Problem 50 - Calculus Early Transcendentals

Question

Find an equation for the tangent line to the graph at the specified value of \(x .\) $$ y=\frac{x}{\sqrt{1-x^{2}}}, x=0 $$

Step-by-Step Solution

Answer

The equation of the tangent line is \(y = x\).

Step 1: Understand the Problem

We need to find an equation for the tangent line to the function \(y = \frac{x}{\sqrt{1-x^2}}\) at the value \(x = 0\). This involves finding the slope of the tangent line at \(x = 0\) and using the point-slope form of a linear equation.

Step 2: Find the Derivative

To find the slope of the tangent line, we first need the derivative \(y'\) of the function \(y = \frac{x}{\sqrt{1-x^2}}\). Use the quotient rule: \(\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}\), where \(u = x\) and \(v = \sqrt{1-x^2}\).

Step 3: Apply Quotient Rule

\(u' = 1\) and \(v = \sqrt{1-x^2}\) with \(v' = \frac{-x}{\sqrt{1-x^2}}\). The derivative is \(y' = \frac{1 \cdot \sqrt{1-x^2} - x \cdot (-x/\sqrt{1-x^2})}{(\sqrt{1-x^2})^2}\).

Step 4: Simplify the Derivative

Simplify the expression: \(y' = \frac{\sqrt{1-x^2} + \frac{x^2}{\sqrt{1-x^2}}}{1-x^2}\). This simplifies to \(y' = \frac{1}{(1-x^2)^{3/2}}\).

Step 5: Evaluate the Derivative at x=0

Substitute \(x = 0\) into the derivative to find the slope of the tangent line: \(y'(0) = \frac{1}{(1-0^2)^{3/2}} = 1\).

Step 6: Find the Point on the Curve at x=0

Evaluate the function at \(x = 0\): \(y(0) = \frac{0}{\sqrt{1-0^2}} = 0\). The point is \((0, 0)\).

Step 7: Use Point-Slope Form

With the slope \(m = 1\) and point \((0, 0)\), use the point-slope form of a line: \(y - y_1 = m(x - x_1)\). This becomes \(y - 0 = 1(x - 0)\), leading to \(y = x\).

Step 8: Write the Equation of the Tangent

The equation of the tangent line at \(x = 0\) is \(y = x\).

Key Concepts

Derivative

The concept of the derivative is fundamental in calculus. It represents the rate at which a function is changing at any given point. In simpler terms, think of it as the function's instantaneous slope at a specific point. Imagine the graph of the function. The slope of the line tangent to that graph at a particular point is the derivative.

When given a function like \(y = \frac{x}{\sqrt{1-x^2}}\), you need to determine its derivative to find the slope of the tangent line at a certain point. In our case, we're interested in \(x = 0\). When you derive a function, you're determining how the function's outputs (y-values) change with respect to its inputs (x-values). Thus, the derivative \(y'\) gives you essential insights into this rate of change.

• The derivative is a crucial step for calculating tangent lines.
• It tells us how steep or flat our tangent line is at a specific point.
• Using derivatives, we can understand the behavior of complex functions.

Quotient Rule

The quotient rule is a specific method used to find the derivative of a function that is the ratio of two differentiable functions. When you have a function in the form of \(\frac{u}{v}\), where both \(u\) and \(v\) are functions of \(x\), the quotient rule provides a formula to evaluate its derivative.

The quotient rule states: \[\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}.\]To break this down

• \(u'\) is the derivative of the numerator \(u\).
• \(v'\) is the derivative of the denominator \(v\).
• Plug these derivatives into the quotient rule formula to find the combined derivative.

In our exercise, \(u=x\) and \(v=\sqrt{1-x^2}\), both needed to apply the rule correctly.This systematic approach helps avoid potential errors compared to taking derivatives separately and simplifies complex fraction derivatives, like the one in our function.

Point-Slope Form

Once you know the slope of the tangent line from the derivative, you can find the equation of the tangent line using the point-slope form. This form is particularly handy when you know the slope of the line and a single point on that line.

The point-slope formula is:\[y - y_1 = m(x - x_1)\]Where:

• \(m\) is the slope of the line.
• \((x_1, y_1)\) is the known point on the line.

In this problem, we found that the slope at \(x = 0\) is \(1\), and the point on the line is \((0,0)\). When plugged into the point-slope equation, it simplifies to \(y = x\), giving us the equation of the tangent line.

This form is straightforward and removes the complexity of calculating additional points or transformations. Therefore, it's a preferred method when you have the slope and a point ready at hand.