Math Fall First Eam Solutions October, Problem. ( pts.) Consider the points A(, 7, ), B(6,, 4), C(6,, 3), and D(7, 4, 4). (a pts.) Find the area of the parallelogram defined b AB and AC. First we calculate the vectors: AB = OB OA = 6,, 4, 7, = 4,, 3 AC = OC OA = 6,, 3, 7, = 4, 9, Then we use the propert of the cross-product: Area( AB, AC) = AB AC = 4 3 4 9 = 4 7, 8, 36 48 = 3, 4, = 9 + 6 + 44 = 69 = 3 (b pts.) Find the volume of the parallelipiped defined b AB, AC, and The vector AD =, 3, 3, so Volume( AB, AC, 3 3 AD) = 4 3 4 9 = AD ( AB AC) =, 3, 3 3, 4, = + 36 = 39 = 39. AD. (c pts.) Find the distance from the point D to the plane defined b AB and AC. Since Volume(parallelipiped) = Area(base) height, d(d, Π) = height = Volume( AB, AC, AD) Area( AB, AC) = 39 3 = 3 Alternativel, d(d, Π) = comp AB AC AD = ( AB AC) AD AB AC = 3, 4,, 3, 3 3, 4, = 39 3 = 3 Problem. ( pts.) Consider the vectors a =,, 3 and b = 3, 3, 4. (a pts.) Find the vector projection of b onto a. proj a b = (comp a b) a a = ( ) b a a a a = b a a a 6 + 3 + 9 a =,, 3 = 4 + + 9 4,, 3 = 9 7, 9 4, 7 4. (b pts.) Find the scalar projection of a onto b. comp b a = a b b = a b 9 = = 9. b 9 + 9 + 6 34
Problem 3. ( pts.) (a pts.) Find the parametric equation for the line of intersection of the planes 3 6 = 3 and + =. To write down a parametric equation of a line, we need a point on a line and a direction vector. Since our line is the line of intersection of two planes, its direction vector should be perpendicular to normal vectors of both planes. We use the cross-product again: v = 3 6 = 4,,. To find a point on the line, we set one of the coordinates to ero and solve for the other two. For eample, setting =, we get 3 = 3, =, and we see that the point P (,, ) is on both planes. So we get: (t) = + 4t (t) = (t) = (b pts.) Find an equation of the plane containing the lines l : r(t) =,, +, 3, t t t l : r(s) =,, +,, 3 s. Since both of the direction vectors v =, 3, and v =,, 3 must lie in the plane, we can obtain the normal vector for the plane b taking their cross-product: n = v v = 3 = 9, 6, + 3 = 8, 7,. 3 Since the point P (,, ) is obviousl on both lines (and therefore on the plane), we get the following equation: (c pts.) Find the point where the line 8, 7,, +, = (t) = t (t) = 3t (t) = + t 8 7 + = 8 7 ( ) + = intersects the plane + 3 = 6. Plugging in (t), (t), and (t) in the equation of the plane, we get ( t) (3t) + 3( + t) = 6 t 3t + 3 + 3t = t = 6 t =, so the line intersects the plane when value of the parameter t is. Plugging in, we get the point of intersection ( ( P ) (, ) (, )) ( 3 = P, 3, ). Problem 4. ( pts.) An unevenl heated plate has temperature T (, ) in C at the point (, ). If T (, ) = 3 and T (, ) = and T (, ) =, estimate the temperature at the point (.4,.97). Given the above information, the best we can do to estimate the change in temperature from the point (, ) to (.4,.97) is b using the lineariation; that is, instead of the actual change T we use the approimate change T +T. Since =.4 =.4 and =.97 =.3, we get T (.4,.97) T (.) + T + T = 3 +.4 (.3) = 3 +.6 +.4 = 36..
