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. \(x^{2 / 5}+y^{2 / 5}=1\) (a) At (1,0) (b) At (0.1,0.281) (which does not exactly lie on the curve, but is very close).

Short answer

(a) The tangent at (1,0) is vertical: \(x=1\). (b) Use point-slope form at (0.1,0.281) for tangent line equation.

Step-by-step solution

Differentiate Implicitly

Given the equation \(x^{2/5} + y^{2/5} = 1\), we begin by differentiating both sides with respect to \(x\). The left side becomes: \[ \frac{d}{dx}(x^{2/5}) + \frac{d}{dx}(y^{2/5}) = 0 \]. Using the chain rule, this results in \[ \frac{2}{5}x^{-3/5} + \frac{2}{5}y^{-3/5}\frac{dy}{dx} = 0 \].

Solve for \(\frac{dy}{dx}\)

Reorganize the differentiated equation to solve for \(\frac{dy}{dx}\) (the slope of the tangent line): \[ \frac{2}{5}y^{-3/5}\frac{dy}{dx} = -\frac{2}{5}x^{-3/5} \]. Simplify to \[ \frac{dy}{dx} = -\left(\frac{x}{y}\right)^{-3/5} \]. This gives us the formula for the slope at any point \((x, y)\).

Calculate Slope at Point (1, 0)

Substitute \(x = 1\) and \(y = 0\) into the slope formula: \[ \frac{dy}{dx} = -\left(\frac{1}{0}\right)^{-3/5} \]. Here, \(y=0\) results in an undefined slope, indicating a vertical tangent line at this point.

Identify Equation for Vertical Tangent

When the slope is undefined (as with division by zero), the tangent line is vertical. The equation of a vertical line through \(x = 1\) is simply \(x = 1\).

Calculate Slope at Point (0.1, 0.281)

Substitute \(x = 0.1\) and \(y = 0.281\) into the slope formula: \[ \frac{dy}{dx} = -\left(\frac{0.1}{0.281}\right)^{-3/5} \]. Calculate this to find the slope value.

Find Equation of Tangent Line at (0.1, 0.281)

Use the point-slope form of a line: \(y - y_1 = m(x - x_1)\), where \((x_1, y_1) = (0.1, 0.281)\). Substitute the slope calculated previously and the point into the equation to get the equation of the tangent line.

Key concepts

Implicit Differentiation

Implicit differentiation is a method used in calculus to find derivatives when a function is not given explicitly. In problems like these, the relationship between variables is expressed implicitly by an equation involving both variables. The main idea is:

• Differentiate both sides of the equation with respect to one variable, usually \(x\).
• Treat the other variable, like \(y\), as a function of \(x\) (i.e., \(y=y(x)\)).
• Apply the chain rule to terms involving \(y\).

For example, consider the equation \(x^{2/5} + y^{2/5} = 1\). If we differentiate with respect to \(x\), notice that for the term \(y^{2/5}\), you multiply by \(\frac{dy}{dx}\), due to the chain rule, resulting in additional terms that involve \(\frac{dy}{dx}\). Once differentiated, you solve for \(\frac{dy}{dx}\) to find the derivative, which represents the slope of the tangent line at any point \((x, y)\).
This method is particularly useful when functions are complex and cannot be easily rewritten in the form \(y = f(x)\).

Tangent Line

The tangent line to a curve at a given point is the straight line that just "touches" the curve at that point. It represents the instantaneous direction of the curve and has the same slope as the curve at that point. For curves defined implicitly, the tangent line's slope at a point \((x, y)\) is given by \(\frac{dy}{dx}\), which we find using implicit differentiation.

Here's how to find a tangent line:

• First, calculate the slope \(m\) of the curve at the desired point using \(\frac{dy}{dx}\).
• If the slope is defined, use the point-slope form: \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is your point.
• If the slope is undefined, the tangent line is vertical, and its equation is \(x = x_1\).

In this exercise:

• At the point (1, 0), the slope was undefined due to division by zero, hence the tangent line is vertical.
• At the point (0.1, 0.281), the slope was calculated and used to find the specific equation of the tangent line.

Chain Rule

The chain rule is crucial when differentiating composite functions. It allows us to differentiate the outer function and multiply it by the derivative of the inner function. When working with implicit differentiation, the chain rule comes into play with terms involving the dependent variable treated as a function of the independent variable.

For example, in the equation \(y^{2/5}\), as \(y\) is a function of \(x\), the derivative with respect to \(x\) is obtained by:

• First applying the power rule to get \(\frac{2}{5}y^{-3/5}\).
• Then multiply by \(\frac{dy}{dx}\), because of the chain rule.

This step is critical in setting up the equation that includes \(\frac{dy}{dx}\), thereby allowing us to find the slope of the tangent line at a given point on an implicitly defined curve.

Slope of a Curve

The slope of a curve at any given point is simply the value of its derivative at that point. It indicates how steeply the curve is rising or falling at that particular location. In the context of tangent lines, the slope of the curve at a point is the same as the slope of the tangent line at that point.

When dealing with implicitly defined functions, the slope can be found by solving \(\frac{dy}{dx}\) using implicit differentiation.

• If \(\frac{dy}{dx}\) results in a regular numerical value, this number is used directly as the slope of the tangent line.
• If \(\frac{dy}{dx}\) leads to an undefined expression, such as division by zero, the slope is vertical, indicating a vertical tangent line.

For instance, at point (1, 0) on the given curve,\(\frac{dy}{dx}\) became undefined implying the slope at that point was vertical. Understanding the concept of slope helps in visualizing and calculating the exact nature of the tangent line, aiding in sketching graphs and analyzing the curve's behavior.