Answer :
To solve this problem, we want to find the line integrals of the gradient of function [tex]\( f(x, y) \)[/tex] over two different paths [tex]\( C \)[/tex].
### Part (a)
Path: A straight line from [tex]\( (0, 2) \)[/tex] to [tex]\( (2, 3) \)[/tex].
1. Understand the Problem: We're given values of the function [tex]\( f(x, y) \)[/tex] in a table at discrete points. We need to calculate the line integral of the gradient of [tex]\( f \)[/tex] along a straight path [tex]\( C \)[/tex].
2. Use the Fundamental Theorem of Line Integrals: The theorem states that for a conservative vector field, the line integral of a gradient over a path depends only on the values of the scalar potential function [tex]\( f \)[/tex] at the endpoints of the path:
[tex]\[
\int_C \operatorname{grad} f \cdot d \vec{r} = f(x_2, y_2) - f(x_1, y_1)
\][/tex]
3. Apply to the Points:
- Start Point: [tex]\((0, 2)\)[/tex]
- End Point: [tex]\((2, 3)\)[/tex]
4. Look Up Values in the Table:
- [tex]\( f(0, 2) = 60 \)[/tex]
- [tex]\( f(2, 3) = 59 \)[/tex]
5. Compute the Line Integral:
[tex]\[
\int_C \operatorname{grad} f \cdot d \vec{r} = f(2, 3) - f(0, 2) = 59 - 60 = -1
\][/tex]
### Part (b)
Path: A circle of radius 1 centered at [tex]\( (3, 1) \)[/tex] traversed counterclockwise.
1. Understand the Path: The path [tex]\( C \)[/tex] is a closed loop, specifically a circle.
2. Key Property of Conservative Fields: For a closed loop in a conservative vector field, the line integral is zero because the net change in the potential over a closed path is zero.
3. Conclusion for the Integral Over the Circle:
[tex]\[
\int_C \operatorname{grad} f \cdot d \vec{r} = 0
\][/tex]
### Summary
- For path (a), the line integral is [tex]\(-1\)[/tex].
- For path (b), the line integral is [tex]\(0\)[/tex].
These results are based on the properties of line integrals and the specific values given in the table for the function [tex]\( f(x, y) \)[/tex].
### Part (a)
Path: A straight line from [tex]\( (0, 2) \)[/tex] to [tex]\( (2, 3) \)[/tex].
1. Understand the Problem: We're given values of the function [tex]\( f(x, y) \)[/tex] in a table at discrete points. We need to calculate the line integral of the gradient of [tex]\( f \)[/tex] along a straight path [tex]\( C \)[/tex].
2. Use the Fundamental Theorem of Line Integrals: The theorem states that for a conservative vector field, the line integral of a gradient over a path depends only on the values of the scalar potential function [tex]\( f \)[/tex] at the endpoints of the path:
[tex]\[
\int_C \operatorname{grad} f \cdot d \vec{r} = f(x_2, y_2) - f(x_1, y_1)
\][/tex]
3. Apply to the Points:
- Start Point: [tex]\((0, 2)\)[/tex]
- End Point: [tex]\((2, 3)\)[/tex]
4. Look Up Values in the Table:
- [tex]\( f(0, 2) = 60 \)[/tex]
- [tex]\( f(2, 3) = 59 \)[/tex]
5. Compute the Line Integral:
[tex]\[
\int_C \operatorname{grad} f \cdot d \vec{r} = f(2, 3) - f(0, 2) = 59 - 60 = -1
\][/tex]
### Part (b)
Path: A circle of radius 1 centered at [tex]\( (3, 1) \)[/tex] traversed counterclockwise.
1. Understand the Path: The path [tex]\( C \)[/tex] is a closed loop, specifically a circle.
2. Key Property of Conservative Fields: For a closed loop in a conservative vector field, the line integral is zero because the net change in the potential over a closed path is zero.
3. Conclusion for the Integral Over the Circle:
[tex]\[
\int_C \operatorname{grad} f \cdot d \vec{r} = 0
\][/tex]
### Summary
- For path (a), the line integral is [tex]\(-1\)[/tex].
- For path (b), the line integral is [tex]\(0\)[/tex].
These results are based on the properties of line integrals and the specific values given in the table for the function [tex]\( f(x, y) \)[/tex].
