Solução_Calculo_Stewart_6e

F.
TX.10
PROBLEMS PLUS
1. (a) r(t) =R cos ωt i + R sin ωt j ⇒ v = r 0 (t) =−ωR sin ωt i + ωR cos ωt j, sor = R(cos ωt i +sinωt j) and
v = ωR(− sin ωt i +cosωt j). v · r = ωR 2 (− cos ωt sin ωt +sinωt cos ωt) =0,sov ⊥ r. Sincer points along a
radius of the circle, and v ⊥ r, v is tangent to the circle. Because it is a velocity vector, v points in the direction of motion.
(b) In (a), we wrote v in the form ωR u,whereu is the unit vector − sin ωt i +cosωt j. Clearly |v| = ωR |u| = ωR. At
speed ωR, the particle completes one revolution, a distance 2πR,intimeT = 2πR
ωR = 2π ω .
(c) a = dv
dt = −ω2 R cos ωt i − ω 2 R sin ωt j = −ω 2 R(cos ωt i +sinωt j),soa = −ω 2 r. This shows that a is proportional
to r and points in the opposite direction (toward the origin). Also, |a| = ω 2 |r| = ω 2 R.
(d) By Newton’s Second Law (see Section 14.4 [ET 13.4]), F = ma,so|F| = m |a| = mRω 2 = m (ωR)2
R
= m |v|2
R .
3. (a) The projectile reaches maximum height when 0= dy
dt = d dt [(v 0 sin α)t − 1 2 gt2 ]=v 0 sin α − gt; thatis,when
t = v
0 sin α
v0 sin α
and y =(v 0 sin α)
− 1 2
g
g 2 g v0 sin α
= v2 0 sin 2 α
. This is the maximum height attained when
g
2g
the projectile is fired with an angle of elevation α. This maximum height is largest when α = π . In that case, sin α =1
2
and the maximum height is v2 0
2g .
(b) Let R = v0
2 g. We are asked to consider the parabola x 2 +2Ry − R 2 =0which can be rewritten as y = − 1
2R x2 + R 2 .
The points on or inside this parabola are those for which −R ≤ x ≤ R and 0 ≤ y ≤ −1
2R x2 + R 2 .
fired at angle of elevation α, the points (x, y) along its path satisfy the relations x =(v 0 cos α) t and
y =(v 0 sin α)t − 1 2 gt2 ,where0 ≤ t ≤ (2v 0 sin α)/g (as in Example 14.4.5 [ET 13.4.5]). Thus
|x| ≤
v0 cos α 2v0 sin α =
g v2 0
g sin 2α ≤
v2 0
g = |R|. This shows that −R ≤ x ≤ R.
For t in the specified range, we also have y = t v 0 sin α − 1 2 gt = 1 2 gt 2v0 sin α
g
− t ≥ 0 and
When the projectile is
x
y =(v 0 sin α)
v 0 cos α − g 2 x
g
1
= (tan α) x −
2 v 0 cos α
2v0 2 cos2 α x2 = −
2R cos 2 α x2 + (tan α) x. Thus
−1
y −
2R x2 + R
−1
=
2 2R cos 2 α x2 + 1
2R x2 +(tanα) x − R 2
= x2
1 − 1
+ (tan α) x − R 2R cos 2 α
2 = x2 (1 − sec 2 α)+2R (tan α) x − R 2
2R
= −(tan2 α) x 2 +2R (tan α) x − R 2
2R
=
− [(tan α) x − R]2
2R
We have shown that every target that can be hit by the projectile lies on or inside the parabola y = − 1
2R x2 + R 2 . 161
≤ 0