Applications of partial differentiation

In mathematics, a partial derivative of a function of several variables is its derivative with respect to one of those variables, with the others held constant (as opposed to the total derivative, in which all variables are allowed to vary). Partial derivatives are used in vector calculus and differential geometry.

1. Prepared by : Vaibhav Tandel ( 16033010xxxx) From : Electrical department Subject : Calculus Mahatma Gandhi Institute Of Technical Education And Research Center

2. Calculus Applications Of Partial Differentiation

3. Index Taylor’s Theorem for the function of two variable Taylor’s series Jacobians Errors and Approximation Maxima and Minima Lagrange’s Method of Undetermined Multipliers

4. Taylor's Theorem for Two Variable Functions Rather than go through the arduous development of Taylor's theorem for functions of two variables, I'll say a few words and then present the theorem. In the one variable case, the n th term in the approximation is composed of the n th derivative of the function. For functions of two variables, there are n +1 different derivatives of n th order. For example, f xxxx , f xxxy , f xxyy , f xyyy , f yyyy are the five fourth order derivatives .

5. There are actually more, but due to the equality of mixed partial derivatives, many of these are the same. Thus, our formula for Taylor's theorem must incorporate more than one derivative at each order. The formula for a third order approximation to f ( x , y ) near ( x , y ) is The factors of 2 and 3 appearing the second and third order mixed partial terms are due to the fact that there are two equal mixed partials derivatives of second order and a pair of three equal third order mixed partials.

6. Let U ⊆R, and let f : U →R such that the derivative, and all the higher derivatives, of f exist on U. For h∈U , Taylor series

7. This is the Taylor series of f centred at h. If the series is truncated after n+1 terms we get the Taylor polynomial or Taylor approximation of f, of degree n, centred at h. If we replace x with x+h throughout in the Taylor series of f, we get the following alternative way of expressing the Taylor series of f centred at h:

8. The following are the Taylor series of some standard functions. The first three are centred at 0, and are valid for all x ∈ R; the other two are centred at 1, and are valid for x∈R with |x|< 1.

9. Now let U ⊆ R2, and let f : U → R such that the partial derivatives, and all the higher partial derivatives, of f exist and are continuous on U. For (X,Y)∈U,

10. and f( r,s ) denotes the (r + s) th partial derivative of f found by differentiating r times with respect to x and s times with respect to y. This is the Taylor series of f centred at (X,Y). If the series is truncated after n + 1 terms we get the Taylor polynomial or Taylor approximation of f, of degree n, centred at h.

11. Example : Expand x2y + 3y−2 in powers of x−1 and y + 2. Let f( x,y ) = x2y + 3y−2. Putting x = 1 + h and y =−2 + k we can obtain the required expansion by finding the Taylor series of f centered at (1,−2). Now

12. and all higher derivatives are 0. Thus

13. If u, v, w are the function of x, y, z having first order partial derivatives w.r.t. x, y, z then the determinant JACOBIAN is called Jacobian of u, v, w w.r.t. x, y, z

14. Example :   Show that when changing to polar coordinates we have Solution So, what we are doing here is justifying the formula that we used back when we were integrating with respect to polar coordinates .  All that we need to do is use the formula above for dA .   The transformation here is the standard conversion formulas,

15. The Jacobian for this transformation is,

16. Error And Approximation Z = f(x , y) be a continuous function of x and y where f ᵪ & f ᵧ be the errors occurring in the measurement of the value of x & y. Then the corresponding error ¶Z occurs in the estimation of the value of Z. i.e. Z+¶Z = f(x+¶x , y+¶y) Therefore, ¶Z = f(x+¶x , y+¶y) – f(x , y).

17. Expanding by using Taylor’s Series and neglecting the higher order terms of ¶x & ¶y, we get, ¶Z = ¶ x.¶f /¶x + ¶ y.¶f /¶y ¶x is known as Absolute Error in x. ¶x/x is known as Relative Error in x ¶x/x*100 is known as Percentage Error in x.

18. Example : In measurement of radius of base and height of a rigid circular cone are incorrect by -1% and 2%. Calculate Error in the Volume. Solution, Let r be the radius and h be the height of the circular cone and V be the volume of the cone. V = π /3*r^2*h

19. Thus, ¶V = ¶ r.¶V /¶r + ¶ h.¶V /¶h Now, ¶r/r*100 = -1 ¶h/h*100 = 2 Again, ¶V = π /3(2rh)(r/100) + π /3(r*r)2h/100 = 0 So, The Error in the measurement in the Volume is Zero.

20. MAXIMUM & MINIMUM VALUES A function f has an absolute maximum (or global maximum) at c if f ( c ) ≥ f ( x ) for all x in D , where D is the domain of f . The number f ( c ) is called the maximum value of f on D .

21. MAXIMUM & MINIMUM VALUES Similarly, f has an absolute minimum at c if f ( c ) ≤ f ( x ) for all x in D and the number f ( c ) is called the minimum value of f on D . The maximum and minimum values of f are called the extreme values of f .

22. LOCAL MAXIMUM VALUE If we consider only values of x near b —for instance , if we restrict our attention to the interval ( a , c )—then f ( b ) is the largest of those values of f ( x ). It is called a local maximum value of f .

23. LOCAL MINIMUM VALUE Likewise, f ( c ) is called a local minimum value of f because f ( c ) ≤ f ( x ) for x near c— for instance, in the interval ( b , d ). The function f also has a local minimum at e .

24. MAXIMUM & MINIMUM VALUES In general, we have the following definition. A function f has a local maximum (or relative maximum) at c if f ( c ) ≥ f ( x ) when x is near c . This means that f ( c ) ≥ f ( x ) for all x in some open interval containing c . Similarly, f has a local minimum at c if f ( c ) ≤ f ( x ) when x is near c .

25. Lagrange multipliers Let U ⊆ R2 and let f,g : U → R. We now consider the problem of finding the maximum and minimum values of f subject to the constraint g( x,y ) = 0 .

26. Method 1 Suppose we can rewrite g( x,y ) = 0 in the form y = h(x), for some function h. Then we can just find the maximum and minimum values of F(x) = f( x,h (x)) using the method for calculus of functions of one variable .

27. Example : Find the maximum and minimum values of f( x,y ) = x3 + 3xy2 + 2xy, subject to the constraint x + y = 4. Here g( x,y ) = x + y −4, and g( x,y ) = 0 gives y = 4−x. (So h(x) = 4−x.) Thus

28. Differentiating F with respect to x we get F0(x) = 12x2−52x+56, and so For x = 2, y = 2 and F(2) = 40, and for x = 7/3 , y = 5/3 and F(7/3 ) = 39(25/27). Thus the maximum and minimum values of f( x,y ) = x3 + 3xy2 + 2xy, subject to the constraint x + y = 4, are 40 and 3925 27, respectively.

29. Method 2 The local maximum and minimum of f( x,y ) subject to the constraint g( x,y ) = 0 correspond to the stationary points of L( x,y,λ ) = f( x,y )− λg ( x,y ). The variable λ is call a Lagrange multiplier.

30. Example : Consider the problem from Example 11. Here f( x,y ) = x3 + 3xy2 + 2xy and g( x,y ) = x + y−4. So we let L( x,y,λ ) = x3 + 3xy2 + 2xy−λ(x + y−4). Now,

31. From equations ( 1) and (2) we get, 3x2 + 3y2 + 2y = 6xy + 2x . We can now use equation (14) to eliminate y form the previous equation. After simplification this gives 12x2 −52x + 56 = 0 (1) (3) (2)

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