Lesson 6 differentials parametric-curvature

The Differentials

Objectives At the end of the lesson, the student should be able to:

1. Compare the value of the differential, dy, with the actual change in y,

2. Estimate a propagated error using a differential.3. Find the approximate value of roots by using

differentials.4. Find the differential of a function using

differentiation formulas.

Consider a function defined by y=f(x) where x is the independent variable. In the four-step rule we introduced the symbol Δx to denote the increment of x. Now we introduce the symbol dx which we call the differential of x. Similarly, we shall call the symbol dy as the differential of y. To give separate meanings to dx and dy, we shall adopt the following definitions of a function defined by the equation y=f(x).

DEFINITION 1: dx = Δx In words, the differential of the independent variable is equal to the increment of the variable.

DEFINITION 2: dy = f’ (x) dx In words, the differential of a function is equal to its derivative multiplied by the differential of its independent variable.

We emphasize that the differential dx is also an independent variable, it may be assigned any value whatsoever. Therefore, from DEFINITION 2, we see that the differential dy is a function of two independent variables x and dx. It should also be noted that while dx=Δx, dy≠Δy in general. Suppose dx≠0 and we divide both sides of the equation dy = f’ (x) dx

by dx. Then we get x'f

dxdy

Note that this time dy/dx denotes the quotient of two differentials, dy and dx . Thus the definition of the differential makes it possible to define the derivative of the function as the ratio of two differentials. That is,

xof aldifferenti theyof aldifferenti the

dxdy

x'f

The differential may be given a geometric interpretation. Consider again the equation y=f(x) and let its graph be as shown below. Let P(x,y) and Q(x+Δx,f(x)+Δx) be two points on the curve. Draw the

tangent to the curve at P. Through Q, draw a perpendicular to the x-axis and intersecting the tangent at T. Then draw a line through P, parallel to the x-axis and intersecting the perpendicular through Q at R. Let θ be the inclination of the tangent PT.

P

Q

T

Rθ

From Analytic Geometry, we know that slope of PT = tan θBut triangle PRT, we see that

xRT

PRRT

tan

However, Δx=dx by DEFINITION 1 . Hence

dxRT

tan

But the derivative of y=f(x) at point P is equal to the slope of the tangent line at that same point P. slope of PT = f’(x)Hence,

dxRT

x' f

And , RT = f’(x) dxBut, dy = f’ (x) dxHence, RT = dy

We see that dy is the increment of the ordinate of the tangent line corresponding to an increment in Δx in x whereas Δy is the corresponding increment of the curve for the same increment in x. We also note that the derivative dy/dx or f’(x) gives the slope of the tangent while the differential dy gives the rise of the tangent line.

SUMMARY: DEFINITION OF DIFFERENTIALS

• Let x represents a function that is differentiable on an open interval containing x. The differential of x (denoted by dx) is any nonzero real number.

• The differential of y (denoted by dy) is dy = f(x )dx.

DIFFERENTIAL FORMULAS

Let u and v and be differentiable functions of x.

• Constant multiple: d(cu)= c du• Sum or difference: d[u±v] =du±dv• Product: d[uv]=udv+vdu• Quotient: d=

• du ―

EXAMPLE 1: Find dy for y = x3 + 5 x −1.

dx 53xdy

dx5dxx3

1x5xddy

2

2

3

EXAMPLE 2: Find dy for . 1x3

x2y

22

2

1x32dx

dy 1x3

x62x6dy

1x33x221x3

1x3x2

ddy

dx.by itmultiply and equation the of member right the of

derivative the get simply we practice,In:Note

EXAMPLE 3: Find dy / dx by means of differentials if xy + sin x = ln y .

1xy

xcosyydxdy

xcosyydxdy

1xy

xcosyydxdy

dxdy

xy

dxdy

xcosyydxdy

xy

dx1

dydx xcosydxydy xy

dydx xcosydxydy xy

ydyy1

dx xcosdx ydy x

dyy1

dx xcosdx ydy x

2

2

2

2

EXAMPLE 1: Use differentials to approximate the change in the area of a square if the length of its side increases from 6 cm to 6.23 cm.Let x = length of the side of the square. The area may be expressed as a function of x, where A= x2. The differential dA is dxx2dA dxx'fdA

Because x is increasing from 6 to 6.23, you find that Δ x = dx = .23 cm; hence, 

2cm76.2dA

cm23.0cm62dA

.cm 2.8129 is y area in increase

exact the that Note 6.23. to6 from increases length sideits ascm2.76 ely approximatby increasewill squarethe of area The

2

2

EXAMPLE 2: Use the local linear approximation to estimate the value of to the nearest thousandth.

