Solução_Calculo_Stewart_6e

F.
TX.10
CHAPTER 6 PROBLEMS PLUS ¤ 293
(c) kA √ h = π[f(h)] 2 dh
dt .Setdh dt = C: π[f(h)]2 C = kA √ h ⇒ [f(h)] 2 = kA
πC
is, f(y) =
√
h ⇒ f(h) =
kA
πC h1/4 ;that
kA
πC y1/4 . The advantage of having dh = C is that the markings on the container are equally spaced.
dt
13. The cubic polynomial passes through the origin, so let its equation be
y = px 3 + qx 2 + rx. The curves intersect when px 3 + qx 2 + rx = x 2
⇔
px 3 +(q − 1)x 2 + rx =0.Calltheleftsidef(x). Sincef(a) =f(b) =0,
another form of f is
f(x)=px(x − a)(x − b) =px[x 2 − (a + b)x + ab]
= p[x 3 − (a + b)x 2 + abx]
Since the two areas are equal, we must have a
f(x) dx = − b
f(x) dx ⇒
0 a
[F (x)] a 0 =[F (x)]a b
⇒ F (a) − F (0) = F (a) − F (b) ⇒ F (0) = F (b),whereF is an antiderivative of f.
Now F (x) = f(x) dx = p[x 3 − (a + b)x 2 + abx] dx = p 1
4 x4 − 1 3 (a + b)x3 + 1 2 abx2 + C,so
F (0) = F (b) ⇒ C = p 1
4 b4 − 1 3 (a + b)b3 + 1 2 ab3 + C ⇒ 0=p 1
4 b4 − 1 3 (a + b)b3 + 1 2 ab3 ⇒
0=3b − 4(a + b)+6a [multiply by 12/(pb 3 ), b 6= 0] ⇒ 0=3b − 4a − 4b +6a ⇒ b =2a.
Hence, b is twice the value of a.
15. We assume that P lies in the region of positive x. Sincey = x 3 is an odd
function, this assumption will not affect the result of the calculation. Let
P = a, a 3 . The slope of the tangent to the curve y = x 3 at P is 3a 2 ,andso
the equation of the tangent is y − a 3 =3a 2 (x − a) ⇔ y =3a 2 x − 2a 3 .
We solve this simultaneously with y = x 3 to find the other point of intersection:
x 3 =3a 2 x − 2a 3 ⇔ (x − a) 2 (x +2a) =0.SoQ = −2a, −8a 3 is
the other point of intersection. The equation of the tangent at Q is
y − (−8a 3 )=12a 2 [x − (−2a)] ⇔ y =12a 2 x +16a 3 . By symmetry,
this tangent will intersect the curve again at x = −2(−2a) =4a.Thecurveliesabovethefirst tangent, and
below the second, so we are looking for a relationship between A = a
−2a x 3 − (3a 2 x − 2a 3 ) dx and
B = 4a
−2a (12a 2 x +16a 3 ) − x 3 dx. We calculate A = 1
4 x4 − 3 2 a2 x 2 +2a 3 x a
= 3 −2a 4 a4 − (−6a 4 )= 27
4 a4 ,and
B = 6a 2 x 2 +16a 3 x − 1 4 x4 4a
−2a =96a4 − (−12a 4 ) = 108a 4 . We see that B =16A =2 4 A.Thisisbecauseour
calculation of area B was essentially the same as that of area A,witha replaced by −2a, so if we replace a with −2a in our
expression for A,weget 27 4 (−2a)4 =108a 4 = B.