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Solução_Calculo_Stewart_6e

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F.<br />

TX.10<br />

CHAPTER 14 REVIEW ET CHAPTER 13 ¤ 157<br />

39. The tangential component of a is the length of the projection of a onto T,sowesketch<br />

the scalar projection of a in the tangential direction to the curve and estimate its length to<br />

be 4.5 (using the fact that a has length 10 as a guide). Similarly, the normal component of<br />

a is the length of the projection of a onto N,sowesketchthescalarprojectionofa in the<br />

normal direction to the curve and estimate its length to be 9.0. Thus a T ≈ 4.5 cm/s 2 and<br />

a N ≈ 9.0 cm/s 2 .<br />

41. If the engines are turned off at time t, then the spacecraft will continue to travel in the direction of v(t), so we need a t such<br />

that for some scalar s>0, r(t)+s v(t) =h6, 4, 9i. v(t) =r 0 (t) =i + 1 t j + 8t<br />

(t 2 +1) k ⇒<br />

2<br />

<br />

r(t)+s v(t) = 3+t + s, 2+lnt + s t , 7 − 4<br />

t 2 +1 + 8st <br />

⇒ 3+t + s =6 ⇒ s =3− t,<br />

(t 2 +1) 2<br />

so 7 − 4 8(3 − t)t<br />

+<br />

t 2 +1 (t 2 +1) =9 ⇔ 24t − 12t2 − 4<br />

=2 ⇔ t 4 +8t 2 − 12t +3=0.<br />

2 (t 2 +1) 2<br />

It is easily seen that t =1is a root of this polynomial. Also 2+ln1+ 3 − 1<br />

1<br />

=4,sot =1is the desired solution.<br />

14 Review ET 13<br />

1. A vector function is a function whose domain is a set of real numbers and whose range is a set of vectors. To find the derivative<br />

or integral, we can differentiate or integrate each component of the vector function.<br />

2. Thetipofthemovingvectorr(t) of a continuous vector function traces out a space curve.<br />

3. The tangent vector to a smooth curve at a point P with position vector r(t) is the vector r 0 (t). The tangent line at P is the line<br />

through P parallel to the tangent vector r 0 (t). The unit tangent vector is T(t) = r0 (t)<br />

|r 0 (t)| .<br />

4. (a) – (f ) See Theorem 14.2.3 [ET 13.2.3].<br />

5. Use Formula 14.3.2 [ ET 13.3.2], or equivalently, 14.3.3 [ ET 13.3.3].<br />

6. (a) The curvature of a curve is κ =<br />

dT<br />

ds where T is the unit tangent vector.<br />

(b) κ(t) =<br />

T0 (t)<br />

r 0 (t) <br />

(c) κ(t) = |r0 (t) × r 00 (t)|<br />

|f 00 (x)|<br />

|r 0 (t)| 3 (d) κ(x) =<br />

[1 + (f 0 (x)) 2 ] 3/2<br />

7. (a) The unit normal vector: N(t) = T0 (t)<br />

. The binormal vector: B(t) =T(t) × N(t).<br />

|T 0 (t)|<br />

(b) See the discussion preceding Example 7 in Section 14.3 [ ET 13.3].

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