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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 < a < b then the function is deﬁned on the disjoint intervals
0 < x < a and b < x < +∞, 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 xaxis 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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 Spring '14
 The Land

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