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Chapter 2: Problem 23

(a) find two explicit functions by solving the equation for \(y\) in terms of \(x,(\) b) sketch the graph of the equation and label the parts differentiate the explicit functions, and (d) find \(d y / d x\) implicitly and show that the result is equivalent to that of part (c). \(16 y^{2}-x^{2}=16\)

Short Answer

Expert verified
The two explicit functions are \(y_1 = \frac{x}{4} + 1\) and \(y_2 = -\frac{x}{4} - 1\). The derivatives of the functions are \(\frac{dy_1}{dx} = \frac{1}{4}\) and \(\frac{dy_2}{dx} = -\frac{1}{4}\). These results are confirmed by the implicit differentiation of the original equation, which gives \(\frac{dy}{dx}=\frac{x}{16y}\).

Step by step solution

01

Solve the equation for \(y\) in terms of \(x\)

To solve the given equation \(16 y^{2}-x^{2}=16\) for \(y\), rearrange the equation to get: \(16y^{2} = x^{2} + 16\). To find \(y\), take the square root of both sides. To acknowledge both positive and negative roots, the solution can be written as: \(y = \pm\sqrt{\frac{x^{2} + 16}{16}}\) which simplifies to \(y = \pm\frac{x}{4} \pm 1\). So, the two explicit functions are \(y_1 = \frac{x}{4} + 1\) and \(y_2 = -\frac{x}{4} - 1\)
02

Sketch the graph of the equation

The graph consists of two parallel lines \(y_1 = \frac{x}{4} + 1\) and \(y_2 = -\frac{x}{4} - 1\). Both the lines are slanted; the first line slants upwards and the second line slants downwards. Additionally, both lines intercept the y-axis at \(\pm 1\). Label the parts correctly on the graph.
03

Differentiate the explicit functions

Next, take the derivative of each explicit function. For \(y_1 = \frac{x}{4} + 1\), its derivative is: \(\frac{dy_1}{dx} = \frac{1}{4}\). For \(y_2 = -\frac{x}{4} - 1\), its derivative is: \(\frac{dy_2}{dx} = -\frac{1}{4}\).
04

Find \(\frac{dy}{dx}\) implicitly

Differentiate the original equation \(16y^{2} - x^{2} = 16\) implicitly with respect to \(x\): \(\frac{d}{dx}(16y^{2}-x^{2}) = \frac{d}{dx} 16\). This gives: \(32y\frac{dy}{dx} - 2x = 0\). Solve for \(\frac{dy}{dx}\) to get \(\frac{dy}{dx}=\frac{x}{16y}\).Results found in this manner should be equivalent to results of Step 3 confirming that the expressions for \(\frac{dy_1}{dx}\) and \(\frac{dy_2}{dx}\) are indeed equivalent to \(\frac{dy}{dx}\).

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Key Concepts

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

Explicit Functions
Explicit functions are equations where the dependent variable is explicitly expressed in terms of the independent variable. In this exercise, the task is to express the variable \( y \) in terms of \( x \). By manipulating the original equation \( 16y^{2} - x^{2} = 16 \), you find the explicit functions as \( y_1 = \frac{x}{4} + 1 \) and \( y_2 = -\frac{x}{4} - 1 \). This means that for any given value of \( x \), we can easily calculate \( y \) using these formulas.
  • **Immediate Representation**: You can directly substitute \( x \) to find \( y \).
  • **Tiability**: It's easy to graph or perform further calculations.

Explicit functions simplify understanding of complex equations by converting them into a simplified format.
Parallel Lines
Parallel lines are lines that do not meet regardless of how far they are extended. In this solution, the two explicit functions represent parallel lines: \( y_1 = \frac{x}{4} + 1 \) and \( y_2 = -\frac{x}{4} - 1 \). These lines have the same slope, \( \pm\frac{1}{4} \), but different y-intercepts, ensuring they never touch each other.
  • **Unique Characteristic**: Parallel lines have the same slope.
  • **Graph Representation**: Each line moves in the same direction with a constant distance apart.

Parallel lines are crucial in geometry and algebra as they show consistent direction and separation despite changes in position.
Derivative
The derivative represents the rate at which a function is changing at any given point. It provides the slope of the tangent line to the curve of the function at that point. For the explicit functions \( y_1 \) and \( y_2 \), their derivatives are \( \frac{1}{4} \) and \( -\frac{1}{4} \) respectively. These derivatives tell us how steeply each line rises or falls.
  • **Slope Information**: Derivatives,\( \frac{dy_1}{dx} = \frac{1}{4} \) and \( \frac{dy_2}{dx} = -\frac{1}{4} \), provide the slope of each line.
  • **Comparative Analysis**: Helps in comparing changes in different functions.

Understanding derivatives is essential in illustrating how changes in one variable affect another.
Graph Sketching
Graph sketching involves plotting mathematical equations on a coordinate plane to visually understand their structure. In the task, the graph consists of two parallel lines \( y_1 = \frac{x}{4} + 1 \) and \( y_2 = -\frac{x}{4} - 1 \). Here's how you can graph them:
  • **Identify Y-intercepts**: Start plotting each line from their y-intercepts, \( 1 \) and \(-1 \), respectively.
  • **Consistent Slope**: Ensure that each line maintains the slope of \( \pm\frac{1}{4} \), slanting gently across quadrants.

Accurate graph sketching enables you to better visualize relationships and trends, solidifying understanding through conceptual imagery.

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Most popular questions from this chapter

$$\begin{array}{l}{\text { Determining Differentiability In Exercises }} \\\ {77-80, \text { describe the } x \text { -values at which } f \text { is }} \\\ {\text { differentiable. }}\end{array}$$ \(f(x)=(x+4)^{2 / 3}\)

If \(y\) is a differentiable function of \(u,\) and \(u\) is a differentiable function of \(x,\) then \(y\) is a differentiable function of \(x .\)

\(\begin{array}{l}{\text { Using the Alternative Form of the }} \\ {\text { Derivative In Exercises } 69-76, \text { use the the }} \\ {\text { alternative form of the derivative to find the }} \\ {\text { derivative at } x=c, \text { if it exists. }}\end{array}\) \(g(x)=\sqrt{|x|}, c=0\)

In Exercises 93-96, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. $$\begin{array}{l}{\text { The slope of the tangent line to the differentiable function } f \text { at }} \\ {\text { the point }(2, f(2)) \text { is }}\end{array}$$ $$\frac{f(2+\Delta x)-f(2)}{\Delta x}$$

FOR FURTHER INFORMATION For more information on the mathematics of moving ladders, see the article "The Falling Ladder Paradox" by Paul Scholten and Andrew Simoson in The College Mathematics Journal. To view this article, go to MathArticles.com. Construction A construction worker pulls a five-meter plank up the side of a building under construction by means of a rope tied to one end of the plank (see figure). Assume the opposite end of the plank follows a path perpendicular to the wall of the building and the worker pulls the rope at a rate of 0.15 meter per second. How fast is the end of the plank sliding along the ground when it is 2.5 meters from the wall of the building?
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