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Unformatted text preview: CHAPTER 10 HOMEWORK SOLUTIONS Section 10.1 4. Let A = 2 i 7 j , B = i +6 j , C = √ 3 i π j . Evaluate A + B C = (2 i 7 j )+( i +6 j ) ( √ 3 i π j ) = (2+1 √ 3) i +( 7+6+ π ) j = (3 √ 3) i +( π 1) j 12. If A is the point ( 7 , 8) and B is the point(6 , 11) Then→ AB = B A = (6 ( 7)) i + (11 ( 8)) j = 13 i + 19 j 2 4 6 8 10 12 5 10 15 18. Given the vector→ PQ = 6 i 4 j and Q is the point (3 , 3) find the point P . Let P be the point ( a,b ). Then→ PQ = (3 a ) i + (3 b ) j = 6 i 4 j ⇒ 3 a = 6 and 3 b = 4 ⇒ a = 9 and b = 7 ⇒ P is the point (9 , 7) 24. Find a unit vector in the same direction as i + 3 j . Divide the vector’s components by the vector’s length:   i + 3 j  = p ( 1) 2 + (3) 2 = √ 10 ⇒ u = 1 √ 10 i + 3 √ 10 j 28. Find the unit vectors that are tangent and normal to the curve at the given point (four vectors in all). Then sketch the vectors and curve together. The curve for this problem is y = ∞ X n =0 x n n ! at the point (0 , 1). (Note that y = e x ) y ( x ) = ∞ X n =1 x n 1 ( n 1)! = ∞ X n =0 x n n ! = e x ⇒ y (0) = e = 1 1 2 CHAPTER 10 HOMEWORK SOLUTIONS This means the equation for the tangent line is y 1 = 1 * ( x 0), which gives us a tangent vector of i + j with unit vectors ± ( 1 √ 2 i + 1 √ 2 j ). The normal has the negative reciprocal of the tangent slope with equation y = x + 1. This gives two unit normal vectors ± ( 1 √ 2 i 1 √ 2 j ). 1.0 0.5 0.5 1.0 1.0 1.5 2.0 2.5 30. Find the unit vectors that are tangent and normal to the curve x 2 6 xy +8 y 2 2 x 1 = 0 at the point (1 , 1) (four vectors in all). Use implicit differentiation and solve for dy dx ⇒ d dx ( x 2 6 xy + 8 y 2 2 x 1) = 0 ⇒ 2 x 6 y 6 x dy dx + 16 y dy dx 2 = 0 ⇒ dy dx = 3 y x +1 8 y 3 x At the point (1 , 1) dy dx = 3 5 ⇒ 5 i + 3 j is a tangent vector to the curve with length:  5 i + 3 j  = √ 5 2 + 3 2 = √ 34 ⇒ the two unit tangent vectors are ± ( 5 √ 34 i +...
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This note was uploaded on 10/24/2010 for the course APPM 2350 at Colorado.
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 ADAMNORRIS

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