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Chapter 3: Problem 17

Finding a Differential In Exercises \(11-20\) , find the differential \(d y\) of the given function. $$ y=\sqrt{9-x^{2}} $$

Short Answer

Expert verified
The differential of the function \(y=\sqrt{9-x^{2}}\) is \(dy = -\frac{x}{\sqrt{9 - x^2}} dx .\)

Step by step solution

01

Recognize the inner function

The inner function here is \(u = 9 - x^2\). Recognize that this function is part of a composition of functions.
02

Differentiate using chain rule

Differentiate the function using the chain rule. The chain rule is a formula to compute the derivative of a composite function. The chain rule states that the derivative of a function composition is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. The derivative of \(y = \sqrt{u}\) is \(\frac{1}{2\sqrt{u}}\) and the derivative of \(u = 9 - x^2\) is \(-2x\). By the chain rule, the derivative of \(y = \sqrt{9 - x^2}\) is \(\frac{1}{2\sqrt{9 - x^2}} \cdot -2x = -\frac{x}{\sqrt{9 - x^2}}\).
03

Compute the differential

The differential of the function \(y\), denoted \(dy\), is defined as the derivative of \(y\) with respect to \(x\), \(y'\), times \(dx\), that is, \(dy = y' dx\). Since we know that \(y'\) = -\(\frac{x}{\sqrt{9 - x^2}}\), then \(dy = -\frac{x}{\sqrt{9 - x^2}} dx .\)

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Key Concepts

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

Chain Rule
The Chain Rule is a fundamental concept in differential calculus that helps us find the derivative of composite functions. In simple terms, it allows us to differentiate complex functions by breaking them down into simpler parts.
  • When we encounter a composite function, look at the innermost function鈥攖his becomes your 'inner function'.
  • The outer function will be what remains when the inner function's effects are applied.
  • The Chain Rule says that you differentiate the outer function first and then multiply it by the derivative of the inner function.
This process helps untangle complex layers of functions. For example, in the given function, after identifying the inner function as \(u = 9 - x^2\), we find the derivative of the outer function \(y = \sqrt{u}\) and multiply it by the derivative of the inner function \(u\).
This gives us the complete derivative needed to find the differential.
Composite Function
A Composite Function is formed when one function is nested inside another. This means the output of one function becomes the input of the other.
  • For instance, if you have two functions, \(f(x)\) and \(g(x)\), the composite function could be \(f(g(x))\).
  • It's like doing one process, and then immediately applying a second process on the result of the first.
  • In our exercise, \(y = \sqrt{9 - x^2}\) is a composite function. Here, \(9 - x^2\) is the inner function, and applying the square root is the outer function.
Understanding composite functions is key in calculus because it allows us to handle and simplify complex expressions using the chain rule effectively. Once decomposed into these two layers, each function can be tackled individually which simplifies the differentiation process.
Derivatives
Derivatives measure how a function changes as its input changes. They are at the core of differential calculus, allowing us to understand rates of change and slopes of curves.
  • The derivative is often denoted as \(f'(x)\), \(\frac{dy}{dx}\), or simply \(y'\).
  • The act of differentiating a function involves finding its derivative concerning its input variable, usually \(x\).
  • Finding derivatives uses rules like the power rule, product rule, quotient rule, and importantly, the chain rule.
In the exercise, we differentiated the square root function using the chain rule to handle its composite nature. By identifying the derivative of both the inner and outer functions, we can accurately describe the rate at which \(y\) changes with \(x\). Derivatives are incredibly powerful for understanding the behavior of functions, especially when plotting them or determining maxima and minima.
Differential
The Differential, denoted as \(dy\), represents an infinitesimally small change in the function's output, corresponding to a small change \(dx\) in the input.
  • It is closely related to the derivative, as \(dy = f'(x) \, dx\).
  • The concept emphasizes the change and serves well in approximations.
  • In calculus problems, the differential helps in solving problems of rate of change and constructing slopes of tangent lines.
In the given problem, after finding the derivative \(-\frac{x}{\sqrt{9 - x^2}}\), the differential \(dy\) becomes \(-\frac{x}{\sqrt{9 - x^2}} \, dx\). This expresses how a tiny increase in \(x\) leads to a change in \(y\), serving as an essential tool for approximation and analysis in calculus.

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Most popular questions from this chapter

Comparing \(\Delta y\) and \(d y\) In Exercises \(7-10\) , use the information to evaluate and compare \(\Delta y\) and \(d y .\) $$ \begin{array}{ll}{\text { Function }} & {x \text { -Value }} \\ {y=6-2 x^{2}} & {x=-2}\end{array} \quad \Delta x=d x=0.1 $$

Numerical, Graphical, and Analytic Analysis An exercise room consists of a rectangle with a semicircle on each end. A 200 -meter running track runs around the outside of the room. (a) Draw a figure to represent the problem. Let \(x\) and \(y\) represent the length and width of the rectangle. (b) Analytically complete six rows of a table such as the one below. (The first two rows are shown.) Use the table to guess the maximum area of the rectangular region. $$ \begin{array}{|c|c|c|}\hline \text { Length, } x & {\text { Width, } y} & {\text { Area, } x y} \\ \hline 10 & {\frac{2}{\pi}(100-10)} & {(10) \frac{2}{\pi}(100-10) \approx 573} \\ \hline 20 & {\frac{2}{\pi}(100-20)} & {(20) \frac{2}{\pi}(100-20) \approx 1019} \\ \hline\end{array} $$ (c) Write the area \(A\) as a function of \(x\) . (d) Use calculus to find the critical number of the function in part (c) and find the maximum value. (e) Use a graphing utility to graph the function in part (c) and verify the maximum area from the graph.

Maximum Area Twenty feet of wire is to be used to form two figures. In each of the following cases, how much wire should be used for each figure so that the total enclosed area is maximum? (a) Equilateral triangle and square (b) Square and regular pentagon (c) Regular pentagon and regular hexagon (d) Regular hexagon and circle What can you conclude from this pattern? \(\\{\text {Hint: The }\) area of a regular polygon with \(n\) sides of length \(x\) is \(A=(n / 4)[\cot (\pi / n)] x^{2} . \\}\)

Volume and Surface Area The radius of a spherical balloon is measured as 8 inches, with a possible error of 0.02 inch. (a) Use differentials to approximate the possible propagated (a) Use differentials the volume of the sphere. (b) Use differentials to approximate the possible propagated error in computing the surface area of the sphere. (c) Approximate the percent errors in parts (a) and (b).

Minimum Distance In Exercises \(49-51\) , consider a fuel distribution center located at the origin of the rectangular coordinate system (units in miles; see figures). The center supplies three factories with coordinates \((4,1),(5,6),\) and \((10,3) .\) A trunk line will run from the distribution center along the line \(y=m x\) , and feeder lines will run to the three factories. The objective is to find \(m\) such that the lengths of the feeder lines are minimized. Minimize the sum of the squares of the lengths of the vertical feeder lines (see figure) given by $$S_{1}=(4 m-1)^{2}+(5 m-6)^{2}+(10 m-3)^{2}$$ Find the equation of the trunk line by this method and then determine the sum of the lengths of the feeder lines.
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