Section 4 7 Differentials Definition of a Differential Let y = f (x) represent a function that is differentiable on an open interval containing x. The derivative of f (x) is written as f (x) = We call the separate and differentials The differential and is defined to be = f (x) Finding an expression for given f (x) Example 1! Example 2 If y = x 2 then find an expression for Step 1. Find = 6x Step 2. Solve for! If y = 4x 2x then find an expression for Step 1. Find = 12x2 6x 2 Step 2. Solve for = 6x = ( 12x 2 6x 2 ) Example! Example 4 If y = x then find an expression for Step 1. Find = 1 2 x Step 2. Solve for 1 = 2 x! If y = sin x then find an expression for Step 1. Find = cos x Step 2. Solve for = cos x Math 400 4 7! Page 1! 2016 Eitel
Given the change in x, Δ x, find the change in y Δy Given a function y = f (x) if we plug a value of x into f (x) we get a value for y. These x and y values represent an ordered pair ( x, y ) that is a point on the graph of f (x). If we move to the right of point A a distance Δ x and then move up a distance Δy to intersect the curve at point B the coordinates for ( ) point B are x + Δx, y + Δy The coordinates of Point B are expressed in terms of Δ x and Δy If we add a small value to x ( denoted by Δ x) we get a new x value denoted by x + Δ x. If we plug ( ) x + Δ x into f (x) we get a value for y denoted by f (x + Δx). The new ordered pair x + Δx, f (x + Δx) represents a Point B on the graph of f (x) The coordinates of Point B are expressed in terms of only Δ x Finding Δy for a given given f (x) the coordinates of Point A and the change in x Δ x The graph above shows that the change in y denoted by Δy is the difference in the y coordinates Δy = ( f (x + Δx) f (x)) Example if y = x 2 and Δ x =.1 then find Δy when x = 6. Δy = ( f (x + Δx) f (x)) ( ) Δy = f (6.1) f (6) Δy = (6.1) 2 (6) 2 Δy = 7.21 6 Δy = 1.21 This means that if the value of x changes by.1 then y changes by exactly 1.21 Math 400 4 7! Page 2! 2016 Eitel
The above example was done using algebra. The exact value for the change in y is found by finding f (x + Δx) and subtracting f (x). This finds the exact value for Δy but requires a calculator. In many cases the difference between f (x) and f (x + Δx) can be approximated by the use of differentials. This method gives an approximate answer but allows us to avoid the use of a calculator. That does not seem worth the effort but the differential is also an expression for the relative error in y. That expression can be of use in other settings. The change in y denoted by Δ y can be approximated by the differential for small values of x. The slope of the secant from Point A! The slope of the tangent line at Point A and Point B can be expressed as Δy.! Δx can be expressed as f (x) or. The value for the slope of the tangent line at Point A can differ from the value for the slope of the secant from A to B as shown in the diagram below.! The slope of the secant Δy Δx does not equal the slope of the tangent line The slope of the secant line at x is The slope of the tangent line at x is at Point A Δy Δx Math 400 4 7! Page! 2016 Eitel
The connection between Δy Δx and As the value for Δx approaches 0 the slope of the secant will approach the value of the value for the slope of the tangent line at Point A with a given value of x. lim Δx 0 Δy Δx = f (x) as Δx 0 we have Δy Δx f (x) This means that Δy Δx Δy Δx f (x) can be approximated by the derivative of the function for small values of x Δy Δx f (x) and = f (x) if we solve for = f (x) The change in y denoted by Δ y can be approximated by the differential for small values of x. Finding for a given f (x), a value for x and the change in x Δ x Example 1 If y = x 2, x = 6, = 1 then find find the value of. Step 1. Find = 2x Step 2. Solve for = 2x Step. Substitute = 2(6).1 = 1.2 This means that if the value of x changes from 6 to 6.1 then the change in y is approximately 1.2 Math 400 4 7! Page 4! 2016 Eitel
Example 2 If y = x 2 4x, x = 2 and the change in x is.01 then find the value of. y = x 2 4x, x = 2, =.01 Step 1. Find = 6x 4 Step 2. Solve for = ( 6x 4) Step. Substitute = ( 6(2) 4).01 =.08 If the value of x changes from 2 to 2.01 then the change in y is approximately.08 Example If y = 4x 2 + x, x = and the change in x is.1 then find the value of. y = 4x 2 + x, x =, =.1 Step 1. Find = 8x +1 Step 2. Solve for = ( 8x +1) Step. Substitute = ( 8() +1) (.1) = 2.5 If the value of x changes from to 2.9 then the change in y is approximately 2.5 Math 400 4 7! Page 5! 2016 Eitel
