From Dif To Int

From Differentials to Integrals
José Luis Gómez-Muñoz Tecnológico de Monterrey Campus Estado de México

Differential Approximation to the Increment in Area of a Square
Calculate the “area increment” of a small square when its side increases from 8 cm to 10 cm. Use differentials to give an approximation to that increment. Give a geometrical interpretation of the differential approximation.José Luis Gómez-Muñoz 2

Initial Square
The initial square has a side of 8 cm.

José Luis Gómez-Muñoz

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Initial Area
The initial square has a side of 8 cm.

Therefore it has an area A equal to:

A=82 cm2 = 64 cm2.

There are 64 small squares in the image.

José Luis Gómez-Muñoz

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Final Square
The final square has a side of 10 cm.

José Luis Gómez-Muñoz

5Final Area
The final square has a side of 10 cm.

Therefore it has an area A f equal to: A f =102 cm2 = 100 cm2. There are 100 small squares in the image.

José Luis Gómez-Muñoz

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Area Increment
The area increment is:

A = A f –A = 102 -82 = 100-64 = 36 There are 36 small squares in the image.

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Approximation to the Increment
We will use differentialsto get an approximation to this increment:

A =A f –A = 102 -82 = 100-64 = 36

José Luis Gómez-Muñoz

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Area as a Function of Side
The independent variable is x The dependent variable is A

A(x)=x2

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Differential of the Independent Variable
The independent variable x (side length) has an increment:

∆x=dx

José Luis Gómez-Muñoz

10Differential of the Dependent Variable
The differential of the dependent variable A=x2 (area) is calculated:

José Luis Gómez-Muñoz

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Differential Approximation to the Increment
In this example A=x2, x=8 and x=2:

dA = (dA/dx) dA = 2x x

x

dA = 2(8) (2) dA = 32

José Luis Gómez-Muñoz

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Real Increment and Approximated Increment
The differential approximation to the incrementis: dA = 2x x

dA = 2(8) (2) dA = 32

The real increment in area is A =A f –A = 102 -82 = 100-64 = 36
José Luis Gómez-Muñoz 13

Interpretation of the Approximated Increment
The differential approximation to the increment is: dA = 2 x· x dA = 2 (8×2) dA = 32

The real increment in area is A =A f –A = 102 -82 = 100-64 = 36
José Luis Gómez-Muñoz 14

Geometrical Interpretation of theApproximated Increment
The differential approximation to the increment is: dA = 2 x· x dA = 2 (8×2) dA = 32

The differential approximation gives the 32 yellow squares

José Luis Gómez-Muñoz

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Exercise: Calculate the Differential from the Derivative
1. Calculate the “area increment” of a disk when its radius increases from 12 cm to 15 cm. 2. Use a differential to approximate thatincrement. 3. Give a geometrical interpretation of the differential approximation.

Answers: 1. A= Af -A=254.5

2. dA= 226.2 3. Area of a rectangle which has length equal to the initial perimeter of the disk, and width equal to the increment of the radius.

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Glass

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Glass Generated by the Rotation of a Function
Rotate a functionr(x) about the x-axis in order to generate a glass

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Glass Generated by the Rotation of a Function
Rotate a function r(x) about the x-axis in order to generate a glass

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Recipe for a Non-alcoholic Piña Colada
• Coconut cream to a height of 8 cm. • Pineapple juice to a height of 11 cm.

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Piña ColadaExercise: Calculate the Differential from Geometry (Not from a Derivative!)
• The radius r as a function of the height x is given by r=√x • Coconut cream to a height of 8 cm. • Pineapple juice to a height of 11 cm. • Use one differencial to estimate the amount of pineapple juice in cm3.

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Piña Colada Answer: Calculate the Differential from Geometry (Not from a...