3 Problem. ( pts.) A to plane model was fling according to the rule r(t) = t, t 3 t, + 7t t + t 3 when at t = seconds its engine broke. Ignoring gravit, find when and where it hit the ground. In this problem it is convenient to introduce a new parameter s that is equal to the time after the engine broke, i.e., s = t. Then at the time t = (or s = ) seconds the plane is at the point r(t = ) =, 4, 3 and its instanteneous velocit at this point is r (t = ) =, 3t t, 7 t + 3t =, 8,. If we assume that no forces act on the model, t= the plane will continue to fl along the tangent line to the trajector at t = with the constant velocit r (t = ). The equation of this line is l(s) = r() + sr () =, 4, 3 +, 8, s = + s, 4 + 8s, 3 s, so the plane hits the ground = when s = 3 at the point l(3) =, 8, seconds after it started fling. Remark. To get a more realistic picture we need to take gravit into account. From F = ma = mg we see that the trajector of the plane after the engine broke should satisf the equations Then r (s) =,, g, r (s = ) = v(s = ) =, 8,, r(s = ) =, 4, 3 r (s) =, 8, gs, r(s) = + s, 4 + 8s, gs s + 3, and the plane hits the ground when gs + s 3 =. So s = ± + 6g and taking the positive root we get s = g 6g +.9 seconds (or.9 seconds after the plane started fling), and at this moment the plane will be at g r(.9).9,.37,. 4 3 3
4 Problem 6. ( pts.) Consider the function f(, ) = and the point P (3, ). (a pts.) Find the gradient vector f(3, ). f = f, f =,, f(3, ) =,. (b pts.) Write down an equation of the tangent line to the level curve of f(, ) at the point P (3, ). Since the gradient vector f(p ) is perpendicular to the level curve (since the rate of change of f along the level curve is ero), the equation of the tangent line will be f(, 3) 3, = + = 3 + = 3 or + 6 =. (c pts.) Find the rate of change of f(, ) at P (3, ) in the direction of v =, 3. First, we need to find the unit vector u in the direction of v. Since v =, u = 3,. Then Problem 7. ( pts) (No partial credit) True or False? () For an two vectors a and b, proj a b = proj b a. D u f(3, ) = f(3, ) u =, 3, = 34. False. There are man reasons wh it is false, for eample, proj a b is a vector in the direction of a, and proj b a is a vector in the direction of b. () If comp a b =, then proj b a =. True. Since comp a b = a b a, comp a b = means that a and b are perpendicular, so proj b a =. (3) If comp a b =, then comp b a =. True for the same reasons as the previous question. (4) a (b c) = (a b) c. False. The epression on the left is a number, the epression on the right is meaningless, since a b is a number, and ou can t take a cross-product of a number and a vector. () a (b c) = (a b) c. True. Both are equal to det(a, b, c) = vol(a, b, c) (up to a ± sign). (6) Two lines perpendicular to a given plane must be parallel. True. The direction vectors for both lines are (up to a multiple) equal to the normal vector of the plane, v = λn, v = µn, so v = λ µ v. (7) Two lines parallel to a given plane must be parallel. False. The direction vectors must be both perpendicular to the normal vector of the plane, n v = n v =, but this does not mean that v v = (take two intersecting lines in the -plane, for eample). (8) Two planes parallel to a given line must be parallel. False. The normal vectors for the planes must be both perpendicular to the direction vector of the line, n v = n v =, but this does not mean that n n = (for eample, both and planes are paralell to the line l(t) =,, t ). (9) Two planes perpendicular to a given line must be parallel. True. The direction vector of the line can be taken as a normal vector for both planes.
() An object moving with constant speed has ero acceleration. False. Acceleration corresponds to change in velocit. For motion in three-dimensional (or even two-dimensional) space, velocit is a vector, and so it has both magnitude (called speed) and direction. So if we keep the speed constant, we can still change the direction of motion producing acceleration. For eample, and object moving with constant speed on a circular trajector will have acceleration pointing towards the center of the circle. Problem 8. ( pts.) (No partial credit) (a 8 pts.) Match the graphs of parametric curves with their parametric equations. (I).... We see that this curve lies above the circle in the -plane, and the motion is periodic in as well, so it must be r 3 (t) = cos t, sin t, sin 6t. Note that si peaks correspond to the fact that the curve oscillates in the -direction si times faster than in the and directions. (II) 3 3 We see that this curve lies above the circle in the -plane, so it must be r (t) = cos t, t, sin t. The t term makes this curve look different from the heli though. (III).3.3 This curve stas near the -ais most of the time, and this is due to the eponential deca in the and components of the corresponding equation r 4 (t) = t, 3te t, t e t........ (IV) This curve lies above the line = in the plane, so it must be r (t) = ( + sin t)t, ( + sin t)t, ( + cos t)t. 8 6 4 (b pts.) What is the maimum value of achieved on curve (I)? Either from the plot or from the equation we can see that the maimum value of achieved on this curve is.