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MATHEMATICS 101 TUTORIAL 4 (SOLUTIONS TO DISCUSSIONS PROBLEMS) MODULE 5 – LINEAR AND MATRIX ALGEBRA 1 Exercise 13.5 (page 867) Equation of a plane Question 28 Find the equation of the plane through the point (–1, 6, –5) and parallel to the plane x y z 2 0. Solution The vector equation of the plane x y z 2 0 is r 1, 1, 1 = –2. Since the planes are parallel, they have the same normal, that is, the vector equation of the required plane is r 1, 1, 1 = 1, 6, 5 1, 1, 1 that is, r 1, 1, 1 = (1)(1) (6)(1) (5)(1) 0 . Its Cartesian equation is x y z 0 . x yz20 n=i+j+k x yz 0 Fig. 1 Courtesy Mr. Rajesh Gunesh Question 35 Find the equation of the plane that passes through the point (6, 0, –2) and contains the line x = 4 –2t, y = 3 + 5t, z = 7 + 4t. Solution n P (6, 0, –2) Q (4, 3, 7) Fig. 2 The vector equation of the line given by parametric equations x = 4 –2t, y = 3 + 5t, z = 7 + 4t is r = 4, 3, 7 t 2, 5, 4 . Choose a point Q, for example, (4, 3, 7) on the line. Then, PQ = 4, 3, 7 6, 0, 2 2, 3, 9 . It is clear that the normal n to the plane is orthogonal to both the line and PQ (see Fig. 2). The normal vector to the plane can be easily found by the vector product of the direction vector of the line, 2, 5, 4 , and 2, 3, 9 , given by i j k 2 5 4 2 3 9 = i 5 4 3 9 j 2 4 2 9 k 2 5 2 3 = 33i + 10j + 4k. The equation of the plane is thus r 33, 10, 4 6, 0, 2 33, 10, 4 190 . Its Cartesian equation is 33x 10 y 4 z 190 . Courtesy Mr. Rajesh Gunesh Question 42 Where does the line through (1, 0, 1) and (4, –2, 2) intersect the plane x + y + z = 6? Solution Let P and Q be the points (1, 0, 1) and (4, –2, 2) respectively. One vector equation of line PQ is therefore OP t PQ , given by r = 1, 0, 1 t 4, 2, 2 1, 0, 1 , that is, r = 1, 0, 1 t 3, 2, 1 , The parametric equations of the line are x = 1 + 3t, y = –2t, z = 1 + t. P (1, 0, 1) Q (4, –2, 2) Point of intersection x+y+z=6 r = 1, 0, 1 t 3, 2, 1 Fig. 3 The point of intersection is contained in the plane and, thus, has to satisfy the equations of both line and plane (see Fig. 3). Substituting the parametric equations of the line in the equation of the plane, we have (1 + 3t) + (–2t) + (1 + t) = 6 2t 2 6 and t = 2. The co-ordinates of the point of intersection are (1 + (3)(2), (–2)(2), 1 +(2)), that is, (7, –4, 3). Courtesy Mr. Rajesh Gunesh Question 55 Find an equation for the plane consisting of all points that are equidistant from the points (1, 1, 0) and (0, 1, 1). Solution Let A and B be the points (1, 1, 0) and (0, 1, 1) respectively. The plane consisting of all such points will (naturally) fall ‘between’ the points A and B in space (see Fig. 1.4). A (1, 1, 0) d P n d B (0, 1, 1) Fig. 4 Its normal vector will not only be parallel to the line segment AB but also contain the midpoint this line segment (think about it!). In that case, any point P on the plane would be equidistant to A and B. The direction vector of line segment AB, that is, the normal to the plane is 0, 1, 1 1, 1, 0 1, 0, 1 0 1 11 1 0 1 , , and the midpoint of AB is , 1, 2 2 2 2 1 2 . The vector equation of the plane is therefore r 1, 0, 1 1 2 , 1, 1 2 1, 0, 1 0 . Its Cartesian equation is –x + z = 0 or x = z. Courtesy Mr. Rajesh Gunesh Question 69 Show that the distance ax by cz d 2 0 is between the D parallel d1 d 2 a2 b2 c2 planes ax by cz d1 0 and . Solution 1 : ax by cz d1 0 P l1 D n O l2 Q 2 : ax by cz d 2 0 Fig. 5 The vector equations of planes 1 and 2 (see Fig. 5) are respectively r a, b, c d1 and r a, b, c d 2 Since the planes are parallel, they have the same normal vector. If O is the origin, then it should be clear that the distances l1 and l2 are respectively the scalar projections