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Chapter 14: Problem 36

Find the partial derivative of the function with respect to each variable. \(g(u, v)=v^{2} e^{(2 u / v)}\)

Short Answer

Expert verified
\( \frac{\partial g}{\partial u} = 2v e^{(2u/v)} \) and \( \frac{\partial g}{\partial v} = (2v-2u) e^{(2u/v)} \).

Step by step solution

01

Identify the Function Components

First, we understand the function given: \(g(u, v) = v^2 e^{\left( \frac{2u}{v} \right)}\). This function is a combination of polynomial and exponential expressions.
02

Differentiate with Respect to \(u\)

To find \( \frac{\partial g}{\partial u} \), treat \(v\) as a constant and differentiate the exponential part. \[ \frac{\partial g}{\partial u} = v^2 \cdot \frac{d}{du} \left(e^{\left(\frac{2u}{v}\right)} \right) = v^2 \cdot e^{\left(\frac{2u}{v}\right)} \cdot \frac{2}{v} \] Simplifying gives: \[ \frac{\partial g}{\partial u} = 2v e^{\left(\frac{2u}{v}\right)} \]
03

Differentiate with Respect to \(v\)

To find \( \frac{\partial g}{\partial v} \), use the product rule as the function involves both powers of \(v\) and an exponential term depending on \(v\). The product rule is \( (uv)' = u'v + uv' \). Let \( u = v^2 \) and \( v = e^{(2u/v)} \): - Differentiate \(u = v^2\): \(u' = 2v\) - Differentiate \(v = e^{(2u/v)}\): - Perform substitution: \( v = e^{f(v)} \) where \( f(v) = \frac{2u}{v} \). - Then, \( v' = e^{f(v)} \left( \frac{-2u}{v^2} \right) \).Apply the product rule:\[ \frac{\partial g}{\partial v} = (2v) \cdot e^{\left(\frac{2u}{v}\right)} + v^2 \cdot e^{\left(\frac{2u}{v}\right)} \cdot \left( \frac{-2u}{v^2} \right) \]Simplify:\[ \frac{\partial g}{\partial v} = 2v e^{\left(\frac{2u}{v}\right)} - 2u e^{\left(\frac{2u}{v}\right)} \]Thus: \[ \frac{\partial g}{\partial v} = (2v - 2u) e^{\left(\frac{2u}{v}\right)} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Multivariable Calculus
Multivariable calculus deals with functions that have more than one variable. In this case, the function is \( g(u, v) = v^2 e^{(2u/v)} \). Understanding how variables interact in multivariable functions is important because each variable can influence the outcome in different ways.
For example, to find how \( u \) influences \( g \), we fix \( v \) and look at the change. For \( v \), we fix \( u \) and examine its influence. This idea is at the heart of partial derivatives, which show how a function changes as one of its variables changes while keeping others constant.
Overall, multivariable calculus provides powerful tools for understanding and visualizing functions that depend on multiple inputs, allowing us to explore a wide range of real-world problems in science and engineering.
Exponential Functions
Exponential functions are mathematical expressions with a constant base raised to a variable exponent. In our function \( g(u, v) = v^2 e^{(2u/v)} \), the term \( e^{(2u/v)} \) is an exponential expression.
Exponential functions have unique properties that make differentiation straightforward but sometimes tricky in multivariable settings. They grow rapidly and are key in describing growth processes, like populations or radioactive decay.
More so, when differentiating exponential functions like \( e^{(x)} \), the derivative remains \( e^{(x)} \), which simplifies the differentiation process. That's why for \( e^{(2u/v)} \), that same form appears in every partial derivative we compute.
Product Rule
In calculus, the product rule is essential when differentiating products of functions. It states that if you have \( u(x) \cdot v(x) \), the derivative is \( u'v + uv' \).
In our example \( g(u, v) = v^2 e^{(2u/v)} \), we need to take partial derivatives with respect to \( v \), which involves a product of functions. Here, \( v^2 \) and \( e^{(2u/v)} \) are two intertwining components of a product, each dependent on \( v \).
By applying the product rule:
  • Differentiating \( v^2 \) with respect to \( v \) gives \( 2v \).
  • Differentiating \( e^{(2u/v)} \) involves applying the chain rule, resulting in \( e^{(2u/v)} \cdot (-2u/v^2) \).
The result is a harmonious blend of these derivatives applied through the product rule, leading to a clear and efficient solution.
Differentiation Steps
Differentiation is a foundational tool in calculus used to determine how a function's value changes as its inputs change. Tackling the function \( g(u, v) = v^2 e^{(2u/v)} \), the differentiation process involves several steps to find partial derivatives.
First, identify the function components to understand which rules apply. Then, for the partial derivative with respect to \( u \), treat \( v \) as a constant, simplifying the derivative to \( 2v e^{(2u/v)} \).
For \( v \), utilize the product rule since \( g(u, v) \) combines power and exponential functions based on \( v \). This leads to breakdown steps:
  • Differentiate each component individually.
  • Apply rules like the product and chain rule appropriately.
  • Simplify the expression.
Following these steps ensures you understand each differentiation aspect, making it easier to handle complex multivariable functions.

