Solução_Calculo_Stewart_6e

F.
TX.10
CHAPTER 10 REVIEW ¤ 445
the length of the orbit is
L =
2π
0
r2 +(dr/dθ) 2 dθ = a(1 − e 2 )
2π
0
√
1+e2 − 2e cos θ
(1 − e cos θ) 2 dθ ≈ 3.6 × 10 8 km
This seems reasonable, since Mercury’s orbit is nearly circular, and the circumference of a circle of radius a
is 2πa ≈ 3.6 × 10 8 km.
10 Review
1. (a) A parametric curve is a set of points of the form (x, y) =(f(t),g(t)), wheref and g are continuous functions of a
variable t.
(b)Sketchingaparametriccurve,likesketching the graph of a function, is difficult to do in general. We can plot points on the
curve by finding f(t) and g(t) for various values of t, either by hand or with a calculator or computer. Sometimes, when
f and g are given by formulas, we can eliminate t from the equations x = f(t) and y = g(t) to get a Cartesian equation
relating x and y. It may be easier to graph that equation than to work with the original formulas for x and y in terms of t.
2. (a) You can find dy
dy
as a function of t by calculating
dx dx = dy/dt
dx/dt
[if dx/dt 6= 0].
(b) Calculate the area as b
a ydx= β
α g(t) f 0 (t)dt [or α
β g(t) f 0 (t)dt if the leftmost point is (f(β),g(β)) rather
than (f(α),g(α))].
3. (a) L = β
(dx/dt)2 +(dy/dt)
α
2 dt = β
α [f 0
(t)] 2 +[g 0 (t)] 2 dt
(b) S = β
α 2πy (dx/dt) 2 +(dy/dt) 2 dt = β
α 2πg(t) [f 0 (t)] 2 +[g 0 (t)] 2 dt
4. (a) See Figure 5 in Section 10.3.
(b) x = r cos θ, y = r sin θ
(c) To find a polar representation (r, θ) with r ≥ 0 and 0 ≤ θ