Problem 153
Question
Find \(\quad \lim _{\Delta y \rightarrow 0} \frac{f(x, y+\Delta y)-f(x, y)}{\Delta y}\) \(f(x, y)=-7 x-2 x y+7 y\)
Step-by-Step Solution
Verified Answer
The limit is \(-2x + 7\).
1Step 1: Identify the Function
The given function is \( f(x, y) = -7x - 2xy + 7y \). We are to find the limit of a difference quotient with respect to \( y \).
2Step 2: Determine \( f(x, y + \Delta y) \)
Calculate \( f(x, y + \Delta y) \) by substituting \( y + \Delta y \) into the function: \[f(x, y + \Delta y) = -7x - 2x(y + \Delta y) + 7(y + \Delta y). \]
3Step 3: Simplify \( f(x, y + \Delta y) \)
Expand \( f(x, y + \Delta y) \): \[ f(x, y + \Delta y) = -7x - 2xy - 2x\Delta y + 7y + 7\Delta y. \]
4Step 4: Compute the Difference Quotient
Find the numerator of the difference quotient: \[ f(x, y + \Delta y) - f(x, y) = (-7x - 2xy - 2x\Delta y + 7y + 7\Delta y) - (-7x - 2xy + 7y).\] Simplifying gives \[ -2x\Delta y + 7\Delta y.\] So, the difference quotient is \[ \frac{-2x\Delta y + 7\Delta y}{\Delta y}.\]
5Step 5: Simplify the Difference Quotient
Factor \( \Delta y \) out of the numerator: \[ \frac{\Delta y(-2x + 7)}{\Delta y}.\]Cancel \( \Delta y \) from numerator and denominator: \[ -2x + 7. \]
6Step 6: Evaluate the Limit
As \( \Delta y \to 0 \), the expression \(-2x + 7\) remains unchanged because it does not depend on \(\Delta y\). Thus, \[\lim _{\Delta y \to 0} \frac{f(x, y + \Delta y) - f(x, y)}{\Delta y} = -2x + 7.\]
Key Concepts
Difference QuotientMultivariable CalculusLimit of a Function
Difference Quotient
The difference quotient is a fundamental concept used to understand rates of change in calculus. In the context of multivariable calculus, it helps in finding partial derivatives, which measure how a function changes as one of its variables changes, while keeping other variables constant.
When you have a function of several variables, say two variables like in this exercise, the difference quotient with respect to one variable is expressed as:
When you have a function of several variables, say two variables like in this exercise, the difference quotient with respect to one variable is expressed as:
- \[ \frac{f(x, y + \Delta y) - f(x, y)}{\Delta y} \]
Multivariable Calculus
Multivariable calculus extends the concepts of calculus to functions with more than one variable. Unlike single-variable calculus, which focuses on derivatives and integrals of functions with one variable, multivariable calculus deals with functions like \( f(x, y) \) that depend on two or more variables.
This area of calculus introduces several new elements:
This area of calculus introduces several new elements:
- **Partial Derivatives:** Measures how a function changes as one variable changes, keeping all other variables constant.
- **Gradient:** A multidimensional extension of derivatives, pointing in the direction of greatest increase.
- **Differentials and Partial Derivatives:** Used to understand changes in multivariable functions with accuracy.
Limit of a Function
The limit of a function is a crucial concept in calculus, providing insight into the behavior of a function as its variables approach some value. When dealing with multivariable functions, the concept is essential for defining derivatives and continuity.
In our exercise, we want to find:
By evaluating this limit, we determine the exact partial derivative with respect to \( y \), which is a fundamental aspect of multivariable calculus.
The essence of taking this limit in the context of our problem allows us to understand the instantaneous rate of change in the function concerning one variable, capturing the precise slope or direction of change. Limits help us transition from average rates to exact, instantaneous changes, giving rise to the derivative.
In our exercise, we want to find:
- \[ \lim_{\Delta y \to 0} \frac{f(x, y + \Delta y) - f(x, y)}{\Delta y} \]
By evaluating this limit, we determine the exact partial derivative with respect to \( y \), which is a fundamental aspect of multivariable calculus.
The essence of taking this limit in the context of our problem allows us to understand the instantaneous rate of change in the function concerning one variable, capturing the precise slope or direction of change. Limits help us transition from average rates to exact, instantaneous changes, giving rise to the derivative.
Other exercises in this chapter
Problem 150
Show that \(f(x, y)=\ln \left(x^{2}+y^{2}\right)\) solves Laplace's equation \(\frac{\partial^{2} z}{\partial x^{2}}+\frac{\partial^{2} z}{\partial y^{2}}=0\)
View solution Problem 151
Show that \(z=e^{-t} \cos \left(\frac{x}{c}\right)\) satisfies the heat equation \(\frac{\partial z}{\partial t}=-e^{-t} \cos \left(\frac{x}{c}\right)\)
View solution Problem 154
Find \(\quad \lim _{\Delta x \rightarrow 0} \frac{\Delta f}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{f(x+\Delta x, y)-f(x, y)}{\Delta x}\) for \(f(x, y)=x
View solution Problem 155
Find \(\lim _{\Delta x \rightarrow 0} \frac{\Delta f}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{f(x+\Delta x, y)-f(x, y)}{\Delta x}\) for \(f(x, y)=\sin (x
View solution