UNIT-4

Multivariable Calculus-Differentiation

4.1 Limits and Continuity.

Limit:

A limit is a number that a function approaches as the independent variable of the function approaches a given value. For example, given the function f (x) = 3x , you could say, “The limit of f (x) as x approaches 2 is 6 .” Symbolically, this is written .

Continuity:

Continuity is another far-reaching concept in calculus. A function can either be continuous or discontinuous. One easy way to test for the continuity of a function is to see whether the graph of a function can be traced with a pen without lifting the pen from the paper. For the math that we are doing in pre calculus and calculus, a conceptual definition of continuity like this one is probably sufficient, but for higher math, a more technical definition is needed. Using limits, we’ll learn a better and far more precise way of defining continuity as well. With an understanding of the concepts of limits and continuity, you are ready for calculus.

Example 1:

Find the limit of the following points

Solution:

We apply the value of each x by finding the respective value through appliying limits

f(x)=

f(x)=

Finding limit at x= 1.9

f(x)=

f(1.9)=

=

= 0.3448

Limit at x=1.99

f(1.99)=

=

= 0.33444

Limit at x=1.999

f(1.999)=

=

= 0.33344

Limit at x=2.001

f(2.001)=

=

= 0.33322

Limit at x=2.01

f(2.01)=

=

= 0.3225

From the above table we have to estimate the limit when x tends to 2

Here answer is 0.333....

Example 2:

Determine whether the following is dis-continuous at x=-1,0,

Solution :

Given 

Verifying for continuity at x=-1

Therefore here = f(-1)

f(x)=f(-1)

Hence the function is continuous at x=-1

Verifying at x=0

Therefore here = f(0)

f(x)=f(0)

Hence the function is continuous at x=0

Example 3:

=

=(3)(

=

4.2 Partial Derivatives, Total Derivatives.

Partial Differentiation Partial differentiation is used to differentiate mathematical functions having more than one variable in them. In ordinary differentiation, we find derivative with respect to one variable only, as function contains only one variable. So partial differentiation is more general than ordinary differentiation.

Example 1:

Solution:

Partial derivative for the given equation is,

Example 2:

Find the partial derivative of the following

Z= ,x=st , y=

Solution:

.

Example 3:

Solve the following D.E

Solution :

Approximating over all change in y

  = 0.02w          =        =

Substituting into equation we get,

(-0.03s) + (0.01d) =0.11

Therefore y decreases by approximately 11 percent.

Total derivative: The total derivative is the derivative with respect to of the function that depends on the variable not only directly but also via the intermediate variables . In mathematics, the total derivative of a function {display style f} at a point is the best linear approximation near this point of the function with respect to its arguments. Unlike partial derivatives, the total derivative approximates the function with respect to all of its arguments, not just a single one.

A(u) = f(x1,x2,.....) and u has continuous partial derivatives fx and fy. Here xi = xi(t) where i=0,1,2,3...n[]

Total derivative

Eq(1) can be written in differential form by eliminating dt as,

Example 1:

Suppose   : is given by

( = B +,

Where B is an mn –matrix and .For example if B= and =

f(x,y,z) = + =

We claim that D. = B for all .to check this note

So indeed, D  = B

Example 2:

Find the total derivative of f(x,y)=2x+3y with respect to x,

Given that y= sin-1(x)

Solution:

=

2x+3y+3x.x +

Example 3:

The radius and height of a cylinder are both 2cm.the radius is decreased at 1 cms and the height is increasing at 2 cms. What is the change in volume at this instant

The volume of a right circular cylinder is,

V=

We are taking total derivative of this whole to get,

2(2)(2)(-1)+ = -8 +8 =0

At this moment, the volume of the cylinder is not changing.

4.3 Tangent Plane and Normal Line.

Example 1:

Find the normal line to Find the normal line to

Solution:

We can re-write this equation as:

Now we need to find the gradient of the function on the left side as at value (3,-3,2)

      =

Then the normal plane is,

=+t

Example 2:

Consider the surface z= 10- at (1,-1,7), find a 3d tangent vector that points in the direction of steepest ascent.

Solution:

Let our tangent vector be v= ai+bj+ck.

To find v , we have that,

The direction of steepest ascent of z(x,y) is given by the two-dimensional vector .

First we find the x,y components of v,then we find z component of v

Finding x,y components of v,

As the gradient provides the direction of steepest ascent, we compute it:

Thus, a=-2 and b=4, and are seeking a tangent vector

V=-2i+4j+ck.

