Answer :
Sure, let's go through each part of the question step-by-step.
### (a) Domain of [tex]\( f \)[/tex]
The domain of a function consists of all the possible input values (x, y) for which the function is defined.
For [tex]\( f(x, y) = -\sqrt{x^2 + y^2} \)[/tex], the expression inside the square root [tex]\( x^2 + y^2 \)[/tex] must be non-negative (since the square root of a negative number is not real).
However, [tex]\( x^2 + y^2 \geq 0 \)[/tex] for all real numbers [tex]\( x \)[/tex] and [tex]\( y \)[/tex]. Therefore, the domain of [tex]\( f \)[/tex] is all pairs of real numbers [tex]\( (x, y) \)[/tex].
[tex]\[ \text{Domain of } f = \{(x, y) \in \mathbb{R}^2\} \][/tex]
or more simply:
[tex]\[ \text{Domain of } f = \mathbb{R}^2 \][/tex]
### (b) Range of [tex]\( f \)[/tex]
The range of a function consists of all the possible output values of the function.
Considering [tex]\( f(x, y) = -\sqrt{x^2 + y^2} \)[/tex], note that [tex]\( \sqrt{x^2 + y^2} \)[/tex] is always non-negative.
[tex]\[ \sqrt{x^2 + y^2} \geq 0 \][/tex]
Therefore, [tex]\( - \sqrt{x^2 + y^2} \)[/tex] is always non-positive.
[tex]\[ -\sqrt{x^2 + y^2} \leq 0 \][/tex]
The smallest value of [tex]\(\sqrt{x^2 + y^2}\)[/tex] is 0 (which happens when [tex]\(x = 0\)[/tex] and [tex]\(y = 0\)[/tex]), so the largest value of [tex]\( -\sqrt{x^2 + y^2} \)[/tex] is 0.
Hence, the range of [tex]\( f \)[/tex] is:
[tex]\[ \text{Range of } f = (-\infty, 0] \][/tex]
### (c) Sketch the level curve of [tex]\( f(x, y) = c \)[/tex], for [tex]\( c = 0, 1, 2 \)[/tex]
Level curves are found by setting [tex]\( f(x, y) = c \)[/tex]:
[tex]\[ - \sqrt{x^2 + y^2} = c \][/tex]
This can be rewritten as:
[tex]\[ \sqrt{x^2 + y^2} = -c \][/tex]
Since [tex]\( \sqrt{x^2 + y^2} \)[/tex] is non-negative and [tex]\( -c \)[/tex] is non-positive, [tex]\( -c \)[/tex] must be non-negative. This forces [tex]\( c \)[/tex] to be non-positive: [tex]\( c \leq 0 \)[/tex].
- For [tex]\( c = 0 \)[/tex]:
[tex]\[ \sqrt{x^2 + y^2} = 0 \implies x^2 + y^2 = 0 \implies x = 0 \text{ and } y = 0 \][/tex]
This is just the origin point [tex]\( (0,0) \)[/tex].
- For [tex]\( c = 1 \)[/tex]:
[tex]\[ \sqrt{x^2 + y^2} = -1 \implies \text{No solution, since } \sqrt{x^2 + y^2} \geq 0 \][/tex]
Thus, there is no level curve for [tex]\( c = 1 \)[/tex].
- For [tex]\( c = 2 \)[/tex]:
[tex]\[ \sqrt{x^2 + y^2} = -2 \implies \text{No solution, since } \sqrt{x^2 + y^2} \geq 0 \][/tex]
Thus, there is no level curve for [tex]\( c = 2 \)[/tex].
Hence, the only level curve for the given values of [tex]\( c \)[/tex] that exists is the single point at the origin for [tex]\( c = 0 \)[/tex].
