14.5Ex25-26

# 14.5Ex25-26 - 556 C H A P T E R 14 C A L C U L U S O F VE C...

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556 CHAPTER 14 CALCULUS OF VECTOR-VALUED FUNCTIONS (ET CHAPTER 13) v 0 2 = gd 2 2cos 2 θ ( d tan h ) = 2 sec 2 2 ( d tan h ) For g = 32 ft / s 2 we get: v 2 0 = 16 d 2 sec 2 d tan h v 0 = 4 d sec d tan h . 25. At a certain moment, a moving particle has velocity v = h 2 , 2 , 1 i and a = h 0 , 4 , 3 i .Find T , N , and the decompo- sition of a into tangential and normal components. SOLUTION We go through the following steps: Step 1. Compute T and a T . The unit tangent is the following vector: T = v k v k = h 2 , 2 , 1 i q 2 2 + 2 2 + ( 1 ) 2 = 1 3 h 2 , 2 , 1 i (1) The tangential component of a = h 0 , 4 , 3 i is: a T = a · T = h 0 , 4 , 3 i · 1 3 h 2 , 2 , 1 i = 1 3 ( 0 + 8 3 ) = 5 3 Step 2. Compute a N and N .Since a N N = a a T T ,wehave: a N N = h 0 , 4 , 3 i − 5 3 · 1 3 h 2 , 2 , 1 i = h 0 , 4 , 3 i − ¿ 10 9 , 10 9 , 5 9 À = 1 9 h− 10 , 26 , 32 i (2) The unit normal N is a unit vector, therefore: a N =k a N N k= 1 9 q ( 10 ) 2 + 26 2 + 32 2 = 1 9 · 30 2 = 10 2 3 (3) We compute N , using (3) and (4): N = a N N a N = 1 9 h− 10 , 26 , 32 i 10 2 3 = 1 15 2 h− 5 , 13 , 16 i Step 3. Write the decomposition. Using (1)–(4) we obtain the following decomposition:
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