Question

Find the equation of the tangent line to the graph of the implicitly defined function at the indicated points. As a visual aid, each function is graphed. \(x^{4}+y^{4}=1\) (a) At (1,0) . (b) At \((\sqrt{0.6}, \sqrt{0.8})\) (c) At (0,1)

Short answer

(a) \(x = 1\); (b) \(y - \sqrt{0.8} = -\frac{3\sqrt{0.6}}{4\sqrt{0.8}} (x - \sqrt{0.6})\); (c) \(y = 1\).

Step-by-step solution

Differentiate Implicitly

Given the equation of the curve is \(x^4 + y^4 = 1\). To find the tangent line at a point, the slope (\(m\)) of the tangent is needed, which can be found using implicit differentiation. Differentiate both sides with respect to \(x\): \[\frac{d}{dx}(x^4) + \frac{d}{dx}(y^4) = \frac{d}{dx}(1)\] This gives \(4x^3 + 4y^3 \frac{dy}{dx} = 0\). Solve for \(\frac{dy}{dx}\): \[\frac{dy}{dx} = -\frac{x^3}{y^3}\].

Calculate Slope at (1,0)

Plug the point (1, 0) into the slope formula. Since \(y = 0\), \(\frac{dy}{dx} = -\frac{1^3}{0^3}\) is undefined, indicating a vertical tangent line at (1, 0). The equation of a vertical line with x-coordinate 1 is \(x = 1\).

Calculate Slope at \((\sqrt{0.6}, \sqrt{0.8})\)

Plug the point \((\sqrt{0.6}, \sqrt{0.8})\) into the slope formula: \[\frac{dy}{dx} = -\frac{(\sqrt{0.6})^3}{(\sqrt{0.8})^3} = -\frac{0.6\cdot\sqrt{0.6}}{0.8\cdot\sqrt{0.8}}\]. Simplifying gives \(\frac{dy}{dx} = -\frac{3\sqrt{0.6}}{4\sqrt{0.8}}\).

Write the Equation of the Tangent Line at \((\sqrt{0.6}, \sqrt{0.8})\)

Using the point-slope form \(y - y_1 = m(x - x_1)\) with \(x_1 = \sqrt{0.6}\), \(y_1 = \sqrt{0.8}\) and slope \(m = -\frac{3\sqrt{0.6}}{4\sqrt{0.8}}\), the equation is: \[y - \sqrt{0.8} = -\frac{3\sqrt{0.6}}{4\sqrt{0.8}} (x - \sqrt{0.6})\].

Calculate Slope at (0,1)

Plug the point (0, 1) into the slope formula: \(\frac{dy}{dx} = -\frac{0^3}{1^3} = 0\), indicating a horizontal tangent line. The equation of a horizontal line with y-coordinate 1 is \(y = 1\).

Summary of Tangent Lines

- At (1, 0), the tangent line is vertical: \(x = 1\).- At \((\sqrt{0.6}, \sqrt{0.8})\), the tangent line is \(y - \sqrt{0.8} = -\frac{3\sqrt{0.6}}{4\sqrt{0.8}} (x - \sqrt{0.6})\).- At (0, 1), the tangent line is horizontal: \(y = 1\).

Key concepts

Tangent Line

A tangent line is essentially a straight line that just touches, or is tangent to, a curve at a particular point. This line uniquely highlights the direction in which the curve is heading at that precise location. Tangent lines are a fundamental concept not only in geometry but also in calculus, where they play a crucial role in understanding rates of change and instantaneous motion.

• The point where the tangent touches the curve is called the point of tangency.
• The slope of the tangent line equals the derivative of the function at the point of tangency.

In the context of this exercise, calculating the slope of the tangent line requires implicit differentiation due to the curve being defined by the equation \(x^{4} + y^{4} = 1\). For instance, at the point (1, 0), the tangent is vertical indicating the direction is parallel to the y-axis resulting in the equation \(x = 1\). Conversely, at (0, 1), the tangent is horizontal, suggesting the slope is zero, giving the line equation \(y = 1\). This showcases how the nature of tangent lines can vary based on the point being considered on an implicit curve.

Implicit Function

An implicit function is a function that's not expressed directly as \(y = f(x)\), but rather as an equation relating \(x\) and \(y\). Implicit functions often require special techniques to work with, particularly for differentiation purposes, as illustrating their dynamics is not as straightforward as explicit functions.In the example \(x^{4} + y^{4} = 1\), the function is implicit because \(y\) is not isolated on one side of the equation. This type of curve often represents more complex shapes, requiring a more nuanced approach to calculus. To navigate through this complexity:

• We utilize implicit differentiation to find derivatives of such relationships.
• It's often necessary to interpret constraints, such as knowing the maximum value of \(x\) and \(y\) based on the equation.

Recognizing an implicit curve, as seen in this problem, allows students to apply implicit differentiation to derive meaningful insights such as tangent lines at specific points. This method will reveal critical information that is difficult to discern just from observing the equation itself.

Differentiation

Differentiation is a fundamental tool in calculus used to find rates of change for various types of functions. It's particularly striking in how it allows us to ascertain the slope of a curve at any given point. Differentiation becomes even more interesting and involved when applied to implicit functions.Implicit differentiation is the technique of differentiating entire equations rather than isolated terms, catering to implicit functions like \(x^{4} + y^{4} = 1\). The process follows these core steps:

• Apply the derivative by differentiating each term with respect to \(x\), treating \(y\) as a function of \(x\).
• Utilize the chain rule when differentiating terms involving \(y\), which leads to additional expressions \(\frac{dy}{dx}\).

This approach subsequently solves for \(\frac{dy}{dx}\), which indicates the slope of the tangent line at any given point. Finding \(\frac{dy}{dx}\) allows us to understand how \(y\) changes with \(x\), providing insights into the nature of the curve. Ultimately, differentiation in this context opens up the capability to draw tangent lines and make comprehensive sense of implicit functions.