Example 3:

Determine in the direction of

Solution:

First we calculate gradient of the points,

Here we require a unit vector but by our knowledge it is clear that the given vector is not so we change it into unit vector,

4.4 Maxima, Minima and Saddle Points.

Example 1: Find out the maxima and minima of the function

Solution:

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

Example 2: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 (

Example 3: Find the maximum and minimum value of

Let

Partially differentiating given function with respect to x and y and equate it to zero

  ..(i)

    ..(ii)

On solving (i) and (ii) we get

Thus pair of values are

Now, we calculate

At the point (0,0)

So further investigation is required

On the x  axis y  = 0  , f(x,0)=0

On the line y=x,

At the point

So that the given function has maximum value at

Therefore maximum value of given function

At the point

So that the given function has minimum value at

Therefore minimum value of the given function

Saddle point: A Differentiable function f(x,y) has a saddle point at a critical point (a,b) if in every open disk centered at (a,b) there are domain points (x,y) where f(x,y)> f(a ,b) and domain points where f(x ,y)<f(a ,b). The corresponding point (a,b,f(a,b)) on the surface z = f(x,y) is called a saddle point of the surface.

Example 4: Find the local extreme values of f (x,y) =

Solution: The domain of f is the entire plane ( so there are no boundary points ) and the partial derivative exist everywhere .Therefore ,local extreme values can occur only where

The only possibility is the point (0,2), where the value of f is 5 .Since

f(x,y)= never less than 5 , we see that the critical point (0,2) gives a local minimum.

Example 5: Find the local extreme values (if any) of f (x,y) =

Solution: The domain of f is the entire plane ( so there are no boundary points )and the partial derivatives  exist everywhere .

Therefore local extrema can occur only at the origin (0,0) where

Along the positive x-axis , however , f has the value f(x,0) =

Along the positive y-axis ,f has the value f(0,y) =>0

Therefore ,every open disk in the xy-plane centered  at (0,0) contains points where the function is positive and points where it is negative . The function has a saddle point at the origin and no local extreme values (see the following figure).

4.5 Method of Lagrange Multipliers.

Let be a function of x, y, z which to be discussed for stationary value.

Let be a relation in x, y, z

for stationary values we have,

i.e.    … (1)

Also from we have

    … (2)

Let ‘’ be undetermined multiplier then multiplying equation (2) by and adding in equation (1) we get,

     … (3)

     … (4)

      … (5)

Solving equation (3), (4) (5) & we get values of x, y, z and .

Example 1: Decampere a positive number ‘a’ in to three parts, so their product is maximum

Solution: Let x, y, z be the three parts of ‘a’ then we get.

    … (1)

Here we have to maximize the product

i.e.

By Lagrange’s undetermined multiplier, we get,

       … (2)

       … (3)

        … (4)

Ie.

        … (2)’

        … (3)’

        … (4)

And

From (1)

Thus .

Hence their maximum product is  .

Example 2: Find the point on plane nearest to the point (1, 1, 1) using Lagrange’s method of multipliers.

Solution: Let be the point on sphere which is nearest to the point . Then shortest distance.

Let

Under the condition    … (1)

By method of Lagrange’s undetermined multipliers we have

       … (2)

       … (3)

i.e. &

       … (4)

From (2) we get

From (3) we get

From (4) we get

Equation (1) becomes

i.e.

y = 2

4.6 Gradient, Curl, and Divergence of directional derivatives.

Example 1:Find the gradient of the following:

Solution:    

  y=

    =      .

= 2x+

Example 2:

Find the curl of F(x,y,z) = 3i+2zj-xk

Curl F =

=

= i -

= (0-2)i-(-1-0)j+(0-0)k

= -2i+j

Example 3:

What is the curl of the vector field F= ( x +y +z ,x-y-z,)?

Solution:

Curl F =

=

=

= (2y+1)i-(2x-1)j+(1-1)k

= (2y+1)i+(1-2x)j+0k

= (2y+1, 1-2x,0)

Example 4: Compute   where F= (3x+  and s is the surface of the box such that   0 use outward normal n

Solution: Writing the given vector fields in a suitable manner for finding divergence

Div F =3+2y+x

We use the divergence theorem to convert the surface integral into a triple integral

Where B is the box     0 , 0

We compute the triple integral of div F=3+2y+x over the box B

=

=

= 36+3=39

Example 5: For F= (  use divergence theorem to evaluate where s is the dphere of radius 3 centered at origin.

Solution: Since div F= , the surface integral is equal to the triple integral.

To evaluate the triple integral we can change value of variables to spherical co-ordinates,

The integral is = .For spherical co-ordinates, we know that the jacobian determinant is   dV = .therefore, the integral is

=

=

=