of OP and OQ onto the normal vector n, where P and Q are any two points on 1 and 2 respectively. Recall that these two points have position vectors conventionally denoted by r. Thus, r a, b, c a2 b2 c2 l1 d1 a2 b2 c2 and r a, b, c a2 b2 c2 l2 d2 a2 b2 c2 . Note that the second scalar product above is negative since, from the diagram, lies below the d1 d 2 origin. Therefore, D = l1 – (–l2) = l1 + l2 = . a2 b2 c2 Without any loss of cause, we could have taken both planes to be on the same side as the origin. The result would still have been the same. Courtesy Mr. Rajesh Gunesh MODULE 2 – INTRODUCTORY CALCULUS Exercise 2.5 (page 111) Continuity Question 18 Explain why the function 1 if x 1 f ( x) x 1 2 if x 1 is discontinuous at x = 1. Sketch the function. Solution (i) Note that lim f ( x) and lim f ( x) . Since the left-hand and right-hand limits x 1 x 1 are not the same, we conclude that lim f ( x) does not exist and deduce that the function x1 is discontinuous at x = 1. (ii) y 2 0 2 x Courtesy Mr. Rajesh Gunesh Exercise 3.1 (page 132) The derivative Question 4 If the tangent line to y f (x) at (4, 3) passes through the point (0, 2), find f (4) and f (4) . Solution It is too obvious that f (4) 3 since the point (4, 3) lies on the curve! f (4) is just the gradient of y f (x) at the point (4, 3), that is, the gradient of the tangent line. Gradient of tangent = Therefore, f (4) = 32 1 . 40 4 1 . 4 Question 19 (1 h)10 1 represents the derivative of some function f at some point a. State such an f h 0 h and the value of a. lim Solution Consider the derivative of the function f ( x) x10 from first principles. f ( x) lim h 0 f ( x h) f ( x ) ( x h)10 x10 . lim h 0 h h (1 h)10 1(10) (1 h)10 1 . lim h 0 h 0 h h At x = 1, it is clear that f (1) lim Therefore, f ( x) x10 and a = 1. Courtesy Mr. Rajesh Gunesh Exercise 3.2 (page 144) The derivative Question 25 Find the derivative of g ( x) 1 2 x using the definition of derivative. State the domain of the function and the domain of its derivative. Solution Differentiating from first principles, g ( x) lim h0 = lim h 0 1 2( x h) 1 2 x g ( x h) g ( x) lim h0 h h 1 2( x h) 1 2 x 1 2( x h) h 1 2( x h) 1 2 x 1 2x (rationalising numerator) = lim h 0 = lim h 0 = lim h 0 = 1 2( x h) (1 2 x) h 1 2( x h) 1 2 x 2h h 1 2( x h) 1 2 x 2 1 2( x h) 2 1 2x 1 2x 1 2x = lim h 0 2h h 1 2( x h) 1 2 x since h 0, h 0 2 1 2( x 0) = 2 2 1 2x 1 1 2x 1 2x . Df : { x | x 12 } Df’ : { x | x 12 } Courtesy Mr. Rajesh Gunesh Question 41 Show that the function f ( x) x 6 is not differentiable at 6. Find a formula for f and sketch its graph. Solution Using differentiation from first principles, f ( x) lim ( x h) 6 x 6 h 0 h Using the fact that x 6 ( x 6) for x > 6, we can use h small enough so that (x + h) – 6 > 0. In that case, ( x h) 6 x h 6 . Thus, f ( x) lim ( x h) 6 x 6 h 0 lim h 0 h ( x h) 6 ( x 6) h lim lim 1 1 . h 0 h h h 0 Therefore, f is differentiable for any x > 6. Similarly, x 6 ( x 6) for x < 6. Again, we can use h small enough so that (x + h) – 6 < 0. This time, f ( x) lim ( x h) 6 x 6 h 0 lim h 0 h ( x h) 6 ( x 6) h lim lim (1) 1 . h 0 h 0 h h Therefore, f is differentiable for any x < 6. However, since lim ( x h) 6 x 6 h 0 lim ( x h) 6 x 6 h 0 h h , f (6) does not exist. Thus, f is differentiable at all x except 6. Note that d x d dx dx x dxd x 2 2 1 2 1 2 x 2 12 (2 x) x x 2 1 2 x . x This result can generalised for the function f ( x) x a so that f ( x) ( x a) . xa Courtesy Mr. Rajesh Gunesh ( x a) for x a From definition, x a ( x a) for x a It is therefore easy to conclude that for x < a, f ( x) 1 and that, x > a, f ( x) 1 . The graph of y f (x) (a = 6) is given below. y 1 0 4 8 x –1 Fig. 3 Courtesy Mr. Rajesh Gunesh