Problem 18
Question
Find the first partial derivatives of the function. $$f(x, t)=\sqrt{x} \ln t$$
Step-by-Step Solution
Verified Answer
\(\frac{\partial f}{\partial x} = \frac{1}{2\sqrt{x}} \ln t\) and \(\frac{\partial f}{\partial t} = \frac{\sqrt{x}}{t}\).
1Step 1: Understanding the Exercise
The goal is to find the first partial derivatives of the function \(f(x, t) = \sqrt{x} \ln t\). This means we need to determine how the function changes as each variable, \(x\) and \(t\), changes individually, while keeping the other variable constant.
2Step 2: Finding the Partial Derivative with Respect to x
To compute the partial derivative of \(f(x, t)\) with respect to \(x\), treat \(t\) as a constant. The derivative of \(\sqrt{x}\) is \(\frac{1}{2\sqrt{x}}\). Thus, the partial derivative is:\[\frac{\partial f}{\partial x} = \frac{1}{2\sqrt{x}} \ln t\].
3Step 3: Finding the Partial Derivative with Respect to t
To compute the partial derivative of \(f(x, t)\) with respect to \(t\), treat \(x\) as a constant. The derivative of \(\ln t\) is \(\frac{1}{t}\). Thus, the partial derivative is: \[\frac{\partial f}{\partial t} = \sqrt{x} \cdot \frac{1}{t} = \frac{\sqrt{x}}{t}\].
Key Concepts
Multivariable CalculusDerivativesFunctions of Several Variables
Multivariable Calculus
In multivariable calculus, we extend the concepts of single-variable calculus to functions that don't just depend on one variable, but on two or more. These are called functions of several variables. Such functions can be visualized as surfaces or even higher-dimensional shapes. For example, the function \(f(x, t) = \sqrt{x} \ln t\) is a function of two variables, which are \(x\) and \(t\). "Multivariable calculus" is crucial because it allows us to understand how a function behaves in environments where several factors interact.
- Functions of several variables: Functions that depend on more than one input, resulting often in more complex geometric interpretations.
- Extension of single-variable calculus: Allows for differentiation and integration in higher dimensions.
- Applications: Used in physics, engineering, economics, and any field where relationships between multiple factors need analysis.
Derivatives
Derivatives are central to calculus. They measure how a function changes as its input changes. In the context of single-variable functions, the derivative gives the slope of the tangent line to the curve at a given point. This concept extends into multivariable calculus with partial derivatives.
- Partial Derivatives: These are derivatives taken with respect to one variable while keeping other variables constant.
- Notation and calculation: For a function \(f(x, t) = \sqrt{x} \ln t\), the partial derivative with respect to \(x\) is written as \(\frac{\partial f}{\partial x}\), and similarly for \(t\).
- Geometric interpretation: Partial derivatives indicate the rate of change of the function in the direction of a given variable.
Functions of Several Variables
When dealing with functions of several variables, such as \(f(x, t)\), we study how multiple inputs affect the output. This could be likened to tuning two dials to achieve a desired outcome on a machine.
- Definition: Functions that depend on more than one variable, e.g., \(f(x, t)\).
- Behavior and interpretation: Examining how changing one variable while keeping others fixed affects the output.
- Practical relevance: Such functions are pervasive in analysis of real-world systems.
Other exercises in this chapter
Problem 18
\(17-20\) Use a tree diagram to write out the Chain Rule for the given case. Assume all functions are differentiable. $$\begin{array}{l}{R=f(x, y, z, t), \quad
View solution Problem 18
\(5-22\) Find the limit, if it exists, or show that the limit does not exist. $$\lim _{(x, y) \rightarrow(0,0)} \frac{x y^{4}}{x^{2}+y^{8}}$$
View solution Problem 18
Find and sketch the domain of the function. $$f(x, y)=\arcsin \left(x^{2}+y^{2}-2\right)$$
View solution Problem 19
Show that \(f(x, y)=x^{2}+4 y^{2}-4 x y+2\) has an infinite number of critical points and that \(D=0\) at each one. Then show that \(f\) has a local (and absolu
View solution