UNIT3
Multivariable calculus
Limits The function f(x,y) is said to tend to limit ‘l’ , as x →a and y→bIff the limit is dependent on point (x,y) as x →a and y→b We can write this as,
Example1: evaluate the Sol. We can simply find the solution as follows,
Example2: evaluate Sol. 6.
Example3: evaluate Sol. Conitinuity – At point (a,b) , a function f(x,y) is said to be continuous if, Working rule for continuity Step1: f(a,b) should be well defined. Step2: should exist. Step3:
Example1: Test the continuity of the following function
Sol. (1) the function is well defined at (0,0) (2) check for the second step, That means the limit exists at (0,0) Now check step3:
So that the function is continuous at origin.
Example2: check for the continuity of the following function at origin,
Sol. (1) Here the function is well defined at (0,0) (2) check for second step
Limit f is not unique for different values of m So that the limit does not exists. Therefore the function is not continuous at origin. Steps to check for existence of limit Step1: find the value of f(x,y) along x →a and y→b Step2: find the value of f(x,y) along x →b and y→a Note if the values in step 1 and step2 are same then we can say that the limits exist otherwise not. Step3: if a →0 and b→0 then find the limit along y =mx , if the value does not contain m then limit exist, If it contains m then the limit does not exist. Note1 put x = 0 and y = 0 in f , then find f1 2  Put y = 0 and x = 0 In f then find x2 If f1 and f2 are equal then limit exist otherwise not. 3 put y = mx then find f3 If f1 = f2 ≠f3, limit does not exist. 4 put y = mx² and find f4, If f1 = f2 = f3 ≠f4 , limit does not exist If f1 = f2 = f3 = f4 , limit exist.
Example1: Evaluate
Sol . 1.
2. Here f1 = f2 3. now put y = mx, we get Here f1 = f2 = f3 Now put y = mx² 4. Therefore , F1 = f2 = f3 =f4 We can say that the limit exists with 0.
Example2: evaluate the following Sol. First we will calculate f1 – Here we see that f1 = 0 Now find f2, Here , f1 = f2 Therefore the limit exists with value 0. Partial derivatives First order partial differentiation Let f(x , y) be a function of two variables. Then the partial derivative of this function with respect to x can be written as and defined as follows:
Now the partial derivative of f with respect to f can be written as and defined as follows: Note: a. while calculating partial derivatives treat all independent variables, other than the variable with respect to which we are differentiating , as constant. b. we apply all differentiation rules. Higher order partial differentiation Let f(x , y) be a function of two variables. Then its secondorder partial derivatives, third order partial derivatives and so on are referred as higher order partial derivatives. These are second order four partial derivatives: (a) = (b) = (c) = (d) = b and c are known as mixed partial derivatives. Similarly we can find the other higher order derivatives. Example1: Calculate and for the following function f(x , y) = 3x³5y²+2xy8x+4y20 sol. To calculate treat the variable y as a constant, then differentiate f(x,y) with respect to x by using differentiation rules, = [3x³5y²+2xy8x+4y20] = 3x³]  5y²] + [2xy] 8x] +4y]  20] = 9x²  0 + 2y – 8 + 0 – 0 = 9x² + 2y – 8 Similarly partial derivative of f(x,y) with respect to y is: = [3x³5y²+2xy8x+4y20] = 3x³]  5y²] + [2xy] 8x] +4y]  20] = 0 – 10y + 2x – 0 + 4 – 0 = 2x – 10y +4. Example2: Calculate and for the following function f( x, y) = sin(y²x + 5x – 8) sol. To calculate treat the variable y as a constant, then differentiate f(x,y) with respect to x by using differentiation rules, [sin(y²x + 5x – 8)] = cos(y²x + 5x – 8)(y²x + 5x – 8) = (y² + 50)cos(y²x + 5x – 8) Similarly partial derivative of f(x,y) with respect to y is, [sin(y²x + 5x – 8)] = cos(y²x + 5x – 8)(y²x + 5x – 8) = 2xycos(y²x + 5x – 8)
Example3: Obtain all the second order partial derivative of the function: f( x, y) = ( x³y²  xy⁵) sol. 3x²y²  y⁵, 2x³y – 5xy⁴, = = 6xy² = 2x³  20xy³ = = 6x²y – 5y⁴ = = 6x²y  5y⁴
Example4: Find Sol. First we will differentiate partially with repsect to r, Now differentiate partially with respect to θ, we get Example5: if, then find.
Sol
Example6: if , then show that Sol. Here we have, u = …………………..(1) now partially differentiate eq.(1) w.r to x and y , we get = Or ………………..(2) And now,
= ………………….(3) Adding eq. (1) and (3) , we get = 0 Hence proved.
Total derivative When we measure the rate of change of the dependent variable owing to any change in a variable on which it depends, when none of the variable is assumed to be constant. Let the function, u = f( x, y), such that x = g(t) , y = h(t) ᵡ Then we can write, This is the total derivative of u with respect to t. Change of variable If w = f (x, y) has continuous partial variables fx and fyand if x = x (t), y = y (t) are Differentiable functions of t, then the composite function w = f (x (t), y (t)) is a Differentiable function of t. In this case, we get, fx(x (t), y (t)) x’(t)+ fy(x(t), y (t)) y’(t).
Example:1 let q = 4x + 3y and x = t³ + t² + 1 , y = t³  t²  t Then find . Sol: . = Where, f1 = , f2 =
In this example f1 = 4 , f2 = 3 Also, 3t² + 2t , 4(3t² + 2t) + 3( = 21t² + 2t – 3
Example2: Find if u = x³y⁴ where x = t³ and y = t².
Sol. as we know that by definition, = 3x²y⁴3t² + 4x³y³2t = 17t¹⁶.
