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(a) find two explicit functions by solving the equation for in terms of (b) sketch the graph of the equation and label the parts given by the corresponding explicit functions, (c) 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 explicit functions obtained from the given equation are \(y = \sqrt{\frac{x^2}{16} + 1}\) and \(y = -\sqrt{\frac{x^2}{16} + 1}\) and their respective derivatives are \(dy/dx = \frac{x}{16\sqrt{\frac{x^2}{16} + 1}}\) and \(dy/dx = -\frac{x}{16\sqrt{\frac{x^2}{16} + 1}}\), which is the same when the equation is implicitly differentiated.

Step by step solution

01

Solve the equation for y in terms of x

Let’s start by rearranging the equation: \(16y^2 - x^2 = 16\). Add \(x^2\) on both sides to get \(16y^2 = x^2 + 16\), then divide by 16 to isolate \(y^2\): \(y^2 = \frac{x^2}{16} + 1\). Since \(y^2\) is something squared equals something, we get two roots: \(y = \sqrt{\frac{x^2}{16} + 1}\) and \(y = -\sqrt{\frac{x^2}{16} + 1}\). Thus, we have obtained two explicit functions.
02

Sketch the graph and label the parts

Since there's a positive and negative solution, these equations would create two semi-ellipses on a graph, centered at the y-axis (since it's symmetric about the y-axis). One ellipse is above the x-axis for the positive solution and the other is below the x-axis for the negative solution.
03

Differentiate the explicit functions

Now, we have the two explicit functions, which are \(y = \sqrt{\frac{x^2}{16} + 1}\) and \(y = -\sqrt{\frac{x^2}{16} + 1}\). Differentiating each will give: \(dy/dx = \frac{x}{16\sqrt{\frac{x^2}{16} + 1}}\) for the first function, and \(dy/dx = -\frac{x}{16\sqrt{\frac{x^2}{16} + 1}}\) for the second function.
04

Conduct implicit differentiation and compare to the results from step 3

Differentiate the given equation implicitly: \(32y \cdot dy/dx - 2x = 0\). Rearranging the terms, we obtain: \(dy/dx = \frac{x}{16y}\). When we substitute \(y = \sqrt{\frac{x^2}{16} + 1}\) into this equation, we get \(dy/dx = \frac{x}{16\sqrt{\frac{x^2}{16} + 1}}\). When we substitute \(y = -\sqrt{\frac{x^2}{16} + 1}\) into this equation, we get \(dy/dx = -\frac{x}{16\sqrt{\frac{x^2}{16} + 1}}\). Hence, the results from the explicit differentiation match the results from the implicit differentiation.

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Key Concepts

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

Explicit Functions
In mathematics, an explicit function is one where the dependent variable, typically represented as \( y \), is isolated on one side of the equation. This means the function is expressed clearly in terms of the independent variable \( x \). For the given exercise, we started with the equation \(16y^2 - x^2 = 16\). By manipulating this equation, we derived two explicit forms of \( y \):
  • \( y = \sqrt{\frac{x^2}{16} + 1} \)
  • \( y = -\sqrt{\frac{x^2}{16} + 1} \)
These represent the positive and negative roots respectively, illustrating that for some equations, multiple explicit functions can be derived. These explicit forms allow for straightforward graphing and differentiation, providing insight into the function’s behavior.
Understanding explicit functions is crucial for simplifying complex mathematical tasks, making it easier to visualize and calculate changes in functions.
Graph Sketching
Graph sketching is an essential skill in mathematics that helps visualize the behavior of functions. In the case of our exercise, where we derived two explicit functions, each one represents a part of the graph of the original equation.
  • The equation \( y = \sqrt{\frac{x^2}{16} + 1} \) represents the upper half of the ellipse.
  • The equation \( y = -\sqrt{\frac{x^2}{16} + 1} \) represents the lower half of the ellipse.
When sketching, it's important to:
  • Identify key points, such as intercepts and vertices.
  • Note symmetry, which tells us that the graph is mirrored along the y-axis in this instance.
  • Smoothly connect points to outline the shape, considering whether the function is continuous.
A neat sketch provides an overview of a function's characteristics and informs further analysis, like differentiation.
Differentiation
Differentiation is a core concept in calculus used to find the rate at which a function changes. When differentiating the explicit functions from our exercise, we calculated the derivative \(dy/dx\) for both:
  • For \( y = \sqrt{\frac{x^2}{16} + 1} \), the derivative is \( \frac{x}{16\sqrt{\frac{x^2}{16} + 1}} \).
  • For \( y = -\sqrt{\frac{x^2}{16} + 1} \), the derivative changes sign: \( -\frac{x}{16\sqrt{\frac{x^2}{16} + 1}} \).
These derivatives tell us how \( y \) changes with \( x \) for each explicit function and provide insights into the slope of the tangent at any point on the corresponding semi-ellipse.
By understanding differentiation, you gain insights into dynamic properties of functions, enabling the calculation of slopes, velocities, and other rates of change, which are pivotal in numerous applications.
Quadratic Equation
Quadratic equations are expressions that can be written in the form \(ax^2 + bx + c = 0\) and are fundamental in algebra. The original equation, \(16y^2 - x^2 = 16\), is a rearranged form of a quadratic when solved for \(y\). Unlike standard quadratic equations, here we solve to express \(y\) explicitly, resulting in square roots.
The solutions to these types of equations often involve:
  • Rearranging terms to isolate the squared component.
  • Applying the square root to both sides to solve for the variable of interest.
  • Considering both positive and negative roots to capture all potential solutions.
It's crucial to recognize that quadratic equations may present two solutions and commonly appear in various scientific fields, symbolizing balance and physical phenomena. Understanding and solving quadratics open the door to numerous practical and theoretical applications.

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