# A 0 x c 2 x 2 b 1 x 1 d x let

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Unformatted text preview: = 0. By the quadratic formula 2y x= (c) 2(a + b) ± 4(a + b)2 − 4 · 3ab 1 = a + b ± (a2 + b2 − ab)1/2 . 6 3 y = ± x(x − a)(x − b). The square root is only deﬁned for nonnegative arguments, so it is necessary that all three of the factors x, x − a, x − b be nonnegative, or that two of them be nonpositive. If, for example, 0 &lt; a &lt; b then the function is deﬁned on the disjoint intervals 0 &lt; x &lt; a and b &lt; x &lt; +∞, so there are two parts. Exercise Set 4.3 35. 110 (a) (b) y ±1.1547 2 -2 2 x 2 (c) dy y − 2x dy dy dy − y + 2y = 0. Solve for = . If = 0 then dx dx dx 2y − x dx 2 y − 2x = 0 or y = 2x. Thus 4 = x2 − xy + y 2 = x2 − 2x2 + 4x2 = 3x2 , x = ± √ . 3 Implicit diﬀerentiation yields 2x − x 36. √ du 1 −1/2 du 1 −1/2 u u +v = −√ = 0 so 2 dv 2 dv v 37. 4a 3 da da da 2t3 + 3a2 da − 4t3 = 6 a2 + 2at , solve for to get =3 dt dt dt dt 2a − 6at 38. 1 = (cos x) dx 1 dx so = = sec x dy dy cos x 39. 2a2 ω b2 λ dω dω + 2b2 λ = 0 so =− 2 dλ dλ aω 40. Let P (x0 , y0 ) be the required point. The slope of the line 4x − 3y + 1 = 0 is 4/3 so the slope of the tangent to y 2 = 2x3 at P must be −3/4. By implicit diﬀerentiation dy/dx = 3x2 /y , so at P , 2 3x2 /y0 = −3/4, or y0 = −4x2 . But y0 = 2x3 because P is on the curve y 2 = 2x3 . Elimination of y0 0 0 0 4 3 3 gives 16x0 = 2x0 , x0 (8x0 − 1) = 0, so x0 = 0 or 1/8. From y0 = −4x2 it follows that y0 = 0 when 0 x0 = 0, and y0 = −1/16 when x0 = 1/8. It does not follow, however, that (0, 0) is a solution because dy/dx = 3x2 /y (the slope of the curve as determined by implicit diﬀerentiation) is valid only if y = 0. Further analysis shows that the curve is tangent to the x-axis at (0, 0), so the point (1/8, −1/16) is the only solution. 41. The point (1,1) is on the graph, so 1 + a = b. The slope of the tangent line at (1,1) is −4/3; use 2xy 2 4 dy =− 2 so at (1,1), − = − , 1 + 2a = 3/2, a = 1/4 and implicit diﬀerentiation to get dx x + 2ay 1 + 2a 3 hence b = 1 + 1/4 = 5/4. 42. Use implicit diﬀerentiation to get dy/dx = (y − 3x2 )/(3y 2 − x), so dy/dx = 0 if y = 3x2 √Substitute . √ this into x3 − xy + y 3 = 0 to obtain 27x6 − 2x3 = 0, x3 = 2/27, x = 3 2/3 and hence y = 3 4/3. 43. Let P (x0 , y0 ) be a point where a line through the origin is tangent to the curve x2 − 4x + y 2 + 3 = 0. Implicit diﬀerentiation applied to the equation of the curve gives dy/dx = (2 − x)/y . At P the slope of the curve must equal the slope of the line so 2 2 (2 − x0 )/y0 = y0 /x0 , or y0 = 2x0 − x2 . But x2 − 4x0 + y0 + 3 = 0 because (x0 , y0 ) is on the curve, 0 0 2 2 and elimination of y0 in the latter two equations gives x0 − 4x0 + (2x0 − x2 ) + 3 = 0, x0 = 3/2 which 0 √ 2 2 when substituted into y0 = 2x0 − x2 yields y0 = 3/4, √ y0 = ± 3/2. The slopes of the lines are so 0 √ √ √ (± 3/2)/(3/2) = ± 3/3 and their equations are y = ( 3/3)x and y = −( 3/3)x. 44. By implicit diﬀerentiation, dy/dx = k/(2y ) so the slope of the tangent to y 2 = kx at (x0 , y0 ) is k 2 (x − x0 ), or 2y0 y − 2y0 = kx − kx0 . k/(2y0 ) if y0 = 0. The tangent line in this case is y − y0 = 2y0 2 But y0 = kx0 because (x0 , y0 ) is on the curve y 2 = kx, so the equation of the tangent line becomes 2y0 y − 2kx0 = kx − kx0 which gives y0 y = k (x + x0 )/2. If y0 = 0, then x0 = 0; the graph of y 2 = kx has a vertical tangent at (0, 0) so its equation is x = 0, but y0 y = k (x + x0 )/2 gives the same result when x0 = y0 = 0. 111 Chapter 4 dy dy dt = . Use implicit diﬀerentiation on 2y 3 t + t3 y = 1 to get dx dt dx 1 dy dt dy 2y 3 + 3t2 y 2y 3 + 3t2 y =− = so =− . , but 2 + t3 dt 6ty dx cos t dx (6ty 2 + t3 ) cos t 45. By the chain rule, 46. 2x3 y 47. 2xy 48. 49. 3x2 y 2 dx dy dx dy dy + 3x2 y 2 + = 0, =− 3 dt dt dt dt 2x y + 1 dt dx dy 3 cos 3x − y 2 dx dy dx = y2 = 3(cos 3x) , = dt dt dt dt 2xy dt 4 1/3 4 x , f (x) = x−2/3 3 9 7 28 1/3 28 −2/3 x , f (x) = x (b) f (x) = x4/3 , f (x) = 3 9 27 (c) generalize parts (a) and (b) with k = (n − 1) + 1/3 = n − 2/3 (a) f (x) = y = rxr−1 , y = r(r − 1)xr−2 so 3x2 r(r − 1)xr−2 + 4x rxr−1 − 2xr = 0, 3r(r − 1)xr + 4rxr − 2xr = 0, (3r2 + r − 2)xr = 0, 3r2 + r − 2 = 0, (3r − 2)(r + 1) = 0; r = −1, 2/3 50. y = rxr−1 , y = r(r − 1)xr−2 so 16x2 r(r − 1)xr−2 + 24x rxr−1 + xr = 0, 16r(r − 1)xr + 24rxr + xr = 0, (16r2 + 8r + 1)xr = 0, 16r2 + 8r + 1 = 0, (4r + 1)2 = 0; r = −1/4 51. 52. 53. 54. We shall ﬁnd when the curves intersect and check that the slopes are negative reciprocals. For the intersection solve the simultaneous equations x2 + (y − c)2 = c2 and (x − k )2 + y 2 = k 2 to obtain x−k y−c 1 =− . cy = kx = (x2 + y 2 ). Thus x2 + y 2 = cy + kx, or y 2 − cy = −x2 + kx, and 2 x y dy x dy x−k Diﬀerentiating the two families yields (black) =− , and (gray) =− . But it was dx y−c dx y proven that these quantities are negative reciprocals of each other. dy dy Diﬀerentiating, we get the equations (black) x + y = 0 and (gray) 2x − 2y = 0. Th...
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