Example3: if w = x² + y – z + sintand x + y = t, find
(a) y,z
(b) t, z
Sol. With x, y, z independent, we have t = x + y, w = x²+ y  z + sin (x + y). Therefore, y,z = 2x + cos(x+y)(x+y) = 2x + cos (x + y) With x, t, z independent, we have Y = tx, w= x² + (tx) + sin t Thus t, z = 2x  1
Example4: If u = u( y – z , z  x , x – y) then prove that = 0
Sol. Let,
Then,
By adding all these equations we get,
= 0 hence proved.
Example5: if φ( cx – az , cy – bz) = 0 then show that ap + bq = c Where p = q =
Sol. We have, φ( cx – az , cy – bz) = 0 φ( r , s) = 0 where,
We know that,
Again we do,
By adding the two results, we get
Example6: If z is the function of x and y , and x = , y = , then prove that,
Sol. Here , it is given that, z is the function of x and y & x , y are the functions of u and v. So that, ……………….(1) And, ………………..(2)
Also there is,
x = and y = , now,
, , ,
From equation(1) , we get
……………….(3)
And from eq. (2) , we get
…………..(4) Subtracting eq. (4) from (3), we get
= ) – (
= x Hence proved. 
Key takeaways
The function f(x,y) is said to tend to limit ‘l’ , as x →a and y→bIff the limit is dependent on point (x,y) as x →a and y→b We can write this as, 2. Conitinuity – At point (a,b) , a function f(x,y) is said to be continuous if, (a) = (b) = (c) = (d) = 
Equation of a tangent plane And Equation of the normal to the plane
Example: Find the equation of the tangent plane and normal line to the surface Sol. Here, At point (1, 2, 1)
Therefore the equation of the tangent plane at (1, 2, 1) is Equation of normal line is 
As we know that the value of a function at maximum point is called maximum value of a function. Similarly the value of a function at minimum point is called minimum value of a function. The maxima and minima of a function is an extreme biggest and extreme smallest point of a function in a given range (interval) or entire region. Pierre de Fermat was the first mathematician to discover general method for calculating maxima and minima of a function. The maxima and minima are complement of each other. Maxima and Minima of a function of one variables If f(x) is a single valued function defined in a region R then Maxima is a maximum point if and only if Minima is a minimum point if and only if Maxima and Minima of a function of two independent variables 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
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 (
Example3 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 
Key takeaways
Maxima is a maximum point if and only if Minima is a minimum point if and only if 3. A point is a saddle point of a function of two variables if at the point.

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 .
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) i.e. … (2)’ … (3)’ … (4) And From (1) Thus . Hence their maximum product is . 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 
Suppose f(x, y, z) be the scalar function and it is continuously differentiable then the vector Is called gradient of f and we can write is as grad f. So that Here is a vector which has three components Properties of gradient Property1: Proof: First we will take left hand side L.H.S = = = = Now taking R.H.S, R.H.S. = = = Here L.H.S. = R.H.S. Hence proved. Property2: Gradient of a constant ( Proof: Suppose Then We know that the gradient = 0
Property3: Gradient of the sum and difference of two functions If f and g are two scalar point functions, then Proof: L.H.S Hence proved
Property4: Gradient of the product of two functions If f and g are two scalar point functions, then
Proof:
So that Hence proved.
Property5: Gradient of the quotient of two functions If f and g are two scalar point functions, then
Proof: So that
Example1: If , then show that 1. 2. Sol. Suppose and Now taking L.H.S, Which is Hence proved.
2. So that
Example: If then find grad f at the point (1,2,1). Sol.
Now grad f at (1 , 2, 1) will be
Example: If then prove that grad u , grad v and grad w are coplanar.
Sol. Here
Now Apply
Which becomes zero. So that we can say that grad u, grad v and grad w are coplanar vectors.
Divergence (Definition) Suppose is a given continuous differentiable vector function then the divergence of this function can be defined as Curl (Definition) Curl of a vector function can be defined as Note Irrotational vector If then the vector is said to be irrotational.
Vector identities: Identity1: grad uv = u grad v + v grad u Proof: So that graduv = u grad v + v grad u
Identity2: Proof: Interchanging , we get We get by using above equations
Identity3 Proof: So that
Identity4 Proof: So that,
Identity5 curl (u Proof: So that curl (u
Identity6: Proof: So that
Identity7: Proof: So that
Example1: Show that 1. 2. Sol. We know that
2. We know that
= 0 Example2: If then find the divergence and curl of . Sol. we know that Now
Example3: Prove that Note here is a constant vector and Sol. here and So that Now So that 
Key takeaways
If f and g are two scalar point functions, then 5. If f and g are two scalar point functions, then 6. If f and g are two scalar point functions, then 7. 8. 9. If then the vector is said to be irrotational. 10. grad uv = u grad v + v grad u 11. 12. curl (u 
Directional derivative Let ϕ be a scalar point function and let ϕ(P) and ϕ(Q) be the values of ϕ at two neighbouring points P and Q in the field. Then,
,are the directional derivative of ϕ in the direction of the coordinate axes at P. The directional derivative of ϕ in the direction l, m, n=l + m+ The directional derivative of ϕ in the direction of =
Example: Find the directional derivative of 1/r in the direction where
Sol. Here Now, And We know that So that Now, Directional derivative =
Example: Find the directional derivative of At the points (3, 1, 2) in the direction of the vector .
Sol. Here it is given that Now at the point (3, 1, 2)
Let be the unit vector in the given direction, then at (3, 1, 2) Now,
Example: Find the directional derivatives of at the point P(1, 1, 1) in the direction of the line Sol. Here Direction ratio of the line are 2, 2, 1 Now directions cosines of the line are Which are Directional derivative in the direction of the line 
Key takeaways

References
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