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Let f be a differentiable function on an interval (a, b) Let c (a, b) Then the slope of the tangent line to the graph of f at c is f (c )
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EXAMPLE 211
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Calculate the instantaneous velocity at time t = 5 of an automobile whose position at time t seconds is given by g(t) = t 3 + 4t 2 + 10 feet
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SOLUTION We know that the required instantaneous velocity is g (5) We calculate g (5) = lim g(5 + h) g(5) h 0 h [(5 + h)3 + 4(5 + h)2 + 10] [53 + 4 52 + 10] = lim h 0 h
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= lim
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115h + 19h2 + h3 h 0 h = lim 115 + 19h + h2 = lim
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= 115
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((125 + 75h + 15h2 + h3 ) + 4 (25 + 10h + h2 ) + 10) h (125 + 100 + 10) h
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We know that this last limit (the same as ( )) is the derivative of f at c We have learned the following:
CHAPTER 2 Foundations of Calculus
We conclude that the instantaneous velocity of the moving body at time t = 5 is g (5) = 115 ft/sec Math Note: Since position (or distance) is measured in feet, and time in seconds, then we measure velocity in feet per second EXAMPLE 212
Calculate the slope of the tangent line to the graph of y = f (x) = x 3 3x at x = 2 Write the equation of the tangent line Draw a gure illustrating these ideas
SOLUTION We know that the desired slope is equal to f ( 2) We calculate f ( 2) = lim f ( 2 + h) f ( 2) h 0 h [( 2 + h)3 3( 2 + h)] [( 2)3 3( 2)] = lim h 0 h [( 8 + 12h 6h2 + h3 ) + (6 3h)] + [2] = lim h 0 h 3 6h2 + 9h h = lim h 0 h = lim h2 6h + 9
= 9 We conclude that the slope of the tangent line to the graph of y = x 3 3x at x = 2 is 9 The tangent line passes through ( 2, f ( 2)) = ( 2, 2) and has slope 9 Thus it has equation y ( 2) = 9(x ( 2)) The graph of the function and the tangent line are exhibited in Fig 29 You Try It: Calculate the tangent line to the graph of f (x) = 4x 2 5x + 2 at the point where x = 2 EXAMPLE 213
A rubber balloon is losing air steadily At time t minutes the balloon contains 75 10t 2 + t cubic inches of air What is the rate of loss of air in the balloon at time t = 1
Foundations of Calculus
Fig 29
SOLUTION Let (t) = 75 10t 2 + t Of course the rate of loss of air is given by (1) We therefore calculate (1 + h) (1) (1) = lim h 0 h [75 10(1 + h)2 + (1 + h)] [75 10 12 + 1] = lim h 0 h 2 ) + (1 + h)] [66] [75 (10 + 20h + 10h = lim h 0 h 19h 10h2 = lim h 0 h = lim 19 10h
= 19 In conclusion, the rate of air loss in the balloon at time t = 1 is (1) = 19 ft 3 /sec Observe that the negative sign in this answer indicates that the change is negative, ie, that the quantity is decreasing You Try It: The amount of water in a leaky tank is given by W (t) = 50 5t 2 + t gallons What is the rate of leakage of the water at time t = 2 Math Note: We have noted that the derivative may be used to describe a rate of change and also to denote the slope of the tangent line to a graph These are really two different manifestations of the same thing, for a slope is the rate of change of rise with respect to run (see Section 14 on the slope of a line)
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