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1 Math171Spring2002-copyright Joe Kahlig Math 171 Exam I Answers February 14, 2002 √ − → → − − 1. → a =< 4, 2 > so a vector that is orthogonal is b =< −2, 4 >. Now | b | = 20 3 − −6 12 → Answer: √ b = √ , √ 20 20 20 → − 2. The vector that represents the path of the object is d =< 3, 1 >. Since → → − − Work = F · d = 15 + 6 = 21J 3. Remember that j =< 0, 1 >. so the vector projectin of < 5, 9 > onto j is < 0, 9 >. 4. Solve x = 1+cos(t) and y = 1+sin2 (t) for cos(t) and sin2 (t) and subsitute that into the equation sin2 (t) + cos2 (t) = 1. Answer: (y − 1) + (x − 1)2 = 1 or y = 1 + 2x − x2 5. (a) r(2) = 6i − 2j (b) If the particle goes through the point (4, 8), then there is some value of t such that 8 = t − 4 which implies that t = 12. Since 122 + 12 6= 4, the particle can not go through the point (4, 8). f (x + h) − f (x) . Look at the class notes and the h→0 h material posted on the web to find the solution. 6. The definition of the derivative is f ′ (x) = lim 7. If this function is to be continuous at x=1 and x=2, then we need: lim f (x) = lim f (x) = f (1) and lim f (x) = lim f (x) = f (2) x→1+ x→1− x→2+ x→2− This means that 2 = c + d and 4c + d = 8. Now solving for c and d gives c = 2 and d = 0. 8. (30 points) Find the exact values of the following limits: (a) lim t→1 * t2 + t − 2 3 , t − 4t t2 − t − 2 + = * t2 + t − 2 lim 2 , lim t3 − 4t t→1 t − t − 2 t→1 + = h0, −3i x+1 = +∞ notice that x = 3 is a vertical asymptote. so plug in values of x − 3) x→3 slightly above 3 to see which direction the function goes. (b) lim+ x2 (x (x − 4)(x + 2) x+2 6 x2 − 2x − 8 = lim = lim = = −3 x→4 (x − 4)(x − 6) x→4 x − 6 x→4 x2 − 10x + 24 −2 (c) lim 1 (d) lim cos(x) = 0 since the cos(x) function is always between -1 and 1 and the function x→∞ x gets closer to zero as x → ∞. (e) lim (x + x→−∞ p x2 + 4x + 1) = lim (x + x→−∞ p x2 √ (x − x2 + 4x + 1) √ + 4x + 1) ∗ = (x − x2 + 4x + 1) −4x − 1 x2 − (x2 + 4x + 1) √ √ = lim ∗ x→−∞ (x − x2 + 4x + 1) x→−∞ (x − x2 + 4x + 1) lim −4 − x1 −4 − x1 √ q = lim −1 x→−∞ (1 − √ x2 + 4x + 1) x→−∞ (1 + 1 + 4 + 2 lim x x 1 ) x2 1 x 1 x −4 − x1 √ = x→−∞ (1 − 1 x2 + 4x + 1) x = lim = 1 x −4 √ = −2 1+ 1 Math171Spring2002-copyright Joe Kahlig 9. (a) lim f (x) = 2 x→9 (b) lim f (x) = 4 x→−3+ (c) lim f (x) = −∞ x→7− (d) For what values of x is the function not differentiable? x = −3, 3, 7, 9 (e) For what values of x is the function not continuous? x = −3, 7, 9 (f) Approximate the value of f ′ (4). If not possible, explain why. Draw the tangent line in the graph and then approximate the slope. My answer is f ′ (4) ≈ 1.35 2