Differentials Differentials (Definitions) Recall that the derivative is defined by
If we drop the limit and assume that Dx is small we have:
we can rearrange this equation to get:
Applications Suppose that a die is manufactured so that each side is .5 inches .01 inches. Then its volume is V = x^{3} So that V' = 3x^{2} = 3(.5)^{2} = .75 and Dy @ (.75)(.01) = .0075 cu inches So that the volume of the die is in the range (.5)^{3} .0075 = .125 .0075 or between .1175 and .1375 cubic inches.
Example We can use differentials to approximate
We have f(x) = x^{1/2} Since f(1 + Dx)  f(1) @ f '(1) Dx We have f(1 + Dx) @ f'(1)Dx + f(1) f(1) = 1 f'(1) = 1/2 Dx = .01 we have f(1 + Dx) @ 1/2 (.01) + 1 = 1.005 (The true value is 1.00499)
Exercise: A spherical bowl is full of jellybeans. You count that there are 25 1 beans that line up from the center to the edge. Give an approximate error of the number of jelly beans in the jar for this estimate.
Example: The level of sound in decibels is equal to V = 5/r^{3} Where r is the distance from the source to the ear. If a listener stands 10 feet 0.5 feet for optimal listening, how much variation will there be in the sound? What is the relative and percent error.
Solution V' = 15r^{4} = 15/10,000 = 0.0015 V' D v = (0.0015)(.5) = 0.00075 V = 0.005 0.00075 We have a percent error of 0.00075 / 0.005 = 15%
Marginal Analysis We define the marginal revenue as the additional revenue from selling an additional unit of a product. If D x = 1 then the marginal revenue follows DR @ R'(x)D x.
Example Suppose that the demand equation for a bicycle is p = 1000  2x Use differentials to approximate the change in revenue as sales increase from 100 to 101 units. Solution: We have R = px = 1000x  2x^{2} R' = 1000  4x R'(100) = 600 Hence DR @ 600(1) = 600 Note that the true marginal revenue is R(101)  R(100) = 598
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