### (d) Sketch the graph of [tex]\( f \)[/tex]
The function [tex]\( f(x, y) = -\sqrt{x^2 + y^2} \)[/tex] represents a surface in three-dimensional space. To sketch this, remember:
- The function reaches its maximum value of 0 when [tex]\( x = 0 \)[/tex] and [tex]\( y = 0 \)[/tex].
- As [tex]\( x^2 + y^2 \)[/tex] increases, [tex]\( \sqrt{x^2 + y^2} \)[/tex] increases, making [tex]\( -\sqrt{x^2 + y^2} \)[/tex] decrease (become more negative).
This surface is a downward-opening cone with its tip at the origin (0,0,0). Every cross-section of this surface parallel to the [tex]\( xy \)[/tex]-plane (for fixed [tex]\( z \)[/tex]) is a circle.
### (e) Find [tex]\( f_x (x, y) \)[/tex] and [tex]\( f_y (x, y) \)[/tex]
To find the partial derivatives, we differentiate [tex]\( f(x, y) = -\sqrt{x^2 + y^2} \)[/tex] with respect to [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
[tex]\[ f_x (x, y) = \frac{\partial}{\partial x} \left( -\sqrt{x^2 + y^2} \right) \][/tex]
Using the chain rule:
[tex]\[ f_x (x, y) = - \frac{x}{\sqrt{x^2 + y^2}} \][/tex]
Similarly, for [tex]\( y \)[/tex]:
[tex]\[ f_y (x, y) = \frac{\partial}{\partial y} \left( -\sqrt{x^2 + y^2} \right) \][/tex]
Using the chain rule:
[tex]\[ f_y (x, y) = - \frac{y}{\sqrt{x^2 + y^2}} \][/tex]
Thus, the partial derivatives are:
[tex]\[ f_x (x, y) = - \frac{x}{\sqrt{x^2 + y^2}}, \quad f_y (x, y) = - \frac{y}{\sqrt{x^2 + y^2}} \][/tex]
These derivatives describe the rate of change of [tex]\( f \)[/tex] in the [tex]\( x \)[/tex] and [tex]\( y \)[/tex] directions respectively.
### (a) Domain of [tex]\( f \)[/tex]
The domain of a function consists of all the possible input values (x, y) for which the function is defined.
For [tex]\( f(x, y) = -\sqrt{x^2 + y^2} \)[/tex], the expression inside the square root [tex]\( x^2 + y^2 \)[/tex] must be non-negative (since the square root of a negative number is not real).
However, [tex]\( x^2 + y^2 \geq 0 \)[/tex] for all real numbers [tex]\( x \)[/tex] and [tex]\( y \)[/tex]. Therefore, the domain of [tex]\( f \)[/tex] is all pairs of real numbers [tex]\( (x, y) \)[/tex].
[tex]\[ \text{Domain of } f = \{(x, y) \in \mathbb{R}^2\} \][/tex]
or more simply:
[tex]\[ \text{Domain of } f = \mathbb{R}^2 \][/tex]
### (b) Range of [tex]\( f \)[/tex]
The range of a function consists of all the possible output values of the function.
Considering [tex]\( f(x, y) = -\sqrt{x^2 + y^2} \)[/tex], note that [tex]\( \sqrt{x^2 + y^2} \)[/tex] is always non-negative.
[tex]\[ \sqrt{x^2 + y^2} \geq 0 \][/tex]
Therefore, [tex]\( - \sqrt{x^2 + y^2} \)[/tex] is always non-positive.
[tex]\[ -\sqrt{x^2 + y^2} \leq 0 \][/tex]
The smallest value of [tex]\(\sqrt{x^2 + y^2}\)[/tex] is 0 (which happens when [tex]\(x = 0\)[/tex] and [tex]\(y = 0\)[/tex]), so the largest value of [tex]\( -\sqrt{x^2 + y^2} \)[/tex] is 0.
