Question

In Exercises 17-26, find the lines that are (a) tangent and (b) normal to the curve at the given point. $$y=2 \sin (\pi x-y), \quad(1,0)$$

Short answer

The equation of the tangent line is \(y= \frac{2\pi}{1+2\pi}(x-1)\) and the equation of the normal line is \(y = -\frac{1+2\pi}{2\pi}(x-1)\)

Step-by-step solution

Find the Derivative

Differentiate the equation implicitly: \nDerive \(y=2 \sin (\pi x-y)\) \nUse the chain rule for \(\sin\) and the product rule for \(\pi x - y\). The derivative becomes: \[\frac{dy}{dx} = 2 \pi \cos(\pi x - y)(1 - \frac{dy}{dx})\]

Solve For dy/dx

Solve the equation derived from the derivative above for \(dy/dx\): \[\frac{dy}{dx} = \frac{2 \pi \cos(\pi x - y)}{1+2\pi \cos(\pi x - y)}\]This will be the slope of the tangent line at any point (x, y).

Evaluate at Given Point

To find the slope of the tangent line at the point (1,0), substitute x = 1 and y = 0 into the derivative equation, giving: \[\frac{dy}{dx}= \frac{2 \pi}{1+2\pi}\]

Find Equation of Tangent Line

Use the point slope formula \(y-y1=m(x-x1)\) where m is the slope at the point and (x1, y1) is the point (1, 0) to find the equation for the tangent line. It results in: \(y= \frac{2\pi}{1+2\pi}(x-1)\)

Find Slope of Normal Line

The normal line is perpendicular to the tangent line, so the slope of the normal line is the negative reciprocal of the slope of the tangent line, which is: \[-\frac{1+2\pi}{2\pi}\]

Find Equation of Normal Line

Use the point slope formula again to find the equation of the normal line: \(y = -\frac{1+2\pi}{2\pi}(x-1)\)