sol12-3

# sol12-3 - c with respect to q = derivative of c c = 0 4 Â 2...

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SOLUTIONS CHAPTER 12.3 MATH 132 WI01 6. Proof. (a) s (1) = - 3 · 1 2 + 2 · 1 + 1 = - 3 + 2 + 1 = 0 (b) We also need s (1 . 25) = - 3 · (1 . 25) 2 + 2 · 1 . 25 + 1 = - 4 . 6875 + 2 . 5 + 1 = - 1 . 1875; average velocity is computed dividing diference in positions (that is, distance travelled) by diference in times (that is, time elapsed): average velocity = s (1 . 25) - s (1) 1 . 25 - 1 = - 1 . 1875 - 0 . 25 = - 1 . 1875 0 . 25 = - 4 . 75 (c) For velocity at a given t -value we need the derivative s 0 = ( - 3) · 2 t + 2 = - 6 t + 2 - and we just plug in t = 1: velocity at t = 1 = - 6 · 1 + 2 = - 6 + 2 = - 4 ± 12. Proof. The problem asks rate of change of A with respect to r ! this can be translated as derivative of A ! simple as that. A 0 = π · 2 r = 2 πr and plug in r = 3 A 0 (3) = 2 π · 3 = 6 π ± 22. Proof. Since c is the average cost , the total cost c equals c · q = (2+ 1000 q ) · q = 2 q + 1000. Marginal cost is the derivative of c , so it’s c 0 = 2 in which we have to plug in q = 235; but c 0 is constant! so anything we plug in it gives us 2! so the result is c 0 (235) = 2 ± Date : 01/21/2000. 1

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2 MATH 132 WI01 34. Proof. Again, notice the formulation of the question: rate of change of
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Unformatted text preview: c with respect to q = derivative of c ! c = 0 . 4 Â· 2 q + 4 = 0 . 8 q + 4 in which we plug in q = 2 and we get c (2) = 0 . 8 Â· 2 + 4 = 1 . 6 + 4 = 5 . 6 As for Î” c Î” q we donâ€™t use the derivative at all! ; what we need is Î” c = c (3)-c (2) = (0 . 4 Â· 3 2 + 4 Â· 3 + 5)-(0 . 4 Â· 2 2 + 5 Â· 2 + 5) = = 3 . 6 + 12 + 5-1 . 6-10-5 = 4 Î” q = 3-2 = 1 and so Î” c Î” q = 4 1 = 4 Â± 36. Proof. (a) rate of change of y with respect to x = derivative of y = y = f ( x ) = (4-2 x ) =-2 (b) relative rate of change= y y =-2 4-2 x (c) plug in x = 3 in the formula in (a): y (3) =-2 (itâ€™s a constant! so anything we plug in will give us-2) (d) plug in x = 3 in the formula in (b): y y (3) =-2 4-2 Â· 3 =-2 4-6 =-2-2 = 1 (e) just express the result in (d) in percentage form: 1 = 100 100 % = 100% Â±...
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sol12-3 - c with respect to q = derivative of c c = 0 4 Â 2...

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