Unit - 3

Applications of Partial differentiation

3.1 Tangent plane and normal line to a surface

The tangent is a line which passes through a single point of the curve. Also known as slope of the curve.

Let be the curve in the XY plane. Then the equation of the tangent at a point P(x,y) in the XY plane  is given by

Intercepts: The equation of the intercept that meet the x axis is  (Y=0)

And the equation of the intercept that meet the y axis is  (X=0).

Example1:   Find the equation of the tangent to the curve at the point ?

Given Curve 

Differentiating both side with respect to t.

By Chain Rule 

Therefore the equation of the tangent

Or   …(i)

At the point  

Also  

Substituting all these values in (i) we get

Hence the equation of the tangent is .

Example2:  At what point is the tangent to the curve parallel to the chord joining the points (0,0) and (0,1).

Given curve is

Equation of the chord joining the points (0,0) and (0,1) is

Or i.e.  Parallel toY-axis    …(i)

Given curve is

Differentiating both side with respect to x.

Equation of tangent parallel to Y-axis is  

Or    (as above) 

Since curve is

At    (ii)

From (i) and (ii) we conclude that the point tangent parallel to the chord is .

Hence the point tangent parallel to the chord is .

Example3: Prove that touches the curve at the point where the curve crosses the axis of y.

Given curve is

It will cross the y-axis  at x=0, we get

Hence the point of the curve on the Y-axis is

Since curve is

Differentiating both side with respect to x, we get

.

At point (0,b)   we have     …(i)

Again given

Differentiating  both side with respect to x, we get

Again at the point (0,b) we have …..(ii)

As we know that two curve touches each other then they have Same Lope at the point of intersection.

From (i) and (ii) we conclude that both the curve touches each other on Y-axis.

Normal:

A normal is  a line which passes  through a point and is perpendicular to the  normal at there.

Let be the curve in the XY plane. Then the equation of the normal at a point P(x,y) in the XY plane  is given by

Angle of intersection of two curves:

Angle of intersection of two curves is formed at the intersection point of tangents to the curve.

Procedure:  1. Solve the given equations of the curve to get the point of intersection let it be M.

2. Find out the slopes of the given curves (i.e. ) at point M. Let the slope of first curve be and second be .

3. Let the angle at the point M be . Then

In case it means the curves are orthogonal to each other.

Lengths of Tangent and Normal

The length of the following is given by formulas:

1. Tangent =
2. Normal =
3. Sub tangent  =
4. Sub normal= .

Example1: Find the equation of the normal of the curve at.

Given curve 

Differentiating both side with respect to t.

By Chain Rule

At  the point ,

Also

Therefore equation of the normal at

Hence the equation of the normal at .

Example2: Prove that the curve and  will cut orthogonally if ?

Given curves      …(i)

And       …(ii)

Let    be a intersection point of curve (i) and (ii)

Then 

And 

On solving we get

Or    …(iii)

Differentiating (i) with respect to x.

At the point slope of (i) is given by

Differentiating (ii) with respect to x.

At the point slope of (ii) is given by

For orthogonal intersection, we have

Substituting the values of slopes we get,

Or

Putting values from (iii) we get

Or

Hence proved.

Example3: Find the lengths of the tangent, normal, sub tangent and subnormal for the cycloid:

The equation of the cycloid

Differentiating both with respect to t .

By Chain Rule

Length of tangent is

Length of normal is

Length of sub tangent =

=

Length of sub normal =

3.2 Maxima and Minima

Let be a defined function of two independent variables.

Then the point is said to be a maximum point of if

Or   =

For all positive and negative values of h and k.

Similarly the point is said to be a minimum point of if

Or   =

For all positive and negative values of h and k.

Saddle point: Critical points of a function of two variables are those points at which both partial derivatives of the function are zero. A critical point of a function of a single variable is either a local maximum, a local minimum, or neither. With functions of two variables there is a fourth possibility - a saddle point.

A point is a saddle point of a function of two variables if

At the point.

