**Problem Statement:**1. **Find the gradient of $ f(x, y) = \frac{14 - x^2 - y^2}{9} $.** ![Placeholder for answer]2. **Find the gradient at point $ P(1, 4) $.** ![Placeholder for answer]**Explanation:**- The gradient of a function $ f(x, y) $ is a vector containing the partial derivatives with respect to $ x $ and $ y $, respectively.- **Gradient Formula:** For a function $ f(x, y) $: \[ \nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \]- **Steps to Solve:** 1. Calculate the partial derivative of $ f $ with respect to $ x $. 2. Calculate the partial derivative of $ f $ with respect to $ y $. 3. Evaluate the gradient at the given point $ P(1, 4) $.- **Partial Derivatives Calculation:** For $ f(x, y) = \frac{14 - x^2 - y^2}{9} $, let's compute: 1. \[ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left(\frac{14 - x^2 - y^2}{9}\right) \] 2. \[ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left(\frac{14 - x^2 - y^2}{9}\right) \]- **Evaluating at Point $ P(1, 4) $:** Once the partial derivatives are found, substitute $ x = 1 $ and $ y = 4 $ into the gradient components to find the value at $ P(1, 4) $.**Note:** Graphs or diagrams are not provided in this image. The solution involves vector calculus concepts and manual computation of derivatives.

Answered: Image /qna-images/answer/1899e12e-6d83-4fbb-957e-8f711b1f8fcf.jpg

Answered: 14 – x2 – y2 Find the gradient of f(x,…

Math

Calculus

14 – x2 – y2 Find the gradient of f(x, y) Find the gradient at point P(1, 4).

Calculus For The Life Sciences

2nd Edition

ISBN:9780321964038

Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Chapter3: The Derivative

Section3.5: Graphical Differentiation

Problem 2E

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Question

**Problem Statement:**

1. **Find the gradient of $ f(x, y) = \frac{14 - x^2 - y^2}{9} $.**
![Placeholder for answer]

2. **Find the gradient at point $ P(1, 4) $.**
![Placeholder for answer]

**Explanation:**

- The gradient of a function $ f(x, y) $ is a vector containing the partial derivatives with respect to $ x $ and $ y $, respectively.
- **Gradient Formula:**

For a function $ f(x, y) $:
\[
\nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)
\]
- **Steps to Solve:**

1. Calculate the partial derivative of $ f $ with respect to $ x $.
2. Calculate the partial derivative of $ f $ with respect to $ y $.
3. Evaluate the gradient at the given point $ P(1, 4) $.

- **Partial Derivatives Calculation:**

For $ f(x, y) = \frac{14 - x^2 - y^2}{9} $, let's compute:

1. \[
\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left(\frac{14 - x^2 - y^2}{9}\right)
\]
2. \[
\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left(\frac{14 - x^2 - y^2}{9}\right)
\]
- **Evaluating at Point $ P(1, 4) $:**

Once the partial derivatives are found, substitute $ x = 1 $ and $ y = 4 $ into the gradient components to find the value at $ P(1, 4) $.

**Note:** Graphs or diagrams are not provided in this image. The solution involves vector calculus concepts and manual computation of derivatives.

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