Math Problem Statement
do only the third ijne with details
Solution
We will focus on the third line: calculating for the function
\begin{cases} \frac{\cos\left(\sqrt{x^2 + y^2 - z^2}\right) - 1}{y}, & \text{if } y \neq 0, \\ 0, & \text{if } y = 0. \end{cases}$$ --- ### Step-by-Step Solution: #### 1. Gradient Definition: The gradient of $$f(x, y, z)$$ is: $$\nabla f(x, y, z) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right).$$ We compute each partial derivative separately at the point $$(0, 0, 0)$$. --- #### 2. Partial Derivative with Respect to $$x$$: When $$y \neq 0$$: $$f(x, y, z) = \frac{\cos\left(\sqrt{x^2 + y^2 - z^2}\right) - 1}{y}.$$ The derivative with respect to $$x$$ is: $$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left( \frac{\cos\left(\sqrt{x^2 + y^2 - z^2}\right) - 1}{y} \right).$$ Using the chain rule: $$\frac{\partial f}{\partial x} = \frac{-\sin\left(\sqrt{x^2 + y^2 - z^2}\right) \cdot \frac{x}{\sqrt{x^2 + y^2 - z^2}}}{y}.$$ At $$(0, 0, 0)$$, $$y = 0$$, so the limit needs to be considered carefully. From the definition of $$f(x, y, z)$$, if $$y = 0$$, $$f(x, y, z) = 0$$. Thus: $$\frac{\partial f}{\partial x}(0, 0, 0) = 0.$$ --- #### 3. Partial Derivative with Respect to $$y$$: For $$y \neq 0$$, we compute: $$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left( \frac{\cos\left(\sqrt{x^2 + y^2 - z^2}\right) - 1}{y} \right).$$ Using the quotient rule: $$\frac{\partial f}{\partial y} = \frac{y \cdot \frac{\partial}{\partial y} \left( \cos\left(\sqrt{x^2 + y^2 - z^2}\right) - 1 \right) - \left(\cos\left(\sqrt{x^2 + y^2 - z^2}\right) - 1\right)}{y^2}.$$ The derivative of $$\cos\left(\sqrt{x^2 + y^2 - z^2}\right)$$ with respect to $$y$$ is: $$-\sin\left(\sqrt{x^2 + y^2 - z^2}\right) \cdot \frac{y}{\sqrt{x^2 + y^2 - z^2}}.$$ So: $$\frac{\partial f}{\partial y} = \frac{y \cdot \left( -\sin\left(\sqrt{x^2 + y^2 - z^2}\right) \cdot \frac{y}{\sqrt{x^2 + y^2 - z^2}} \right) - \left(\cos\left(\sqrt{x^2 + y^2 - z^2}\right) - 1\right)}{y^2}.$$ At $$(0, 0, 0)$$, we analyze the limit of this expression. Since $$f(x, y, z) = 0$$ when $$y = 0$$, we find: $$\frac{\partial f}{\partial y}(0, 0, 0) = 0.$$ --- #### 4. Partial Derivative with Respect to $$z$$: For $$y \neq 0$$: $$\frac{\partial f}{\partial z} = \frac{\partial}{\partial z} \left( \frac{\cos\left(\sqrt{x^2 + y^2 - z^2}\right) - 1}{y} \right).$$ Using the chain rule: $$\frac{\partial f}{\partial z} = \frac{-\sin\left(\sqrt{x^2 + y^2 - z^2}\right) \cdot \frac{-z}{\sqrt{x^2 + y^2 - z^2}}}{y}.$$ At $$(0, 0, 0)$$, when $$y = 0$$, $$f(x, y, z) = 0$$. Thus: $$\frac{\partial f}{\partial z}(0, 0, 0) = 0.$$ --- ### Final Result: $$\nabla f(0, 0, 0) = (0, 0, 0).$$ --- Would you like further clarifications or details on the calculations? --- ### Related Questions: 1. Why does $$f(0, 0, 0) = 0$$ affect the continuity of the function? 2. How can we rigorously compute limits for multivariable functions? 3. What is the role of the chain rule in the gradient computation? 4. Why do we analyze cases where $$y = 0$$ separately? 5. How does the quotient rule apply to partial derivatives? ### Tip: Always verify the continuity of a piecewise-defined function before computing derivatives.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Partial derivatives
Multivariable calculus
Gradient
Formulas
f(x, y, z) = \begin{cases} \frac{\cos(\sqrt{x^2 + y^2 - z^2}) - 1}{y}, & \text{if } y \neq 0, \\ 0, & \text{if } y = 0 \end{cases}
Gradient formula: \nabla f = (\partial f / \partial x, \partial f / \partial y, \partial f / \partial z)
Chain rule for derivatives
Theorems
Definition of partial derivatives
Continuity and differentiability in multivariable functions
Suitable Grade Level
Undergraduate level (Calculus 3 or higher)
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