Question

Find the equation of the line tangent to the graph of \(f\) at the indicated \(x\) value. \(f(x)=\sin ^{-1} x \quad\) at \(\quad x=\frac{\sqrt{2}}{2}\).

Short answer

The tangent line equation is \(y = \sqrt{2}x - 1 + \frac{\pi}{4}\).

Step-by-step solution

Understand the Function

The function given is the inverse sine function, denoted as \(f(x) = \sin^{-1}(x)\). We need to find the tangent line at \(x = \frac{\sqrt{2}}{2}\).

Find the Point on the Curve

Calculate the \(y\)-coordinate of the point on the curve by evaluating \(f\left(\frac{\sqrt{2}}{2}\right)\). Since \(f(x) = \sin^{-1}(x)\), this is \(\sin^{-1}\left(\frac{\sqrt{2}}{2}\right)\), which corresponds to \(\frac{\pi}{4}\). Thus, the point is \(\left(\frac{\sqrt{2}}{2}, \frac{\pi}{4}\right)\).

Calculate the Derivative

The derivative of the inverse sine function is \(f'(x) = \frac{1}{\sqrt{1-x^2}}\). We will use this to find the slope of the tangent line at the given \(x\)-value.

Evaluate the Derivative at the Given Point

Substitute \(x = \frac{\sqrt{2}}{2}\) into the derivative: \(f'\left(\frac{\sqrt{2}}{2}\right) = \frac{1}{\sqrt{1-\left(\frac{\sqrt{2}}{2}\right)^2}} = \frac{1}{\sqrt{1-\frac{1}{2}}} = \frac{1}{\sqrt{\frac{1}{2}}} = \sqrt{2}\). Thus, the slope of the tangent line is \(\sqrt{2}\).

Write the Equation of the Tangent Line

The equation of a line is given by the formula \(y - y_1 = m(x - x_1)\), where \(m\) is the slope, and \((x_1, y_1)\) is a point on the line. Substitute \(m = \sqrt{2}\), \(x_1 = \frac{\sqrt{2}}{2}\), and \(y_1 = \frac{\pi}{4}\) into the formula: \(y - \frac{\pi}{4} = \sqrt{2}\left(x - \frac{\sqrt{2}}{2}\right)\).

Simplify the Equation

Expand and simplify the equation to get the tangent line in slope-intercept form, if needed. It becomes \(y = \sqrt{2}x - 1 + \frac{\pi}{4}\).

Key concepts

Inverse Sine Function

The inverse sine function, often expressed as \( \sin^{-1} x \) or \( \arcsin(x) \), is the function that reverses the effect of the sine function. While the sine function inputs an angle and outputs a ratio, the inverse sine function does the opposite, taking a ratio and returning an angle. This function is defined only for ratios between -1 and 1, inclusive, because the sine of any angle must also lie within these bounds.
Understanding the range of the inverse sine function is crucial. Its output, or range, is limited to angles between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\) radians. Therefore, any ratio plugged into \( \sin^{-1} x \) will yield an angle within these bounds.

Derivative

The derivative of a function provides a formula for calculating its instantaneous rate of change. For the inverse sine function, the derivative is \( f'(x) = \frac{1}{\sqrt{1-x^2}} \). This formula is derived from the necessity to find a tangent, a line that just touches the curve and shares its gradient at a specific point.
To comprehend why this derivative formula works, consider the unit circle definition of sine. Based on this definition, as \( x \) approaches the bounds of -1 or 1, the value under the square root in the denominator tends to zero, making the derivative large. This behavior aligns with the sharp turns or asymptotic behavior near the boundaries of the inverse sine function's domain.

Slope of Tangent Line

The slope of a tangent line to a curve at a given point reflects how steep the curve is at that point. For any function, including inverse trigonometric functions, this is determined by evaluating the function's derivative at the desired point.
In our exercise, we evaluated the derivative of the inverse sine function at \( x = \frac{\sqrt{2}}{2} \). This gave us \( \sqrt{2} \) as the slope. Intuitively, it tells us that the function is increasing and the graph is upward sloping at that specific point. Understanding this slope concept helps not only with this particular calculation but also with analyzing curves more generally.

Equation of a Line

Knowing how to formulate the equation of a line is fundamental in both algebra and calculus. The general form of a line is \( y = mx + b \), where \( m \) represents the slope and \( b \) the y-intercept. However, when dealing with tangent lines, we often use the point-slope form: \( y - y_1 = m(x - x_1) \).
In solving the exercise, we used point-slope form because we knew both the slope and a specific point \( (x_1, y_1) \) on our tangent line. By substituting \( m = \sqrt{2} \), \( x_1 = \frac{\sqrt{2}}{2} \), and \( y_1 = \frac{\pi}{4} \) into the formula, we derived the equation of the tangent line: \( y = \sqrt{2}x - 1 + \frac{\pi}{4} \). This form allows you to see how changes in \( x \) directly affect \( y \) given the specific curve's behavior at that point.