Question

Let \(P=(x, y)\) be a point on the graph of \(y=\frac{1}{x}\) (a) Express the distance \(d\) from \(P\) to the origin as a function of \(x\). (b) Use a graphing utility to graph \(d=d(x)\). (c) For what values of \(x\) is \(d\) smallest? (d) What is the smallest distance?

Short answer

For \(x = 1\) and \(x = -1\), d is smallest at \(\sqrt{2}\).

Step-by-step solution

- Express distance as a function of x

The distance formula from a point (\(x_1, y_1\)) to the origin (\(0,0\)) is: \[d = \sqrt{x_1^2 + y_1^2}\]. Here, \(P = (x, y)\) and since \(y = \frac{1}{x}\), we replace \(y\) in the distance formula: \[d = \sqrt{x^2 + \left(\frac{1}{x}\right)^2}\].

- Simplify the function

Simplify the expression for \(d\): \[d = \sqrt{x^2 + \frac{1}{x^2}} = \sqrt{\frac{x^4 + 1}{x^2}} = \frac{\sqrt{x^4 + 1}}{x}\].

- Using a graphing utility

Graph the function \(d(x) = \frac{\sqrt{x^4 + 1}}{x}\) using any graphing utility. Observe the graph to determine the behavior of \(d(x)\).

- Determine the minimum distance

From the graph, observe where the function reaches its minimum. Alternatively, find the derivative \(d'(x)\) and solve for \(d'(x) = 0\) to find critical points. Let’s find \(d'(x)\) using the quotient rule: \[d'(x) = \frac{x^2(4x^3) - (x^4 + 1)(1)}{x^2\sqrt{x^4+1}}\]. Simplifying it and solving for 0 will give \[x^3 - x^{-3} = 0 \Rightarrow x^6 = 1 \Rightarrow x =1\] or \(x = -1\).

- What is the smallest distance?

Substitute \(x = 1\) into \(d(x)\) : \[d(1) = \frac{\sqrt{1^4 + 1}}{1} = \sqrt{2}\].

Key concepts

Distance Formula

The distance formula is used to determine the distance between two points in a coordinate plane. It is derived from the Pythagorean theorem and is expressed as: \ \ \ \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \) \ \ \ In our problem, we aim to find the distance from point \( P = (x, \frac{1}{x}) \) to the origin \((0, 0)\). Using the distance formula: \ \ \ \( d = \sqrt{x^2 + \left(\frac{1}{x}\right)^2} \) \ \ \ This equation gives us the distance in terms of \( x \), thereby expressing the distance as a function of \( x \).

Functions

In mathematics, a function is a relation between a set of inputs and a set of permissible outputs. In this case, the function is the distance from the point \( P \) to the origin in terms of the variable \( x \). The function we have is: \ \ \ \( d(x) = \sqrt{x^2 + \left(\frac{1}{x}\right)^2} \) \ \ \ Simplifying this, we get: \ \ \ \( d(x) = \sqrt{x^2 + \frac{1}{x^2}} \) \ \ \ \( d(x) = \frac{\sqrt{x^4 + 1}}{x} \) \ \ \ This represents a function where each \( x \) value will provide a corresponding distance \( d \)from the origin.

Derivatives

Derivatives in calculus are measures of how a function changes as its input changes — they give us the rate of change. To find when the distance function is at its minimum, we need to find its derivative and set it to zero to identify critical points. The derivative of \( d(x) = \frac{\sqrt{x^4 + 1}}{x} \) can be found using the quotient rule: \ \ \ The quotient rule is: \ \ \ \( \left(\frac{u}{v}\right)' = \frac{vu' - uv'}{v^2} \) \ \ \ Applying this, let \( u = \sqrt{x^4 + 1} \) and \( v = x \), we get: \ \ \ \( d'(x) = \frac{x \cdot 4x^3(\sqrt{x^4 + 1}) - (x^4 + 1)(1)}{x^2 (\sqrt{x^4 + 1})} \) \ \ \ Simplifying and setting it to zero we solve and find that \( x = 1 \) or \( x = -1 \) are potential critical points.

Minimum Distance

The minimum distance occurs at the critical points from our derivative calculations. Substituting \( x = 1 \) into our distance function: \ \ \ \( d(1) = \frac{\sqrt{1^4 + 1}}{1} = \sqrt{2} \) \ \ \ Thus, we find the minimum distance from the point \( P \) to the origin is \( \sqrt{2} \) at \( x = 1 \). Using a graphing utility to graph \( d(x) \) helps visualize this minimum point. The graph should show that the minimum value of the function is indeed \( \sqrt{2} \).