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Chapter 8: Problem 13

Find the partial derivatives in Problems. The variables are restricted to a domain on which the function is defined. $$\frac{\partial A}{\partial h} \text { if } A=\frac{1}{2}(a+b) h$$

Short Answer

Expert verified
\( \frac{\partial A}{\partial h} = \frac{1}{2}(a+b) \)

Step by step solution

01

Identify the function

The function given is \[ A = \frac{1}{2}(a+b)h \] where \(a\), \(b\), and \(h\) are variables.
02

Apply the definition of partial derivative

The partial derivative of a function with respect to a variable involves differentiating while keeping the other variables constant. Here, we take the partial derivative of \(A\) with respect to \(h\).
03

Differentiate with respect to h

Differentiate the function \(A = \frac{1}{2}(a+b)h\) with respect to \(h\), treating \(\frac{1}{2}(a+b)\) as a constant.\[ \frac{\partial A}{\partial h} = \frac{d}{dh} \left( \frac{1}{2}(a+b)h \right) = \frac{1}{2}(a+b) \times \frac{d}{dh}(h) \]
04

Simplify the derivative

Differentiate \(h\) with respect to \(h\), which gives 1. Therefore, the expression becomes:\[ \frac{\partial A}{\partial h} = \frac{1}{2}(a+b) \times 1 = \frac{1}{2}(a+b) \]

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Key Concepts

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

Differentiation
Differentiation is a fundamental concept in calculus involving finding how a function changes as its input changes. When you derive a function with respect to one of its variables, you're essentially understanding its rate of change or slope.
This process allows you to determine the function's behavior, like its increasing or decreasing trends. With functions of one variable, you apply basic rules such as the power rule, product rule, among others.
In the case of multivariable functions, partial differentiation focuses on differentiating one variable while treating others as constant. This technique is crucial in exploring and analyzing functions where multiple variables play a role.
Multivariable Functions
Multivariable functions involve multiple input variables impacting a single output. These functions are common in areas like physics, economics, and engineering where several factors influence a result. To work effectively with these functions, understanding their domain and range is vital. The domain refers to all possible input values, whereas the range is all possible output values these inputs can produce.
Partial derivatives, like in our exercise, help to dissect how each variable uniquely affects the function's result. This ability to isolate and comprehend the input's influence ensures detailed insights into complex systems. Each variable's derivative illustrates its contribution to the overall change, offering a fine-grained understanding of the function's dynamics.
Calculus Problem Solving
Solving calculus problems typically involves a structured approach:1. **Identify the Function**: Understand what you're working with, as we did with \( A = \frac{1}{2}(a+b)h \). - Clearly identifying variables and constants is essential.2. **Apply Relevant Calculus Concepts**: Here, partial differentiation was key. Recognizing the need for this allows you to tackle problems effectively.
3. **Evaluate the Derivative**: Perform differentiation according to rules, paying heed to constants and changing variables. Simplifying results is crucial to get final clear solutions.
Applying such procedures helps in dealing with real-world problems where multivariable functions frequently occur, making calculus a powerful tool for analysis and prediction.

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Most popular questions from this chapter

The power \(P\) produced by a windmill is proportional to the square of the diameter \(d\) of the windmill and to the cube of the speed \(v\) of the wind. \(^{5}\) (a) Write a formula for \(P\) as a function of \(d\) and \(v\) (b) A windmill generates \(100 \mathrm{kW}\) of power at a certain wind speed. If a second windmill is built having twice the diameter of the original, what fraction of the original wind speed is needed by the second windmill to produce \(100 \mathrm{kW} ?\) (c) Sketch a contour diagram for \(P\).

Use Lagrange multipliers to find the maximum or minimum values of \(f(x, y)\) subject to the constraint. $$f(x, y)=x^{2}+3 y^{2}+100, \quad 8 x+6 y=88$$

Find the partial derivatives in Problems. The variables are restricted to a domain on which the function is defined. $$f_{x} \text { and } f_{y} \text { if } f(x, y)=x^{2}+2 x y+y^{3}$$

For Problems calculate all four second-order partial derivatives and confirm that the mixed partials are equal. $$f(x, y)=x^{2} y$$

A steel manufacturer can produce \(P(K, L)\) tons of steel using \(K\) units of capital and \(L\) units of labor, with production costs \(C(K, L)\) dollars. With a budget of \(\$ 600,000,\) the maximum production is 2,500,000 tons, using \(\$ 400,000\) of capital and \(\$ 200,000\) of labor. The Lagrange multiplier is \(\lambda=3.17\) (a) What is the objective function? (b) What is the constraint? (c) What are the units for \(\lambda ?\) (d) What is the practical meaning of the statement \(\lambda=\) \(3.17 ?\)
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