Question

In Exercises 35 and 36, find an equation of the tangent line to the graph of the equation at the given point. $$ \arcsin x+\arcsin y=\frac{\pi}{2}, \quad\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right) $$

Short answer

The equation for the tangent line is \(y - \frac{\sqrt{2}}{2} = -1\left(x - \frac{\sqrt{2}}{2}\right)\).

Step-by-step solution

Rewrite the equation in terms of y

First, we rewrite the given equation as \(y = \sin \left( \frac{\pi}{2} - \arcsin x \right)\). Using the identity \( \sin(u - v) = \sin u \cos v - \cos u \sin v \), we simplify it down to \(y = \sqrt{1 - x^2}\).

Calculate the derivative of y

Next, we calculate the derivative of the equation we found in Step 1. So, \(\frac{dy}{dx} = \frac{d}{dx}(\sqrt{1 - x^{2}})\). The derivative is \(\frac{-x}{\sqrt{1 - x^{2}}}\).

Substitute the given point into the derivative to find the slope

Substitute the value \(x = \frac{\sqrt{2}}{2}\) into \(\frac{-x}{\sqrt{1 - x^{2}}}\) to calculate the slope of the tangent line at the point. The slope becomes \(\frac{-\sqrt{2}/2}{\sqrt{1 - (\sqrt{2}/2)^{2}}}\) which simplifies to -1.

Write the equation for the tangent line

Now that we have the slope, we can write the equation of the tangent line. With the slope (m = -1) and the given point \( (\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}) \), we use the point-slope formula \(y - y_1 = m(x - x_1)\) to find our equation: \(y - \frac{\sqrt{2}}{2} = -1\left(x - \frac{\sqrt{2}}{2}\right)\).