Problem 55 - Thomas Calculus

Question

In Exercises \(47-56,\) verify that the given point is on the curve and find the lines that are (a) tangent and (b) normal to the curve at the given point. $$ y=2 \sin (\pi x-y), \quad(1,0) $$

Step-by-Step Solution

Answer

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

Step 1: Verify the point is on the curve

Substitute the point \((1, 0)\) into the equation \(y = 2 \sin (\pi x - y)\). We have \((0) = 2 \sin (\pi \cdot 1 - 0) = 2 \sin \pi = 0\). Thus, the point \((1,0)\) satisfies the equation, confirming that it is on the curve.

Step 2: Differentiate to find the derivative

First, implicitly differentiate the equation \(y = 2 \sin (\pi x - y)\) with respect to \(x\). Use the chain rule: \[\frac{dy}{dx} = 2\cos(\pi x - y)(\pi - \frac{dy}{dx})\].Expand and rearrange to solve for \(\frac{dy}{dx}\):\[\frac{dy}{dx} + 2\cos(\pi x - y)\frac{dy}{dx} = 2\pi\cos(\pi x - y)\].\[\frac{dy}{dx}(1 + 2\cos(\pi x - y)) = 2\pi\cos(\pi x - y)\].\[\frac{dy}{dx} = \frac{2\pi\cos(\pi x - y)}{1 + 2\cos(\pi x - y)}\].

Step 3: Evaluate the derivative at the point

Substitute \((1, 0)\) into the derivative:\[\frac{dy}{dx}\bigg|_{(1,0)} = \frac{2\pi\cos(\pi \cdot 1 - 0)}{1 + 2\cos(\pi \cdot 1 - 0)}\].Since \(\cos\pi = -1\), we have:\[\frac{dy}{dx}\bigg|_{(1,0)} = \frac{2\pi(-1)}{1 + 2(-1)} = \frac{-2\pi}{-1} = 2\pi\].

Step 4: Find the equation of the tangent line

The slope of the tangent line is \(2\pi\). Use the point-slope form of a line equation: \[y - y_1 = m(x - x_1)\], where \((x_1, y_1) = (1, 0)\) and \(m = 2\pi\).Thus, the equation is:\[y - 0 = 2\pi(x - 1)\].Simplified, it becomes:\[y = 2\pi x - 2\pi\].

Step 5: Find the equation of the normal line

For the normal line, the slope is the negative reciprocal of the tangent line slope, which is \(-\frac{1}{2\pi}\).Using the point-slope form, the equation becomes:\[y - 0 = -\frac{1}{2\pi}(x - 1)\].Simplified, it is:\[y = -\frac{1}{2\pi}x + \frac{1}{2\pi}\].

Key Concepts

Implicit Differentiation

Implicit differentiation is a method used to find the derivative of an equation that defines a function implicitly rather than explicitly. When dealing with equations where the dependent variable (like \(y\)) is intermixed with the independent variable (like \(x\)), implicit differentiation becomes handy. For an equation like \(y = 2 \sin(\pi x - y)\), directly solving for \(y\) in terms of \(x\) can be challenging or even impossible. \[ \text{Lack of separability means that explicit differentiation won't work.}\]Instead, we assume \(y\) is a function of \(x\) and differentiate both sides with respect to \(x\), using the chain rule when necessary. In our problem, after differentiating, we rearranged terms to solve for \(\frac{dy}{dx}\), which is the derivative of \(y\) with respect to \(x\).

• Use the chain rule: Differentiate \(\sin(\theta)\) where \(\theta=\pi x-y\), this gives \(\cos(\theta)(\pi - \frac{dy}{dx})\)
• Rearrange to isolate \(\frac{dy}{dx}\)

Trigonometric Functions

Trigonometric functions often appear in calculus problems to model wave-like phenomena. In this exercise, the sine function, \(\sin(\theta)\), is used. Sine functions are periodic and range between -1 and 1, making them ideal for modeling oscillations. The trigonometric identity that helped simplify the problem was knowing that: \[ \sin \pi = 0\quad \text{and}\quad \cos \pi = -1\]. When using trigonometric functions, understand the angles involved. The term \(\pi x - y\) is treated as an angle which is input for the sine and cosine functions. Here, recognizing \(\pi x\)'s influence was key to verifying the point \((1,0)\) was on the curve.

• \(\sin(\alpha)\) when \(\alpha = n\pi\) results in 0
• Trigonometric patterns repeat, indicating periodic nature
• To compute derivative, \(\cos\) function aids in finding slope of tangent

Point-Slope Form

The point-slope form is a convenient way to write the equation of a line when you know the slope and a point on the line. It's expressed as:\[y - y_1 = m(x - x_1)\], where \((x_1, y_1)\) is a point on the line, and \(m\) is the slope. To find the tangent line at the point \((1,0)\) in our example, we used the derivative \(\frac{dy}{dx}\) evaluated at this point to represent the slope of the tangent line. For the normal line, the slope is the negative reciprocal of the tangent slope, reflecting how normals are perpendicular to tangents.

• Tangent line's slope: evaluate derivative at specific point
• Normal line's slope: negative reciprocal of tangent slope (flip and negate!)
• Helps write equations quickly and accurately from known slope and points

Derivative Evaluation

Evaluating the derivative at a specific point tells us about the instantaneous rate of change of the function at that point, crucial for drawing tangent lines. Calculating \(\frac{dy}{dx}\) involves substituting the coordinates of the point \((1, 0)\) into the derived expression for the derivative. This gives the slope of the tangent. The derivative \(\frac{2\pi \cos(\pi x - y)}{1 + 2\cos(\pi x - y)}\) was evaluated at \((1, 0)\), simplifying with \(\cos \pi = -1\). This resulted in a slope of \(2\pi\), giving routes to find both tangent and normal lines.

• Substitute specific \((x, y)\) into \(\frac{dy}{dx}\) to get slope
• Simplify using known trigonometric values
• Confirmed by reaching a clear steepness of slope at that point