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22.2 Find y if y3 xy = 1
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Differentiation of (0), using the power chain rule for Dx y3 and the product rule for Dx (xy), gives 3y2 y (xy + y) = 0 3y2 y xy y = 0 (3y2 x)y y = 0 Next, differentiate (1), (3y2 x)Dx y + y Dx (3y2 x) y = 0 (3y2 x)y + y (6yy 1) y = 0 (3y2 x)y + y ((6yy 1) 1) = 0 (3y2 x)y + y (6yy 2) = 0 [by the product rule] [by the power chain rule] [factor y ] (2) (1)
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Now solve (1) for y , y = Finally, substitute into (2) and solve for y , (3y2 x)y + (3y2 x)3 y + (3y2 x) y 3y2 x y 3y2 x 6y2 2 =0 3y2 x 6y2 2 =0 3y2 x multiply by (3y2 x)2 a b c = b ca a y 3y2 x
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(3y2 x)3 y + y (6y2 2(3y2 x)) = 0 (3y2 x)3 y + y(6y2 6y2 + 2x) = 0 (3y2 x)3 y + 2xy = 0 y =
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22.3 If y is a function of x such that x 3 2xy + y3 = 8 and such that y = 2 when x = 2 [note that these values satisfy (0)], nd the values of y and y when x = 2.
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Proceed as in Problem 22.2. x 3 2xy + y3 = 8 Dx (x 3 2xy + y3 ) = Dx (8) 3x 2 2(xy + y) + 3y2 y = 0 3x 2 2xy 2y + 3y2 y = 0 Dx (3x 2 2xy 2y + 3y2 y ) = Dx (0) 6x 2(xy + y ) 2y + 3(y2 y + y (2yy )) = 0 6x 2xy 2y 2y + 3y2 y + 6y(y )2 = 0 Substitute 2 for x and 2 for y in (1), 12 4y + 4 + 12y = 0 Substitute 2 for x, 2 for y, and 1 for y in (2), 12 4y + 2 + 2 + 12y + 12 = 0 or 8y + 28 = 0 or y = 7 28 = 8 2 or 8y + 8 = 0 or y = 1 (2) (1)
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22.4 Let s = t 3 9t 2 + 24t describe the position s at time t of an object moving on a straight line. (a) Find the velocity and acceleration. (b) Determine when the velocity is positive and when it is negative. (c) Determine when the acceleration is positive and when it is negative. (d) Describe the motion of the object.
(a) v = ds = 3t 2 18t + 24 = 3(t 2 6t + 8) = 3(t 2)(t 4) dt dv = 6t 18 = 6(t 3) a= dt (b) v is positive when t 2 > 0 and t 4 > 0 or when t 2 < 0 and t 4 < 0; that is, when t > 2 and t > 4 or t < 2 and t < 4
which is equivalent to t > 4 or t < 2. v = 0 if and only if t = 2 or t = 4. Hence v < 0 when 2 < t < 4.
HIGHER-ORDER DERIVATIVES
[CHAP. 22
(c) a > 0 when t > 3, and a < 0 when t < 3. (d) Assuming that s increases to the right, positive velocity indicates movement to the right, and negative velocity movement to the left. The object moves right until, at t = 2, it is at s = 20, where it reverses direction. It then moves left until, at t = 4, it is at s = 16, where it reverses direction again. Thereafter, it continues to move right (see Fig. 22-2).
Fig. 22-2
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