t) be the position vector of a particle at time t. Find the scalar tangential and normal components of acceleration. Solution: From above we have s = 13t and (t) = 9/13. 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Otherwise if z0 = 0, we have x0 = -4 , y0 x0 = -8 , x0 x2 = 0 -8 . Also, x0 y0 = -4 and 2x0 = y0 so x2 = -2. It follows that x4 = 16 so x0 = 2, and 0 0 the points on g(x, y, z) = 0 satisfying this are P3 (2, -2, 0) and P4 (-2, 2, 0). Therefore the points closest to the origin on the suface z 2 = xy + 4 are P1 and P2 . 4.) Find the area of the region shared by r = 2 + 2cos and r = 2 - 2cos. Solution: A=4 =2 1 2 0 /2 /2 r2 d = 2 0 /2 (2 + 2 cos )2 d /2 0 /2 4d - 2 0 8 cos d + 2 0 4 cos2 d We have the identity cos2 = cos 2 + 1 . Hence 2 /2 /2 /2 A = 4 - 16 sin 0 + 2 sin 2 0 + 4 0 = 6 - 16 ...
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## This note was uploaded on 04/30/2008 for the course C CMPE 150 taught by Professor Tuna during the Spring '08 term at Boğaziçi University.

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