Stationary Value

The value is said to be a stationary value of if

i.e. the function is a stationary at (a , b).

Rule to find the maximum and minimum values of

1. Calculate.
2. Form and solve , we get the value of x and y let it be pairs of values
3. Calculate the following values :

4.     (a) If

(b) If

(c) If

(d) If

Example1 Find out the maxima and minima of the function

Given     …(i)

Partially differentiating (i) with respect to x we get

    ….(ii)

Partially differentiating (i) with respect to y we get

    ….(iii)

Now, form the equations

Using (ii) and (iii) we get

          using above two equations

Squaring both side we get

Or 

This show that

Also we get

Thus we get the pair of value as

Now, we calculate

Putting above values in

At point (0,0)  we get

So, the point (0,0) is a saddle point.

At point we get

So the point is the minimum point where

In case 

So the point is the maximum point where

Example2 Find the maximum and minimum point of the function

Partially differentiating given equation with respect to and x and y then equate them to zero                 

On solving above we get 

Also

Thus, we get the pair of values (0,0), (,0) and (0,

Now, we calculate

At the point (0,0)

So function has saddle point at (0,0).

At the point (

So the function has maxima at this point (.

At the point (0,

So the function has minima at this point (0,.

At the point (

So the function has an saddle point at (

3.3 Langrage’s method of multiplier

Generally to calculate the stationary value of a function with some relation by converting the given function into the least possible independent variables and then solve them. When this method fail, we use Lagrange’s method.

This method is used to calculate the stationary value of a function of several variables which are all not independent but are connected by some relation.

Let  be the function in the variable x, y and z which is connected by the relation

Rule: a) Form the equation

Where is a parameter.

b) Form the equation using partial differentiation is

(We always try to eliminate)

c) Solve the all above equation with the given relation

These give the value of

These value obtained when substituted in the given function will give the stationary value of the function.

Example1 Divide 24 into three parts such that the continued product of the first, square of second and cube of third may be maximum.                                    

Let first number be x, second be y and third be z.

According to the question

Let the given function be  f

And the relation

By Lagrange’s Method

   ….(i)

Partially differentiating (i) with respect to x,y and z and  equate  them  to zero

   ….(ii)

   ….(iii)

   ….(iv)

From (ii),(iii)  and (iv) we get

On solving

Putting it in given relation we get

Or 

Or

Thus the first number is 4 second is 8 and third is 12

Example2 The temperature T at any point   in space is  .Find  the highest temperature on the surface of the unit sphere.

Given function is  

On the surface of unit sphere given     [ is an equation of unit sphere in 3   dimensional space]

By Lagrange’s Method

     ….(i)

Partially differentiating (i) with respect to  x, y and z and equate them to zero

or           …(ii)

or           …(iii)

          …(iv)

Dividing (ii) and (ii) by (iv) we get

Using given relation 

Or       

Or               

So that

Or 

Thus points are

The maximum temperature is

Example3  If ,Find the value of x and y for  which is maximum.

Given function is

And relation is

By Lagrange’s Method

[]     ..(i)

Partially differentiating (i) with respect to  x, y and z and equate them to zero

Or       …(ii)

Or       …(iii)

Or       …(iv)

On solving (ii),(iii) and (iv) we  get

Using the given relation  we get

So that

Thus the point for the maximum value of the given function is

Example4: Find the points on the surface nearest to the origin.

Let be any point on the surface, then its distance from the origin is

Thus the given equation will be

And relation is

By Lagrange’s Method

     ….(i)

Partially differentiating (i) with respect to  x, y and z and equate them to zero

Or        …(ii)

Or        …(iii)

Or

Or

On solving equation  (ii) by (iii) we get

And     

On subtracting we  get

Putting in above

Or 

Thus

Using the given relation we  get

= 0.0 +1=1

Or 

Thus point on the surface nearest to the origin is

3.4 Jacobian

Jacobians, Errors and Approximations, maxima and minima

Jacobians

If u and v be continuous and differentiable functions of two other independent variables x and y such as

, then we define the determine

as Jacobian of u, v with respect to x, y

Similarly ,

JJ’ = 1

Actually Jacobins are functional determines

Ex.