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Most popular questions from this chapter

In Exercises \(43-46,\) find the linearization \(L(x, y, z)\) of the function \(f(x, y, z)\) at \(P_{0}\) . Then find an upper bound for the magnitude of the error \(E\) in the approximation \(f(x, y, z) \approx L(x, y, z)\) over the region \(R\) . $$ \begin{array}{l}{f(x, y, z)=x^{2}+x y+y z+(1 / 4) z^{2} \quad \text { at } \quad P_{0}(1,1,2)} \\ {R : \quad|x-1| \leq 0.01, \quad|y-1| \leq 0.01, \quad|z-2| \leq 0.08}\end{array} $$

The Wilson lot size formula The Wilson lot size formula in economics says that the most economical quantity \(Q\) of goods (radios, shoes, brooms, whatever) for a store to order is given by the formula \(Q=\sqrt{2 K M / h},\) where \(K\) is the cost of placing the order, \(M\) is the number of items sold per week, and \(h\) is the weekly holding cost for each item (cost of space, utilities, security, and so on). To which of the variables \(K, M,\) and \(h\) is \(Q\) most sensitive near the point \(\left(K_{0}, M_{0}, h_{0}\right)=(2,20,0.05) ?\) Give reasons for your answer.

Use a CAS to perform the following steps implementing the method of Lagrange multipliers for finding constrained extrema: a. Form the function \(h=f-\lambda_{1} g_{1}-\lambda_{2} g_{2},\) where \(f\) is the function to optimize subject to the constraints \(g_{1}=0\) and \(g_{2}=0 .\) b. Determine all the first partial derivatives of \(h\) , including the partials with respect to \(\lambda_{1}\) and \(\lambda_{2},\) and set them equal to \(0 .\) c. Solve the system of equations found in part (b) for all the unknowns, including \(\lambda_{1}\) and \(\lambda_{2} .\) d. Evaluate \(f\) at each of the solution points found in part (c) and select the extreme value subject to the constraints asked for in the exercise. Minimize \(f(x, y, z)=x^{2}+y^{2}+z^{2}\) subject to the constraints \(x^{2}-x y+y^{2}-z^{2}-1=0\) and \(x^{2}+y^{2}-1=0\)

a. Maximum on line of intersection Find the maximum value of \(w=x y z\) on the line of intersection of the two planes \(x+y+z=40\) and \(x+y-z=0\) b. Give a geometric argument to support your claim that you have found a maximum, and not a minimum, value of \(w\) .

Each of Exercises \(59-62\) gives a function \(f(x, y)\) and a positive number \(\epsilon .\) In each exercise, show that there exists a \(\delta>0\) such that for all \((x, y),\) $$ \sqrt{x^{2}+y^{2}}<\delta \Rightarrow|f(x, y)-f(0,0)|<\epsilon $$ $$ f(x, y)=x^{2}+y^{2}, \quad \epsilon=0.01 $$
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