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:

Dy @ f '(x) Dx

Applications

Suppose that a die is manufactured so that each side is .5 inches .01 inches.  Then its volume is

        V  =  x³ 

So that 

        V'  =  3x²   =  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.

Another Ball Example

 

Definition   The relative error is defined as the error/total, while the percent error is defined by error/total x 100%

Example:  

The level of sound in decibels is equal to

        V = 5/r³

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 - 2x²

        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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