3 55.26

2.9830.0167-3 55.26 601

355.26 therefore ;327 that less

601

tely approxima be will 55.26 that implies which

0167.0601

10045

271

45.0273

1dy , Hence

0.45dxx then 26.55, to 27 from decreasing is x Because

dx x3

1 dx x

31

dy

xxf dx xf'dy

is dy aldifferenti The 27.x namely 26.55, to close relatively is and cube perfect a is that xof value

convenient a choose ,xxf is applying are you function the Because

3

33

3

32

32

32

31

3

EXAMPLE 3: If y = x3 + 2x2 – 3, find the approximate value of y when x = 2.01.

2.xof value original an to 0.01dxxof increment an applyingof result the as 2.01 gconsiderin are we then 0.01,22.01 write weif that Note dy. y

for solve shall we then value, eapproximat the find to asked simply are we since but y y is value exact The

20.1320.013dyyis ionapproximat required the,therefore

20.001.0812dythen ,01.0dx and 2x when and13388y then ,2x when

dxx4x3dy then

3x2xy Since2

23

EXAMPLE 4: Use an appropriate local linear approximation to estimate the value of cos 310.

8573.0008725.0866.0dyyis ionapproximat required the,therefore

008725.001745.05.0dy180

130 sindy

then ,01745.0180

1dx and 30x when and

866.030cosy then ,30x when

dx x sindy then x cosy Let

0

00

0

00

00

Derivative of Parametric Equations• The first derivative of the parametric

equations y=f(u) and x=g(u) with respect to x is equal to the ratio of the first derivative of y with respect u divided by the derivative of x with respect to u, i.e.

• The second derivative of the equations in parametric form is

Find the derivatives of the following parametric equations :

3t cot3sin3t-

3cos3t

dtdxdtdy

dxdy

tcosdtdy

and tsindtdx

:Solution

3t sin y 3t, cos x .

3333

1

EXAMPLE :

tsintsintcostcos

tsintsintcostcos

tsintsintcostcos

dtdxdtdy

dxdy

tcostcosdtdy

and tsintsindtdx

:Solution

5sin4t-t 8sin y 5cos4t, 8cost x .

452

452

4524

4524

4208

4208

4208

4208

2

2

2213

dxyd

find ,tty ,txIf . 3

2t31t2

dtdxdtdy

dxdy

52

2

24

2

2

2

22

2

t91t2

dxyd

t31

t9t61t22t3

dxyd

dxdt

t31t2

dtd

dxyd

2

2

414dx

yd find , cos y , sin 2xIf .

tan2cos2sin4

ddxddy

dxdy

3

2

2

2

2

2

2

2

2

2

2

secdx

yd

secsecdx

ydcos21

sec2dx

yddxd

tan2dd

dxyd

.0,4 at ty and ttx

:curve parametric the to s line tangent the Find .25 34

5

125tt2

125tt

2t

t125t2t

dtdxdtdy

dxdy

line. tangent the of slopethe get can we

that so,dxdy

find and derivative the find to have We

22224

4x81

yx81

4-y is line tangent secondof equation the thus

81

m is 0,4 at line tangent the of slopethe Therefore

81

122522

dxdy

,2t at ,Now

4x81

yx81

4-y is line tangent of equation the thus

81

m is 0,4 at line tangent the of slopethe Therefore

81

122522

dxdy

2, t at

defined not is dxdy

0,t at

2t 0,t 04t ,0t

04ttt4t0

becomes curve the of equation c parametrithe 4,0 at ,Now

2

2

23

2335

Other Examples:• Use differentials to approximate the value of

the following expression.

• Find the differential dy of the given function.

Find the first and second derivative of y with respect to x from the given parametric equations.

Differential of Arc LengthLet y=f(x)be a continuous function. Let P(x,y) and Qbe on the curve of f(x). Denote be the arc length from P to Q. The rate of the arc s from P to Q per unit change in x and the rate of change per unit change in y are given by

P(x,y)

When the curve is given in parametric equations , the rate of change of s with respect to time t is given by

CurvatureThe curvature K (the measure of how sharply a curve bends) of a curve y=f(x), at any point P on it is the rate of change in direction, i.e. the angle of inclination of the tangent line at P per unit of arc length s.

K= =curvature at P

𝜶+∆𝜶

∆ 𝒔 ∆𝜶

The curvature at a point P of the curve y=f(x) is

When the equation of the curve is given in parametric form x=f(u) and y=g(u)

Radius of CurvatureConsider the curve y=f(x) having the tangent line L at P and curvature K. Consider a circle which lies on the side of the curve and tangent to line L at P with curvature K. this is called the circle of curvature with radius of curvature R defined to R=

K

L

P

y

XO

ExampleFind the curvature and radius of curvature of the plane curve at the given value of x.

2. X=sin