Problem 129
Question
Evaluate the partial derivatives at point \(P(0,1)\) Find \(\frac{\partial z}{\partial x}\) at (0,1) for \(z=e^{-x} \cos (y)\).
Step-by-Step Solution
Verified Answer
The value of \( \frac{\partial z}{\partial x} \) at (0,1) is \(-\cos(1)\).
1Step 1: Identify the function for differentiation
The function given is \( z = e^{-x} \cos(y) \). We are asked to find the partial derivative of \( z \) with respect to \( x \) at the point \( P(0, 1) \).
2Step 2: Differentiate with respect to x
To find the partial derivative \( \frac{\partial z}{\partial x} \), differentiate the function \( z = e^{-x} \cos(y) \) with respect to \( x \). Treat \( y \) as a constant during this differentiation.The derivative of \( e^{-x} \) with respect to \( x \) is \( -e^{-x} \), and \( \cos(y) \) remains unchanged:\[\frac{\partial z}{\partial x} = -e^{-x} \cos(y).\]
3Step 3: Evaluate the partial derivative at P(0,1)
Substitute \( x = 0 \) and \( y = 1 \) into the partial derivative equation:\[\frac{\partial z}{\partial x} = -e^{0} \cos(1) = -1 \cdot \cos(1) = -\cos(1).\]This simplifies the expression completely.
Key Concepts
DifferentiationPartial Derivative EvaluationFunctions of Multiple Variables
Differentiation
Differentiation is a fundamental concept in calculus. It involves computing the derivative of a function, which measures how a function changes as its input changes. In essence, it gives the rate of change or the slope of the function at any given point. When working with single-variable functions, differentiation can be straightforward, as it involves just one independent variable. However, things can become more intricate when dealing with functions of multiple variables.
For functions of a single variable, such as \(f(x)\), differentiation provides the derivative \(f'(x)\). This tells us how \(f\) changes as \(x\) changes. The process of differentiation can uncover important properties of functions, such as identifying local maxima, minima, and points of inflection.
For functions of a single variable, such as \(f(x)\), differentiation provides the derivative \(f'(x)\). This tells us how \(f\) changes as \(x\) changes. The process of differentiation can uncover important properties of functions, such as identifying local maxima, minima, and points of inflection.
Partial Derivative Evaluation
Partial derivatives are used when we have functions of more than one variable. They represent the rate of change of the function with respect to one of the variables, holding the other variables constant. This is particularly useful in fields like physics and engineering, where systems often depend on multiple changing variables.
In the exercise, the function given is \( z = e^{-x} \cos(y) \), which depends on two variables, \(x\) and \(y\). The partial derivative \( \frac{\partial z}{\partial x} \) tells us how \( z \) changes as \( x \) changes, while keeping \( y \) constant. We can find this by differentiating \( z \) with respect to \( x \) and treating \( y \) as a constant.
As shown, the partial derivative \( \frac{\partial z}{\partial x} = -e^{-x} \cos(y) \) is evaluated at the point \((0,1)\), leading to the result \(-\cos(1)\).
In the exercise, the function given is \( z = e^{-x} \cos(y) \), which depends on two variables, \(x\) and \(y\). The partial derivative \( \frac{\partial z}{\partial x} \) tells us how \( z \) changes as \( x \) changes, while keeping \( y \) constant. We can find this by differentiating \( z \) with respect to \( x \) and treating \( y \) as a constant.
- The derivative of \( e^{-x} \) with respect to \( x \) is \( -e^{-x} \).
- \( \cos(y) \) is treated as a constant, so it remains unchanged during differentiation.
As shown, the partial derivative \( \frac{\partial z}{\partial x} = -e^{-x} \cos(y) \) is evaluated at the point \((0,1)\), leading to the result \(-\cos(1)\).
Functions of Multiple Variables
Functions of multiple variables are common in real-world scenarios. They involve inputs that can change independently, affecting the output in different ways. For example, the function \( z = e^{-x} \cos(y) \) has two variables, \(x\) and \(y\), which can both change and impact the value of \(z\).
Understanding these functions involves not only knowing how each variable individually affects the output but also how they might interact when both are changing. Partial derivatives help by providing a way to look at the influence of one variable at a time, fixing the others. This selective look at the variables is crucial for both theoretical and practical understanding.
By mastering the concept of functions with multiple variables, we can better model and predict outcomes in complex systems.
Understanding these functions involves not only knowing how each variable individually affects the output but also how they might interact when both are changing. Partial derivatives help by providing a way to look at the influence of one variable at a time, fixing the others. This selective look at the variables is crucial for both theoretical and practical understanding.
- In economics, these types of functions can represent cost structures dependent on different input factors.
- In engineering, they might model stresses on a component affected by varying forces.
By mastering the concept of functions with multiple variables, we can better model and predict outcomes in complex systems.
Other exercises in this chapter
Problem 127
Let \(f(x, y)=\arctan \left(\frac{y}{x}\right)\). Evaluate \(f_{x}(2,-2)\) and \(f_{y}(2,-2)\),
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Let \(f(x, y)=\frac{x y}{x-y}\). Find \(f_{x}(2,-2)\) and \(f_{y}(2,-2)\).
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The area of a parallelogram with adjacent side lengths that are \(a\) and \(b\), and in which the angle between these two sides is \(\theta,\) is given by the f
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Express the volume of a right circular cylinder as a function of two variables: a. its radius \(r\) and its height \(h\). b. Show that the rate of change of the
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