Question

Find all points on the graph of \(y=100 / x^{5}\) where the tangent line is perpendicular to the line \(y=x\).

Short answer

The points are \(\left(\pm \frac{1}{(500)^{1/6}}, 100 \times (500^{5/6})\right)\).

Step-by-step solution

Understand the Problem

We need to find points on the graph of \(y = \frac{100}{x^5}\) where the tangent line is perpendicular to the line \(y = x\). A line perpendicular to \(y = x\) has slope \(-1\). Thus, we want the slope of the tangent to be \(-1\).

Differentiate the Function

Find the derivative of \(y = \frac{100}{x^5}\). This derivative will represent the slope of the tangent line at any point \((x, y)\) on the curve. Differentiate using the power rule: \(y' = \frac{d}{dx}\left(100x^{-5}\right) = -500x^{-6}\).

Set the Derivative Equal to Slope

We set the derivative \(-500x^{-6}\) equal to \(-1\) because we want the tangent line to have a slope of \(-1\). This gives us the equation: \(-500x^{-6} = -1\).

Solve for x

Solve the equation \(-500x^{-6} = -1\) by dividing both sides by \(-500\), resulting in \(x^{-6} = \frac{1}{500}\). Take the sixth root of both sides to isolate \(x\): \(x = \left(\frac{1}{500}\right)^{-1/6}\) or \(x = \pm \frac{1}{(500)^{1/6}}\).

Calculate Corresponding y-values

Substitute \(x = \pm \frac{1}{(500)^{1/6}}\) back into the original equation \(y = \frac{100}{x^5}\) to find corresponding \(y\) values. For each \(x\), calculate \( y \).

Find Points

For \(x = \frac{1}{(500)^{1/6}}\), \(y = \frac{100}{\left(\frac{1}{(500)^{1/6}}\right)^5} = 100 \times (500^{5/6})\). Similarly, for \(x = -\frac{1}{(500)^{1/6}}\), \(y = 100 \times (500^{5/6})\). The points are \(\left(\frac{1}{(500)^{1/6}}, 100 \times (500^{5/6})\right)\) and \(\left(-\frac{1}{(500)^{1/6}}, 100 \times (500^{5/6})\right)\).

Key concepts

Derivative Calculation

The derivative is a fundamental concept in calculus. In simple terms, it helps in finding the rate at which a function is changing at any given point. For our function, we have:
\[ y = \frac{100}{x^5} \]
To find the derivative, apply the rules of differentiation. The derivative of a function gives you the slope of the tangent line to the curve at any point \((x, y)\).
In this exercise, by rewriting \(y = \frac{100}{x^5}\) as \(100x^{-5}\), the power rule of differentiation can be easily applied. This rule states:

• For a function \(f(x) = x^n\), the derivative \(f'(x) = nx^{n-1}\).

Applying this to our function:

• \(y' = \frac{d}{dx}(100x^{-5}) = -500x^{-6}\)

This result illustrates how derivatives streamline the calculation of slopes of tangent lines.

Perpendicular Lines

Understanding perpendicular lines is crucial for solving geometric problems in calculus. Two lines are perpendicular if the product of their slopes is -1. The line \(y=x\) has a slope of 1, so any line perpendicular to it must have a slope of-1.For this exercise, we needed the tangent line's slope of the graph to be perpendicular to the line \(y = x\), giving a required slope of -1. By setting the derivative equal to -1, we found the points at which the tangent line is perpendicular. This leads us tosolve:

• \(-500x^{-6} = -1\)

This methodology underscores the importance of the slope relationship when identifying perpendicular lines.

Graph Analysis

Graph analysis involves understanding and analyzing the behavior of functions on a graph. In our problem, graph analysis helps visualize where the tangent line to \(y = \frac{100}{x^5}\) is perpendicular to the line \(y = x\).
Considering the equation \(-500x^{-6} = -1\), solving for \(x\) gives us the points of interest where the slope conditions are met. At these points, both the value of \(x\) and the corresponding \(y\) coordinates are crucial for complete graph analysis.

• Solve \(x^{-6} = \frac{1}{500}\) gives \(x = \pm \frac{1}{(500)^{1/6}}\).
• Substitute back to get \(y = 100 \times (500^{5/6})\).

Ultimately, these points,\(\left(\frac{1}{(500)^{1/6}}, 100 \times (500^{5/6})\right)\) and \(\left(-\frac{1}{(500)^{1/6}}, 100 \times (500^{5/6})\right)\), are where graph analysis confirms perpendicular tangents exist.

Power Rule Differentiation

Power rule differentiation is a key technique in calculus. It simplifies finding derivatives for functions involving powers of \(x\). This rule, briefly explained, is:

• If \(f(x) = x^n\), then \(f'(x) = nx^{n-1}\).

In our problem, we dealt with the function \(y = \frac{100}{x^5}\). By expressing it as \(y = 100x^{-5}\), it became straightforward to apply power rule differentiation:

• Find the derivative: \(y' = -500x^{-6}\).

This straightforward process illustrates how power rule differentiation is essential for tackling more complex differentiation problems. Using this rule efficiently allows us to identify critical points where certain conditions, like perpendicular slopes, are met. By converting different forms of functions, the power rule facilitates easier calculations and interpretations of derivatives.