Section 3.9 - Differentials

Local LinearityIf a function is differentiable at a point, it is at least locally linear. Differentiable

Local LinearityIf a function is differentiable at a point, it is at least locally linear. Not differentiable.

The function is NOT smooth at this point.

Local LinearityIf a function is differentiable at a point, it is at least locally linear. The function is

differentiable at every point.The function is always smooth.

How to Approximate a Function

A line that best approximates the

graph of a function is the tangent line (if

the values are near the point of tangency).

How to Approximate a FunctionPoint of Tangency

( c , f (c) )

,c f c

Tangent Liney – f (c) = f '(c) (x – c)

A point (x,y) on the tangent line

approximates the value of the function (must be

near the point of tangency)

y = f (c) + f '(c) (x – c)

,x y

Linear Approximation

The linear approximation or tangent line approximation of a function f at x = c:

'L x f c f c x c

This is also referred to as the linearization of f at x = c.

Our Book uses T(x) instead of L(x)

Example 1Find the local linear approximation of at x0 = 1.

f x x

'L x f c f c x c Use the linearization formula if c = x0 = 1.

f c 1f 1 1 'f x 1

2 x

' 1f 12 1

12

121 1L x x

'f c

Example 2Use the linearization of at π/6 to approximate sin 0.5.

sinf x x

Use the linearization formula if c = π/6.

f c 6f 6sin 12

'f x cos x

6'f 6cos 32

312 2 6L x x

Evaluate the linearization formula at x = 0.5.

0.5L 312 2 60.5 0.479563

'f c

Error in Local Linear ApproximationsAs a general rule, the accuracy of the linearization to f (x) at c deteriorates as x gets further away from c. The size of the error of the approximation is simply the vertical gap between the graph and the tangent line:

,c f c

,c f c

Error

Under Estimate Over Estimate

Error L x f x

Example 2 (Continued)Use the linearization of at π/6 to approximate sin 0.5.

sinf x x

0.5L 312 2 60.5 0.479563

Since sin 0.5 0.479426Our approximation is an over estimate and the error is:

0.479563 0.479426 0.000137

The further away from π/6 the worse the

approximation:

x Linearization Actual Error0.5 0.479563 0.479426 0.000137

0.4 0.39296 0.389418 0.003542

0.01 0.05521 .01000 0.045211

Example 3Use linearization to approximate . Is the approximation an overestimate or underestimate.

4.05

Since the square root of 4 is well known, let c = 4.

f c 4f 4 2 142 4L x x

Evaluate the linearization formula at x = 0.5.

4.05L 142 4.05 4

Let the function be: f x x

'f x 12 x

' 4f 12 4

14 'f c 2.0125

Since 4.05 2.01246 Our approximation is an over estimate

Introduction to DifferentialsSo far we associated the following items to be equivalent:

' ' dydxy f x

Now we will consider the differentials “dy” and “dx” to represent two different real numbers. Thus dy/dx becomes a real number ratio. Also, if f is differentiable at x, then becomes

'dydx f x

'dy f x dx

Introduction to DifferentialsSlope of tangent line:

' dydxf x

x

dy

dxx x

Error Solve for the differential dy:

'dy f x dx

Find the change in y: y f x x f x f x

f x x

Thus the two vertical differences are almost equal:

y dy

y

Thus the horizontal differences are always equal:

x dx

x

Error is not the only “differen”ce we can

calculate

DifferentialsIf y = f (x), where f is a differentiable function, then the differential of x, dx, is an independent variable; that is, dx can be given the value of any real number. The differential of y, dy, is then defined in terms of dx by the equation:

So dy is the is a dependent variable; it depends on the values of x and dx.

'dy f x dx

Remember:&dy y dx x

y f x x f x

Example 1Compare the values of Δy and dy if f (x) = x3 + x2 – 2x + 1 and dx = 0.05.

y 2 0.05 2f f 9.718 9 0.718

'f x 23 2 2x x

dy ' 2f dx 14 0.05

Thus, Δy and dy are approximately equal (with Δy being slightly larger)

0.7

Example 2

Find the differential of 2

2

11

xyx

'dy f x dx Use the differential formula for dy.

'y

2 2

22

1 2 1 2

1

x x x x

x

22

4

1

x

x

22

4

1

xdy dxx

Example 3Find the differential dy of x y xy

d x y d xy

Since this is an implicit equation, you can either solve for y or use a method similar to the differentiation process in related

rates/implicit differentiation. The following shows the latter:

dx dy Now solve for dy:dy x dy y dx dx

1dy x y dx dx

1y dx dx

xdy

x dy y dx

Example 4The radius r of a circle increases from 10 m to 10.1 m. Use differentials to estimate the increase in the circles area A.

The function for area is: 2A rUse the differential formula

to find dA:

'A 2 rdA 2 r dr

Evaluate the differential formula at r = 10

2 10 10.1 10dA 2dA

The increase in the circles areas is about 2π m2

Three Ways to Describe Change

Actual EstimatedActual

Error (change)Relative Error

(change)Percentage

Error (change)

'dy f c dx y f c x f c

y

f c

dy

f c

100dyf c 100y

f c

As we move from c to a nearby point c + dx, we can describe the change in f in three ways:

Other Vocab dx = Δxdy

Measurement Error:Propagated Error:

Example 1Inflating a bicycle tire changes its radius from 12 inches to 13 inches. Use differentials to estimate the actual change, the relative change, and the percentage change in the perimeter of the tire.

Actual: Relative: Percentage:

f c 12P 2 12 24 'f x 2d

dr r ' 12P 2 'P r

'f c

Calculate necessary information for the formulas:

2

dy ' 12P dr 2 1 2dP dx dr 13 12 1

Estimate the changes (Errors):Actual

dy 2

~ 2 in

Relative224

1

12

~ 0.083

Percentage 2

24 100 25

3

~ 8.333%

( )dy

f c

( ) 100dyf c

Example 2Suppose that the side of a square is measured with a ruler to be 10 inches with a measurement error of at most ± 1/32 in. Estimate the error in the computed area of the square.

The function for area is: 2A sUse the differential formula

to find dA:

'A 2sdA 2s ds

Evaluate the differential formula at s = 10

1322 10dA

58dA

Thus, the propagated error in the area is estimated to be within ± 5/8 in2

Example 3The volume of a sphere is to be computed from a measured value of its radius. Estimate the maximum permissible percentage error in the measurement if the percentage error in the volume must be kept within ± 1.2%.

The function for volume is: 343V r

Thus, the estimated percentage error in the volume is within ± 0.4%

Use the Differential Formula to find dV:

'V 24 rdV 24 r dr

Use the known information to Find % Error of the Radius:

% 100dVVError

2

343

41.2 100r drr

31.2 100dr

r

0.4 100drr

Definition of percentage error