MTH301 ASSIGNMENT NO. 1 FALL 2022 || 100% RIGHT SOLUTION || CALCULUS - II || BY VuTech

MTH301 ASSIGNMENT NO. 1 FALL 2022

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MTH301 ASSIGNMENT NO. 1 FALL 2022 || 100% RIGHT SOLUTION || CALCULUS - II || BY VuTech

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Assignment No. 1                           MTH301 (Fall 2022)

Total Marks: 20

Due Date: 30^th November, 2022

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Question No.1

Check whether the given function is continuous or not at the point (0, 0) .

 

Solution:

We know that A function f of two variables is called continuous at the point `(x_0 , y_­­0)` if

 

`f(x_0 , y_­­0)`  is defined.

` lim_{(x,y) \to (x_0,y_0)} f(x,y)`exists.

` lim_{(x,y) \to (x_0,y_0)} f(x,y) = f(x_0,y_0)`

To determine ,  if f is continuous at (0,0), we need to compare ` lim_{(x,y) \to (x_0,y_0)} f(x,y) ` to  `f(x_0,y_0)`

According to the definition of the function f, we find that

`f(0,0) = 1`

we now consider the limit ` lim_{(x,y) \to (x_0,y_0)} f(x,y)` of the function

`f(x,y) = \frac{x^2 + y^2 - x^2y^2}{x^2 + y^2}`

Let (x,y) approaches to the line x, where y=0

`f(x,y) = \frac{x^2 + 0^2 - x^20^2}{x^2 + 0^2}`

`f(x,y) = \frac{x^2 + 0 - 0}{x^2 + 0}`

`f(x,y) = \frac{x^2}{x^2}`

`f(x,y) = 1                 -  -  -  - (1)`

Let (x,y) approaches to the line y, where x=0

`f(x,y) = \frac{0^2 + y^2 - 0^2y^2}{0^2 + y^2}`

`f(x,y) = \frac{0 + y^2 - 0}{0 + y^2}`

`f(x,y) = \frac{y^2}{y^2}`

`f(x,y) = 1           -  -  -  -  - ( 2 )`

Let (x,y) approaches to the line, where y=x

`f(x,y) = \frac{x^2 + x^2 - x^2x^2}{x^2 + x^2}`

`f(x,y) = \frac{2x^2 - x^4}{2x^2}`

`f(x,y) = \frac{2x^2}{2x^2} - \frac{x^4}{2x^2}`

`f(x,y) = 1 - \frac{x^2}{2}`

By putting `f(x_0 , y_­­0) = (0,0)`

`f(x,y) = 1 - \frac{0^2}{2}`

`f(x,y) = 1           -  -  -  -  - ( 3 )`

From Equation (1),(2) and (3),

It is proved that the limit of the Above function exists and is equal to 1.

`f(x_0 , y_­­0) = 1`

`lim_{(x,y) \to (x_0,y_0} \right)  f(x,y) = 1`

`lim_{(x,y) \to (x_0,y_0} \right)  f(x,y) =f(x_0 , y_­­0) = 1`

Hence,

Its is proved that the given function 

 

is continuous or not at the point (0,0)

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Question No.2

Let  `f(x,y) = xy^2 – x^2`, where `x = t^3` and `y = 3t^2` . Use chain rule to find `\frac{df}{dt}` and express the answer in variable t only

Solution:

Here `f(x,y) = xy^2 – x^2 ,                         x = t^3 and y = 3t^2`

Now we have to find                    

`\frac{df}{dt}` in variable t only

We know that

`\frac{df}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt} -  -  -  -  -  -  -  -  - (1)`

`\frac{dx}{dt} = \frac{d}{dt}(t^3) = 3t^2 -  -  -  -  - (i)`

`\frac{dy}{dt} = \frac{d}{dt}(3t^2) = 6t -  -  -  -  - (ii)`

`\frac{\partial f}{\partial x} = \frac{\partial }{\partial x}f(x,y) = \frac{\partial }{\partial x}(xy^2 - x^2) = \frac{\partial }{\partial x}(xy^2) - \frac{\partial }{\partial x}(x^2) = y^2 - 2x -  -  -  -  - (iii)`

`\frac{\partial f}{\partial y} = \frac{\partial }{\partial y}f(x,y\right) = \frac{\partial }{\partial y}(xy^2 - x^2) = \frac{\partial }{\partial y}(xy^2) - \frac{\partial }{\partial y}(x^2) = 2xy - 0 = 2xy -  -  -  - (iv)`

From equation (i),(ii),(iii) and (iv), (1) becomes

`\frac{df}{dt} = (y^2 - 2x)(3t^2)(2xy)(6t)`

Now by putting the value of x and y,

`\frac{df}{dt} = ((3t^2)^2 - 2(t^3))(3t^2) + (2(t^3)(3t^2))(6t)`

`\frac{df}{dt} = (3t^2+2 - 2t^3)(3t^2) + ({6t^{3 + 2})(6t)`

`\frac{df}{dt} = (3t^4 - 2t^3)(3t^2) + (6t^5)(6t)`

`\frac{df}{dt} = (9t^{4 + 2} - 6t^{3 + 2}) + (36t^{5 + 1})`

`\frac{df}{dt} = (9t^6 - 6t^5) + (36t^6)`

`\frac{df}{dt} = 9t^6 - 6t^5 + 36t^6`

`\frac{df}{dt} = 45t^6 - 6t^5`

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