https://file.aiquickdraw.com/mathbot/file/172767775543388g3qdvo.jpg

This problem involves understanding the relationship between the gradient of a function \( f(x, y) \) and its level curves, as well as partial derivatives. Let's break it down based on what is shown in the image:

### The diagram and level curves:
- The figure represents level curves of \( f(x, y) = c \), where the values of \( f \) are marked on the contours.
- The contour lines represent locations where \( f(x, y) \) is constant, and the values on these lines are given by the numbers 1, 2, 3, and 4.

### The gradient vector \( \nabla f \):
- The gradient of \( f \), \( \nabla f \), points in the direction of the greatest increase of \( f \) and is perpendicular to the level curves. 
- Its magnitude represents the rate of change of \( f \) in that direction.

### Analysis of the given parts:

#### (a) The value of \( \nabla f \cdot \hat{i} \) at point \( P \):
- \( \hat{i} \) is the unit vector in the \( x \)-direction. 
- The gradient \( \nabla f \) at \( P \) will be perpendicular to the level curves. Since \( P \) lies between level curves 2 and 3, we can see that \( \nabla f \) points somewhat diagonally upwards (towards the higher value 3).
- The dot product \( \nabla f \cdot \hat{i} \) measures the component of the gradient in the \( x \)-direction. At \( P \), the gradient points somewhat upwards and to the right, so it has a positive \( x \)-component.
- **Answer: Positive (This is marked as correct in the image).**

#### (b) The value of \( \nabla f \cdot \hat{j} \) at point \( P \):
- \( \hat{j} \) is the unit vector in the \( y \)-direction.
- The gradient at \( P \) points diagonally upwards (again towards the higher value 3), meaning it has a positive \( y \)-component.
- Therefore, \( \nabla f \cdot \hat{j} \) is positive.
- **Answer: Positive (Marked incorrectly as zero in the image).**

#### (c) \( \frac{\partial f}{\partial x} \) at \( Q \):
- \( Q \) is located near the contour for \( f = 4 \), and based on the diagram, the contours are almost vertical at \( Q \).
- Since the contours are vertical, the function \( f \) changes very little as we move horizontally, implying that \( \frac{\partial f}{\partial x} \) is close to zero.
- **Answer: Zero (Marked incorrectly as negative in the image).**

#### (d) \( \frac{\partial f}{\partial y} \) at \( Q \):
- The contours at \( Q \) are vertical, indicating that as we move in the \( y \)-direction (up or down), the function \( f \) changes significantly.
- Moving upward at \( Q \), \( f \) decreases from 4 to a lower value, indicating that \( \frac{\partial f}{\partial y} \) is negative.
- **Answer: Negative (Marked correctly in the image).**

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Would you like more details on any of these answers or have specific questions? Here are 5 related questions for further exploration:

1. What does the dot product of a gradient with a unit vector represent geometrically?
2. How does the spacing between level curves affect the magnitude of the gradient?
3. Why does \( \frac{\partial f}{\partial x} \) become zero when the contours are vertical?
4. Can the gradient \( \nabla f \) ever point in the direction of the level curve? Why or why not?
5. How would the partial derivative \( \frac{\partial f}{\partial y} \) change if the contours were horizontal at \( Q \)?

**Tip:** The gradient \( \nabla f \) is always perpendicular to level curves and points in the direction of the steepest ascent.

The figure below represents the level curves f(x, y) = c; the values of f on each curve are marked. In each of the following parts, decide whether the given quantity is positive, negative, or zero.

Explore how to determine the gradient and partial derivatives of a function using level curves. This example explains the concepts with detailed steps and answers for derivatives in the x and y directions, suitable for advanced high school or college students.

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