Chapter 14: Problem 65
Find the value of \(\partial z / \partial x\) at the point \((1,1,1)\) if the equation $$x y+z^{3} x-2 y z=0$$ defines \(z\) as a function of the two independent variables \(x\) and \(y\) and the partial derivative exists.
Short Answer
Expert verified
The value of \( \partial z / \partial x \) at the point \((1,1,1)\) is -2.
Step by step solution
01
Differentiate the function with respect to x
Differentiate the equation \(x y + z^3 x - 2 y z = 0\) partially with respect to \(x\). Use the implicit differentiation technique. Remember \(z\) is a function of \(x\) and \(y\), so whenever we differentiate terms involving \(z\), apply the chain rule.The differentiation will result in:\[ y + z^3 + x \cdot 3z^2 \frac{\partial z}{\partial x} - 2y \frac{\partial z}{\partial x} = 0 \]
02
Solve for \(\partial z / \partial x\)
Isolate \(\partial z / \partial x\) on one side of the differentiated equation. Rearrange the equation:\[ y + z^3 = - \left(x \cdot 3z^2 \frac{\partial z}{\partial x} - 2y \frac{\partial z}{\partial x}\right) \]This simplifies to:\( y + z^3 = -\frac{\partial z}{\partial x} (x \cdot 3z^2 - 2y) \)Thus, we have:\[ \frac{\partial z}{\partial x} = \frac{-(y + z^3)}{x \cdot 3z^2 - 2y} \]
03
Evaluate at the point (1, 1, 1)
Substitute \(x = 1\), \(y = 1\), and \(z = 1\) into the expression for \(\frac{\partial z}{\partial x}\):\[ \frac{\partial z}{\partial x} = \frac{-(1 + 1^3)}{1 \cdot 3(1)^2 - 2(1)} \]Simplifying further:\[ \frac{\partial z}{\partial x} = \frac{-2}{3 - 2} = \frac{-2}{1} = -2 \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
Partial derivatives are a fundamental concept in multivariable calculus, allowing us to measure how a function changes as just one of the variables changes, keeping the others constant. Imagine you have a function \(z\) that is dependent on two or more independent variables like \(x\) and \(y\). To find the partial derivative of \(z\) with respect to \(x\), denoted as \(\partial z / \partial x\), you are examining how \(z\) changes as you make small changes in \(x\), while \(y\) is held steady.
- Partial derivatives help in analyzing multivariable functions.
- They are denoted by \(\partial\) and not \(d\), to highlight changes in one variable.
- Partial derivative computations often involve implicit differentiation and the chain rule when the function is implicitly defined.
Implicit Differentiation
Implicit differentiation is a technique to find derivatives when the function is not explicitly solved for one of the variables. It's handy when dealing with equations that intertwine variables, like \(xy + z^3x - 2yz = 0\). Instead of trying to rewrite the equation to express \(z\) as \(z = f(x, y)\) and then differentiate, implicit differentiation simplifies the task by treating \(z\) as a function of both \(x\) and \(y\).
- Involves differentiating each term considering \(z\) as a dependent variable.
- Apply partial differentiation across terms that depend on \(x\).
- Needing to apply the chain rule for terms involving derivatives of \(z\).
Chain Rule
The chain rule is crucial in differentiating composite functions and, in our context, when a function is not directly dependent on a variable but indirectly through other variable relationships. When a function like \(z\) in terms of \(x\) and \(y\) isn't isolated, the chain rule becomes indispensable in finding the rate of change through intermediary steps.
- Used when differentiating a function that involves another function.
- Key in problems involving implicit differentiation.
- Looks complex but simplifies differentiation by breaking it into parts.