Problem 75 - Thomas Calculus

Question

a. Given that \(x^{4}+4 y^{2}=1\) , find \(d y / d x\) two ways: \((1)\) by solving for \(y\) and differentiating the resulting functions in the usual way and \((2)\) by implicit differentiation. Do you get the same result each way? b. Solve the equation \(x^{4}+4 y^{2}=1\) for \(y\) and graph the resulting functions together to produce a complete graph of the equation \(x^{4}+4 y^{2}=1 .\) Then add the graphs of the first derivatives of these functions to your display. Could you have predicted the general behavior of the derivative graphs from looking at the graph of \(x^{4}+4 y^{2}=1 ?\) Could you have predicted the general behavior of the graph of \(x^{4}+4 y^{2}=1\) by looking at the derivative graphs? Give reasons for your answers.

Step-by-Step Solution

Answer

The results are consistent with both methods for \(dy/dx\). The derivative graphs show rapid changes near endpoints, evident in the original graph's steepness.

Step 1: Solve for y

Given the equation \(x^{4} + 4y^{2} = 1\), solve for \(y\). The equation simplifies to \(4y^{2} = 1 - x^{4}\). Divide both sides by 4:\[y^{2} = \frac{1 - x^{4}}{4}\]Now, take the square root of both sides. Remember that there are two solutions since we will have both the positive and negative square roots:\[y = \pm \sqrt{\frac{1 - x^{4}}{4}}\]

Step 2: Differentiate Explicitly

Differentiating \(y = \sqrt{\frac{1 - x^{4}}{4}}\) with respect to \(x\): Use the chain rule. Let \(u = \frac{1 - x^{4}}{4}\), then:\[y = \sqrt{u} \Rightarrow \frac{dy}{du} = \frac{1}{2\sqrt{u}}\]And:\[\frac{du}{dx} = -x^{3}\]Hence:\[\frac{dy}{dx} = \frac{1}{2\sqrt{\frac{1 - x^{4}}{4}}} \times \left(-4x^{3}\right) = -\frac{2x^{3}}{\sqrt{1 - x^{4}}}\]For the negative case, \(y = -\sqrt{\frac{1 - x^{4}}{4}}\), the derivative has the same magnitude but opposite sign:\[\frac{dy}{dx} = \frac{2x^{3}}{\sqrt{1 - x^{4}}}\]

Step 3: Implicit Differentiation

Start with the original equation \(x^{4} + 4y^{2} = 1\). Differentiate both sides with respect to \(x\):\[4x^{3} + 8y \frac{dy}{dx} = 0\]Solve for \(\frac{dy}{dx}\):\[8y \frac{dy}{dx} = -4x^{3} \Rightarrow \frac{dy}{dx} = -\frac{4x^{3}}{8y} = -\frac{x^{3}}{2y}\]Suppress \(y\) using the first step where \(y = \pm \sqrt{\frac{1 - x^{4}}{4}}\), hence:\[\frac{dy}{dx} = -\frac{x^{3}}{2\left(\pm \sqrt{\frac{1 - x^{4}}{4}}\right)} = \mp\frac{2x^{3}}{\sqrt{1 - x^{4}}}\]

Step 4: Conclusion on Both Methods

Both methods yield the same result: \(\frac{dy}{dx} = \pm \frac{2x^{3}}{\sqrt{1 - x^{4}}}\). This confirms the correctness of both methods, which are consistent.

Step 5: Graphing Functions and Their Derivatives

The given equation represents an ellipse-like shape. Solving for \(y\) provides two functions, \(y = \pm \sqrt{\frac{1 - x^{4}}{4}}\), which graph as two halves of this shape. The derivatives, \(\pm \frac{2x^{3}}{\sqrt{1 - x^{4}}}\), indicate behavior at the extrema (where the slope approaches infinity) and at \(x = 0\) (where the slope is zero).

Step 6: Analyze Graph Behavior and Derivatives

The graph of \(x^{4} + 4y^{2} = 1\) is symmetrical and has horizontal tangents at its intersections with the y-axis, which corresponds to zero slope. The derivative graphs, showing rapid changes, highlight the steep changes as \(x\) moves away from zero. These derivative traits could be inferred from the original graph, showing how steep the curve gets at the ends (vertical stretches).

Key Concepts

Explicit Differentiation

Explicit differentiation is a method used to find derivatives where functions are written explicitly in terms of the independent variable. For example, if we start with the equation \(x^4 + 4y^2 = 1\), we can isolate \(y\) by solving it explicitly as \(y = \pm \sqrt{\frac{1 - x^4}{4}}\). Once we have this explicit form, we can easily differentiate using common rules.

• For positive \(y\), the derivative \(\frac{dy}{dx}\) is derived using the chain rule, resulting in \(-\frac{2x^3}{\sqrt{1 - x^4}}\).
• For negative \(y\), the chain rule still applies, but we end up with the opposite: \(\frac{2x^3}{\sqrt{1 - x^4}}\).

The explicit differentiation approach gives us a direct calculation of the derivative, showcasing the rate of change of \(y\) with respect to \(x\) by treating \(y\) as a function of \(x\).

Graphing Functions

When graphing functions like \(x^4 + 4y^2 = 1\), we interpret how the solutions form a shape on a coordinate plane. First, solving explicitly for \(y\) gives two functions: \(y = \sqrt{\frac{1 - x^4}{4}}\) and \(y = -\sqrt{\frac{1 - x^4}{4}}\). These represent two symmetrical curves above and below the x-axis.

• Each solution speaks to a half of the entire graph. The positive function appears in the upper half of the plane, while the negative function appears in the lower half.
• Together, these halves form an ellipse-like structure, as implied by the original constraint.

This graphical approach aids in visualizing algebraic solutions, emphasizing the symmetry and overall shape suggested by the original implicit equation.

Derivative Graphs

Derivative graphs are excellent tools for understanding how functions behave, beyond just calculating derivatives. When looking at \(x^4 + 4y^2 = 1\), knowing the derivatives \(\pm \frac{2x^3}{\sqrt{1 - x^4}}\) helps in predicting the curve's direction at any \(x\).

• At \(x = 0\), both derivatives become zero, indicating horizontal tangents.
• As \(x\) moves away from zero, the derivative becomes steeper, indicating the rapid increase or decrease in slope.

What these graphs of derivatives show is how certain features like extrema (where slopes can be incredibly steep or even infinite) correspond to visual behavior in the original function graph.

Symmetrical Curves

The notion of symmetrical curves is fundamental in understanding the geometry and the calculus of the equation \(x^4 + 4y^2 = 1\). Since the original equation doesn't specify a single function of \(x\), both positive and negative square root solutions contribute to symmetry.

• Each solution of \(y\) represents one symmetric half about the x-axis. Together, they form a symmetric whole.
• In addition, symmetry in derivative graphs, such as the steep changes at extreme x-values, reflects the symmetry we observe in the original curve.

This symmetry is crucial when working with implicit differentiation, as it tells us much about the function's geometrical properties and how it should look and behave when analyzed visually or analytically.