Problem 60 - Thomas Calculus

Question

The cissoid of Diocles (from about 200 B.C.) Find equations for the tangent and normal to the cissoid of Diocles \(y^{2}(2-x)=x^{3}\) at \((1,1) .\)

Step-by-Step Solution

Answer

Tangent: \( y = 2x - 1 \); Normal: \( y = -\frac{1}{2}x + \frac{3}{2} \).

Step 1: Differentiate the Equation

First, find the equation for the derivative \( \frac{dy}{dx} \). Start by solving \( y^2 (2-x) = x^3 \) implicitly. Differentiate both sides with respect to \(x\):\[ \frac{d}{dx}(y^2(2-x)) = \frac{d}{dx}(x^3) \]Using the product and chain rules, we have:\[ 2y \cdot \frac{dy}{dx} (2-x) - y^2 = 3x^2 \]

Step 2: Solve for \( \frac{dy}{dx} \)

Rearrange the equation obtained from differentiation to solve for \( \frac{dy}{dx} \).\[ 2y (2-x) \cdot \frac{dy}{dx} = 3x^2 + y^2 \]\[ \frac{dy}{dx} = \frac{3x^2 + y^2}{2y(2-x)} \]

Step 3: Evaluate \( \frac{dy}{dx} \) at (1,1)

Substitute \( x = 1 \) and \( y = 1 \) into the derivative:\[ \frac{dy}{dx} = \frac{3(1)^2 + (1)^2}{2(1)(2-1)} = \frac{3 + 1}{2} = 2 \]

Step 4: Find Equation of the Tangent Line

To find the equation of the tangent line at \( (1, 1) \), use the point-slope form of a line, \( y - y_1 = m(x - x_1) \), where \( m \) is the slope.\[ y - 1 = 2(x - 1) \]Simplifying, the equation becomes:\[ y = 2x - 1 \]

Step 5: Find Equation of the Normal Line

The slope of the normal line is the negative reciprocal of the tangent's slope. Therefore, the normal line slope is \(-\frac{1}{2}\).Using the point-slope form:\[ y - 1 = -\frac{1}{2}(x - 1) \]Simplifying, the equation becomes:\[ y = -\frac{1}{2}x + \frac{3}{2} \]

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique used when dealing with equations where the variables cannot be easily separated. This means that the equation is not explicitly solved for one variable in terms of the other, such as in the equation \( y^2(2-x) = x^3 \). Instead of solving for \( y \) and then differentiating, we differentiate the entire equation.
The process involves differentiating both sides with respect to \( x \), treating \( y \) as a function of \( x \). This creates derivatives that include \( \frac{dy}{dx} \), reflecting the change in \( y \) with respect to \( x \). Implicit differentiation is crucial for equations where solving for one variable is either too complex or not possible.

Tangent Line

The tangent line to a curve at a given point is a straight line that just "touches" the curve at that point, matching the curve's immediate direction. To find the equation of a tangent line, we first need to determine the slope of the curve at the particular point. This is done using the derivative.
For the cissoid of Diocles, after finding \( \frac{dy}{dx} \), we can determine the slope at the point \((1,1)\). With a slope \( m = 2 \), the equation of the line can be derived using the point-slope form:

• Point-slope form: \( y - y_1 = m(x - x_1) \)
• In this example: \( y - 1 = 2(x - 1) \)

By rearranging, we get the tangent line \( y = 2x - 1 \), which closely follows the curvature at \( (1,1) \).

Normal Line

A normal line to a curve at a particular point is perpendicular to the tangent line at that point. Finding the normal line involves computing the negative reciprocal of the tangent slope.
For example, if the slope of the tangent is \( 2 \), then the slope of the normal line is \(-\frac{1}{2}\). Using the point-slope form once again with this new slope, the normal line's equation becomes:

• Point-slope form: \( y - y_1 = m(x - x_1) \)
• For the normal line: \( y - 1 = -\frac{1}{2}(x - 1) \)

Simplifying, we find that the normal line is \( y = -\frac{1}{2}x + \frac{3}{2} \), which intersects the curve perpendicularly at \((1,1)\).

Product Rule

The product rule is used when differentiating products of functions. If you have two functions of \( x \), such as \( u(x) \) and \( v(x) \), the derivative of their product \( uv \) is given by:

• \( (uv)' = u'v + uv' \)

In the context of implicit differentiation, it is often used to differentiate expressions like \( y^2(2-x) \).
Here, we apply the product rule to differentiate this expression with respect to \( x \): the derivative of \( y^2 \) times \( (2-x) \) plus \( y^2 \) times the derivative of \( (2-x) \). This results in:

• \( 2y \cdot \frac{dy}{dx} (2-x) - y^2 \)

Thereby revealing its crucial role in complex differentiations.

Chain Rule

The chain rule is vital when dealing with composed functions, where one function is inside another. For such a scenario, if \( y = f(g(x)) \), the derivative \( y' \) is found by:

• \( y' = f'(g(x)) \cdot g'(x) \)

In implicit differentiation, the chain rule often assists with differentiating terms involving \( y \). It helps us treat \( y \) as a function of \( x \) even if \( y \) is not explicitly solved.
For example, in our problem the \( y^2 \) term requires differentiating \( y \) (in \( 2y \cdot \frac{dy}{dx} \)), and the chain rule manages this composite nature by multiplying the derivative \( 2y \) with \( \frac{dy}{dx} \). This ensures that every dependence on \( x \) is accounted for correctly.