Comparing Δy and Example 1 f (x) = x 2 + Find Δy when x =1 and Δx =.01 f (1) = 4 Δy = f (x + Δx) f (x) Δy = f (1.01) f (1) Δy = (1.01) 2 + (4) Δy = 1.0201 1 Δy =.0201 When x =1 a difference in x of.01 produces a difference in y of.0201! f (x) = x 2 + Find when x =1 and =.01 = 2x = ( 2x) = ( 2).01 =.02 Δy =.02 Example 2 f (x) = x 2 4x Find Δy when x = 2 and Δx =.1 f (2) = 4 Δy = f (x + Δx) f (x) Δy = f (2.1) f (2) [ ] 4 Δy = (2.1) 2 4(2.1) Δy = 4.8 4 Δy =.8 When x = 2 a difference in x of.1 produces a difference in y of.8 f (x) = x 2 4x Find when x = 2 and =.1 = 6x 4 = ( 6x 4) = ( 8).1 =.8 Δy =.8 Math 400 4 7! Page 6! 2016 Eitel
Estimating Function Values Example 1 If y = x then = 1 2 x Approximate the value of 9 We know the exact value of x if x = 6 x = 6 To approximate 9 we let x = 6 and Δx = y = x = 1 2 x = 1 2 x when x = 6 and = = 1 2 6 = 1 4 When x = 6 a difference in in x produces a difference of.25 in y y f (6) + = 6 +.25 = 6.25 the calculator answer for 9 is 6.244997.. Math 400 4 7! Page 7! 2016 Eitel
Example 2 If y = x then = 1 x 2 1 and = x 2 Approximate the value of 124 We know the exact value of x To approximate 124 if x =125 125 = 5 we let x =125 and Δx = 1 y = x = 1 x 2 1 = x 2 when x = 125 and = 1 1 = 125 1 = 1 15 When x =125 a difference of 1 in x produces a difference of 1/15 in y y f (125) + = 125 1 15 4.9... the calculator answer for 27 is 4.9866 Math 400 4 7! Page 8! 2016 Eitel
Example Approximate the value of sin() We know the exact value of sin(x) if x = π then sin(π) = 0 To approximate sin() we let x = π (.14) and Δx =.14 y = sin(x) = cos(x) = cos(x) when x = π and Δx =.14 = cos(π).14 =.14 When x = π a difference of.14 in x produces a difference of.14 in y y f (π) + = sin(π).14 = 0.14 =.14 the calculator answer for sin() is.14112 Note: We increase the error when we use.14 instead ofπ but it still produces a good aproximation Note: Today a calculator can find the correct answer much easier than this method can find an approximate answer. If you stopped at the computation of d y you would have an expression for the relative error in y. That expression can be of use in other settings. The change in y denoted by Δ y can be approximated by the differential for small values of x. We call the relative error in the y Math 400 4 7! Page 9! 2016 Eitel
Applications of finding the relative error in the y Example 1 A 4 foot by 8 foot sheet of plywood has a possible error in length of.1 feet. What is the possible error in the the area of the sheet of the plywood? A = W L Find when W = 4, L = 8 and dl =.01 A = 8L da dl = 8 da = 8dL da = ( 8).01 da =.08 feet Example 2 A square window is measured and the maximum error in measurement of.1 inch. What is the possible error in the the area of the sheet of window when the sides are reported to measure 6 feet? A = S 2 Find da when S = 6 and ds =.01 da ds = 2S da = 2S ds da = 2( 6).01 da =.12 square inches The exact area is y = 6 2 = 6 sq. in. The area may be as low as y = 5.9 2 = 4.81 sq. in or as high as y = 6.1 2 = 7.21 sq. in. The maximum possible change in y is 1.21 sq. in. If Δx = +.1 then Δy = +1.21 Math 400 4 7! Page 10! 2016 Eitel
Example A sphere is made with a radius of inches. The possible error in the measurement of the radius is +.1 What is the possible error in the the volume of the sphere? V = 4 πr Find da when S = 6 and ds =.01 dv dr = 4πR2 dv = 4πR 2 dr dr =.1 dv = 4πR 2 dr dv = 4π 2 (.1) dv =.6π cubic inches dr =.1 dv = 4πR 2 dr dv = 4π 2 (.1) dv =.6π cubic inches What is the possible error in the the volume of the sphere is ±.6π The exact volume would be 6π cubic inches. The volume may be as low as 6π.6π cubic inches. or as high as 6π.6π cubic inches.. Math 400 4 7! Page 11! 2016 Eitel
Error Propagation One field where the approximation of Δy by is engineering. When any measurement is made an error in the measurement can be expected. The difference in the true measurement of an object and the recorded measurement is called the Error in Measurement. If you use this measurement in calculating the values the error is included in the calculations and produces further error in the new calculation. This error is called the Propagated Error. The measure of the radius of a circle is recorded as inches but it may be off by as much a.1 inch. We write this as ±.1 This means the actual length of the radius may be from 2.9 to.1 inches long. If we use the inches value to calculate the area of a circle with that radius we get an area of 2 π or approximately 28.27 sq. in Since the real length of the radius is somewhere between 2.9 to.1 the area of the circle could be between 2.9 2 π and.1 2 π or 26.42 and 0.19 sq. in. An error of.1 inches in the measurement of the radius can produce an error in the calculation of the area between 1.85 sq. in. to 1.92 sq. in. We call the 1.92 value the Propagated Error. If we want to find the percent error we divide the error by the actual value to get 6.79% error in the calculation of the area. Math 400 4 7! Page 12! 2016 Eitel