barcode in ssrs 2008 Copyright 2008, 1997, 1985 by The McGraw-Hill Companies, Inc. Click here for terms of use. in Visual Studio .NET

Creator QR in Visual Studio .NET Copyright 2008, 1997, 1985 by The McGraw-Hill Companies, Inc. Click here for terms of use.

Copyright 2008, 1997, 1985 by The McGraw-Hill Companies, Inc. Click here for terms of use.
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CHAP. 19]
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INSTANTANEOUS RATE OF CHANGE
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x = 0.1, then the new side x + x is 2.1, and the new volume is (2.1)3 = 9.261. So, 1.261 y = = 12.61 x 0.1
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y = 9.261 8 = 1.261,
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If x = 0.01, then the new side x + 8.120 601 8 = 0.120 601, and
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x is 2.01, and the new volume is (2.01)3 = 8.120 601. So, 0.120 601 y = = 12.0601 x 0.01
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x = 0.001, a similar computation yields y = 12.006 001 x
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Let us extend the result of the above example from y = x 3 to an arbitrary differentiable function y = f (x). Consider a small change x in the value of the argument x. The new value of the argument is then x + x, and the new value of y will be f (x + x). Hence, the change y in the value of the function is y = f (x + x) f (x)
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The ratio of the change in the function value to the change in the argument is y f (x + = x x) f (x) x
The instantaneous rate of change of y with respect to x is de ned to be lim y f (x + = lim x 0 x x) f (x) = f (x) x x close to 0, y/ x will be close
The instantaneous rate of change is evaluated by the derivative. It follows that, for to f (x), so that y f (x) x
(19.1)
Solved Problems
19.1 The weekly pro t P, in dollars, of a corporation is determined by the number x of radios produced per week, according to the formula P = 75x 0.03x 2 15 000 (a) Find the rate at which the pro t is changing when the production level x is 1000 radios per week, (b) Find the change in weekly pro t when the production level x is increased to 1001 radios per week.
(a) The rate of change of the pro t P with respect to the production level x is dP/dx = 75 0.06x. When x = 1000, dP = 75 0.06(1000) = 75 60 = 15 dollars per radio dx
INSTANTANEOUS RATE OF CHANGE
[CHAP. 19
(b) In economics, the rate of change of pro t with respect to the production level is called the marginal pro t. According to formula (19.1), the marginal pro t is an approximate measure of how much the pro t will change when the production level is increased by one unit. In the present case, we have P(1000) = 75(1000) 0.03(1000)2 15 000 = 75 000 30 000 15 000 = 30 000 P(1001) = 75(1001) 0.03(1001)2 15 000 = 75 075 30 060.03 15 000 = 30 014.97 P = P(1001) P(1000) = 14.97 dollars per radio which is very closely approximated by the marginal pro t, 15 dollars, as computed in part (a).
19.2 The volume V of a sphere of radius r is given by the formula V = 4 r 3 /3. (a) How fast is the volume changing relative to the radius when the radius is 10 millimeters (b) What is the change in volume when the radius changes from 10 to 10.1 millimeters
dV 4 3 = Dr r dr 3 When r = 10, dV = 4 (10)2 = 400 400(3.14) = 1256 dr (b) 4 4000 (10)3 = 3 3 4 4121.204 4 V (10.1) = (10.1)3 = (1030.301) = 3 3 3 V (10) = V = V (10.1) V (10) = 4121.240 4000 3 3 3.14 = (4121.204 4000) = (121.204) (121.204) 3 3 3 = 126.86 cubic millimeters The change predicted from (19.1) and part (a) is V dV dr r = 1256(0.1) = 125.6 cubic millimeters = 4 (3r 2 ) = 4 r 2 3
19.3 An oil tank is being lled. The oil volume V , in gallons, after t minutes is given by V = 1.5t 2 + 2t How fast is the volume increasing when there is 10 gallons of oil in the tank [Hint: To answer the question, how fast , you must always nd the derivative with respect to time.]
When there are 10 gallons in the tank, 1.5t 2 + 2t = 10 Solving by the quadratic formula, t= 2 4 4(1.5)( 10) 2 4 + 60 2 64 2 8 = = = =2 2(1.5) 3 3 3 or 10 3 or 1.5t 2 + 2t 10 = 0
CHAP. 19]
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