Problem 85 - Calculus Early Transcendentals

Question

Use the gradient rules of Exercise 81 to find the gradient of the following functions. $$f(x, y, z)=\sqrt{25-x^{2}-y^{2}-z^{2}}$$

Step-by-Step Solution

Answer

Question: Calculate the gradient of the function $$f(x, y, z)=\sqrt{25-x^{2}-y^{2}-z^{2}}$$. Answer: The gradient of the function f(x, y, z) is $$\nabla f = \left\langle \frac{-x}{\sqrt{25-x^2-y^2-z^2}}, \frac{-y}{\sqrt{25-x^2-y^2-z^2}}, \frac{-z}{\sqrt{25-x^2-y^2-z^2}} \right\rangle$$.

Step 1: Calculate the partial derivative with respect to x.

To find the partial derivative of f(x, y, z) with respect to x, we will differentiate the function with respect to x while treating y and z as constants. This is represented as: $$\frac{\partial f}{\partial x}$$ Using the chain rule, we have: $$\frac{\partial f}{\partial x} = \frac{-x}{\sqrt{25-x^2-y^2-z^2}}$$

Step 2: Calculate the partial derivative with respect to y.

Next, we will find the partial derivative of f(x, y, z) with respect to y. We differentiate the function with respect to y while treating x and z as constants. This is represented as: $$\frac{\partial f}{\partial y}$$ Using the chain rule, we obtain: $$\frac{\partial f}{\partial y} = \frac{-y}{\sqrt{25-x^2-y^2-z^2}}$$

Step 3: Calculate the partial derivative with respect to z.

Lastly, we will find the partial derivative of f(x, y, z) with respect to z. Differentiate the function with respect to z while treating x and y as constants. This is represented as: $$\frac{\partial f}{\partial z}$$ Using the chain rule, we have: $$\frac{\partial f}{\partial z} = \frac{-z}{\sqrt{25-x^2-y^2-z^2}}$$

Step 4: Determine the gradient vector.

The gradient vector is formed by combining the partial derivatives that we calculated in the previous steps. The gradient of the function f(x, y, z) is: $$\nabla f = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right\rangle$$ Substitute the partial derivatives from Steps 1, 2, and 3: $$\nabla f = \left\langle \frac{-x}{\sqrt{25-x^2-y^2-z^2}}, \frac{-y}{\sqrt{25-x^2-y^2-z^2}}, \frac{-z}{\sqrt{25-x^2-y^2-z^2}} \right\rangle$$