Hence, the range of [tex]\( f \)[/tex] is:
[tex]\[ \text{Range of } f = (-\infty, 0] \][/tex]
### (c) Sketch the level curve of [tex]\( f(x, y) = c \)[/tex], for [tex]\( c = 0, 1, 2 \)[/tex]
Level curves are found by setting [tex]\( f(x, y) = c \)[/tex]:
[tex]\[ - \sqrt{x^2 + y^2} = c \][/tex]
This can be rewritten as:
[tex]\[ \sqrt{x^2 + y^2} = -c \][/tex]
Since [tex]\( \sqrt{x^2 + y^2} \)[/tex] is non-negative and [tex]\( -c \)[/tex] is non-positive, [tex]\( -c \)[/tex] must be non-negative. This forces [tex]\( c \)[/tex] to be non-positive: [tex]\( c \leq 0 \)[/tex].
- For [tex]\( c = 0 \)[/tex]:
[tex]\[ \sqrt{x^2 + y^2} = 0 \implies x^2 + y^2 = 0 \implies x = 0 \text{ and } y = 0 \][/tex]
This is just the origin point [tex]\( (0,0) \)[/tex].
- For [tex]\( c = 1 \)[/tex]:
[tex]\[ \sqrt{x^2 + y^2} = -1 \implies \text{No solution, since } \sqrt{x^2 + y^2} \geq 0 \][/tex]
Thus, there is no level curve for [tex]\( c = 1 \)[/tex].
- For [tex]\( c = 2 \)[/tex]:
[tex]\[ \sqrt{x^2 + y^2} = -2 \implies \text{No solution, since } \sqrt{x^2 + y^2} \geq 0 \][/tex]
Thus, there is no level curve for [tex]\( c = 2 \)[/tex].
Hence, the only level curve for the given values of [tex]\( c \)[/tex] that exists is the single point at the origin for [tex]\( c = 0 \)[/tex].
### (d) Sketch the graph of [tex]\( f \)[/tex]
The function [tex]\( f(x, y) = -\sqrt{x^2 + y^2} \)[/tex] represents a surface in three-dimensional space. To sketch this, remember:
- The function reaches its maximum value of 0 when [tex]\( x = 0 \)[/tex] and [tex]\( y = 0 \)[/tex].
- As [tex]\( x^2 + y^2 \)[/tex] increases, [tex]\( \sqrt{x^2 + y^2} \)[/tex] increases, making [tex]\( -\sqrt{x^2 + y^2} \)[/tex] decrease (become more negative).
This surface is a downward-opening cone with its tip at the origin (0,0,0). Every cross-section of this surface parallel to the [tex]\( xy \)[/tex]-plane (for fixed [tex]\( z \)[/tex]) is a circle.
### (e) Find [tex]\( f_x (x, y) \)[/tex] and [tex]\( f_y (x, y) \)[/tex]
To find the partial derivatives, we differentiate [tex]\( f(x, y) = -\sqrt{x^2 + y^2} \)[/tex] with respect to [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
[tex]\[ f_x (x, y) = \frac{\partial}{\partial x} \left( -\sqrt{x^2 + y^2} \right) \][/tex]
Using the chain rule:
[tex]\[ f_x (x, y) = - \frac{x}{\sqrt{x^2 + y^2}} \][/tex]
Similarly, for [tex]\( y \)[/tex]:
[tex]\[ f_y (x, y) = \frac{\partial}{\partial y} \left( -\sqrt{x^2 + y^2} \right) \][/tex]
Using the chain rule:
[tex]\[ f_y (x, y) = - \frac{y}{\sqrt{x^2 + y^2}} \][/tex]
Thus, the partial derivatives are:
[tex]\[ f_x (x, y) = - \frac{x}{\sqrt{x^2 + y^2}}, \quad f_y (x, y) = - \frac{y}{\sqrt{x^2 + y^2}} \][/tex]
These derivatives describe the rate of change of [tex]\( f \)[/tex] in the [tex]\( x \)[/tex] and [tex]\( y \)[/tex] directions respectively.