1.   Calculate
2. If
3. If

ST

4.      find

5.     If and , find

6.    

7.     If  

8.     If , ,

JJ1 = 1

If ,

JJ1=1

Jacobian of composite function (chain rule)

Then

Ex.

1. If

Where

2.     If   and

Find

3.     If

Find

Jacobian of Implicit function

Let u1, u2 be implicit functions of x1, x2 connected by f1, f2 such there

,

Then

Similarly,

Ex.

If

If

Find

Partial derivative of implicit functions

Consider four variables u, v, x, y related by implicit function.

,

Then

Ex.

If and

Find

If and

Find

Find

If

Find

3.5 Errors and approximations

The approximation error in some data is the discrepancy between an exact value and some approximation to it. An approximation error can occur because:

*  The  measurement of the data is not precise due to the instruments. (e.g., the accurate reading of a piece of paper is 4.5 cm but since the ruler does not use decimals, you round it to 5 cm.) or

*   Approximations  are used instead of the real data (e.g., 3.14 instead of π).

In the mathematical field of numerical analysis, the numerical stability of an algorithm indicates how the error is propagated by the algorithm.

One commonly distinguishes between the relative error and the absolute error.

Given some value v and its approximation v approx, the absolute error is

||

{\displaystyle \epsilon =|v-v_{\text{approx}}|\ ,}where the vertical bars denote the absolute value. If v ≠0 , {\displaystyle v\neq 0,}the relative error is

=   =| | =  | 1 - |{\displaystyle \eta ={\frac {\epsilon }{|v|}}=\left|{\frac {v-v_{\text{approx}}}{v}}\right|=\left|1-{\frac {v_{\text{approx}}}{v}}\right|,}

And the percent error is.,

=100%. = 100 %.=100%.| 1 - |

{\displaystyle \delta =100\%\times \eta =100\%\times {\frac {\epsilon }{|v|}}=100\%\times \left|{\frac {v-v_{\text{approx}}}{v}}\right|.}In words, the absolute error is the magnitude of the difference between the exact value and the approximation. The relative error is the absolute error divided by the magnitude of the exact value. The percent error is the relative error expressed in terms of per 100.

An error bound is an upper limit on the relative or absolute size of an approximation error.

Example 1:

If the exact value is 50 and the approximation is 49.9, then the absolute error is 0.1 and the relative error is 0.1/50 = 0.002 = 0.2%.

Example 2:

If in measuring a 6 mL beaker, the value read was 5 mL. The correct reading being 6 mL, this means the percent error in that particular situation is, rounded, 16.7%.

Example 3:

Compute the error and relative error when xT=0.3320 and xA=0.3321.

Using the formula above, we have that the error of xA from xT is:

(1)

Error (xA)=xT−xA=0.3320−0.3321=−0.0001

Now the relative error of xA from xT is:

(2)

Rel(xA)=xT−xAxT=−0.00010.3320≈−0.00030120

Example 4:

A polynomial p(x) is used to approximate a function f(x) that is difficult to compute at x=0. We find that p(0)=9.1111334 and the true value f(0)=9.1122114. Find the error and relative error of p(0) from f(0).

Using the formula above, we have that the error of p(0) (our approximate value) from f(0) (the actual value) is:

(3)

Error(p(0))=f(0)−p(0)=9.1122114−9.1111334=0.001078

Now the relative error of p(0) from f(0) is:

(4)

Rel(p0)=f(0)−p(0)f(0)=0.0010789.1122114≈0.0001183...

Reference Books

1. M.D. Wier, et. Al., Thomas’ Calculus, 11th Ed., Pearson Education, 2008

2. Stewart James, Calculus Early Transcendental, 5th Ed., Thomson India, 2007

3. Wylie & Barrett, Advanced Engineering Mathematics, Mc graw Hill pub.

4. Greenberg, M.D., Advanced Engineering Mathematics, 2nd ed., Pearson