Question

Show that the normal line to \(x^{3}+y^{3}=3 x y\) at \(\left(\frac{3}{2}, \frac{3}{2}\right)\) passes through the origin.

Short answer

The normal line to the curve at the given point is \( y = x \), which passes through the origin.

Step-by-step solution

Differentiate the given equation

First, we need to find the derivative of the given implicit equation with respect to \( x \). The equation is \( x^3 + y^3 = 3xy \). Differentiate both sides of the equation.\[ \frac{d}{dx}(x^3 + y^3) = \frac{d}{dx}(3xy) \]Using the chain rule and product rule, we get:\[ 3x^2 + 3y^2 \cdot \frac{dy}{dx} = 3y + 3x \cdot \frac{dy}{dx} \]

Solve for the derivative

Rearrange the equation from Step 1 to solve for \( \frac{dy}{dx} \):\[ 3y^2 \cdot \frac{dy}{dx} - 3x \cdot \frac{dy}{dx} = 3y - 3x^2 \]Factor to isolate \( \frac{dy}{dx} \):\[ \frac{dy}{dx} (3y^2 - 3x) = 3y - 3x^2 \]Divide both sides:\[ \frac{dy}{dx} = \frac{3y - 3x^2}{3y^2 - 3x} \]

Evaluate the derivative at the given point

Plug in the point \( \left( \frac{3}{2}, \frac{3}{2} \right) \) into the derivative to find the slope of the tangent line at this point.\[ \frac{dy}{dx} = \frac{3 \left( \frac{3}{2} \right) - 3 \left( \frac{3}{2} \right)^2}{3 \left( \frac{3}{2} \right)^2 - 3 \left( \frac{3}{2} \right)} \]Calculate the values:\[ \frac{dy}{dx} = \frac{\frac{9}{2} - \frac{27}{4}}{\frac{27}{4} - \frac{9}{2}} = \frac{\frac{18}{4} - \frac{27}{4}}{\frac{27}{4} - \frac{18}{4}} = \frac{-\frac{9}{4}}{\frac{9}{4}} = -1 \]

Determine the slope of the normal line

The slope of the normal line is the negative reciprocal of the slope of the tangent line. Since the tangent line has a slope of \(-1\), the normal line will have a slope:\[ m_{normal} = -\frac{1}{-1} = 1 \]

Write the equation of the normal line

Using the point-slope form with the point \( \left( \frac{3}{2}, \frac{3}{2} \right) \) and the slope of the normal line:\[ y - \frac{3}{2} = 1 \cdot \left( x - \frac{3}{2} \right) \]Simplify to get the equation of the normal line:\[ y = x \]

Show the normal line passes through the origin

Since the equation of the normal line is \( y = x \), it clearly passes through the origin \((0, 0)\) because when \( x = 0 \), \( y = 0 \). This confirms that the normal line passes through the origin.

Key concepts

Implicit Differentiation

Implicit differentiation is a useful technique in calculus, especially when dealing with equations where one variable is not isolated. In equations like \(x^3 + y^3 = 3xy\), both \(x\) and \(y\) are intertwined. Unlike explicit functions where \(y\) is isolated (e.g., \(y = f(x)\)), implicit differentiation helps us differentiate these mixed equations directly with respect to \(x\). This involves applying the chain rule to differentiate terms containing \(y\), assuming \(y\) is a function of \(x\).

The key steps in using implicit differentiation are:

• Differentiate both sides of the equation with respect to \(x\).
• Use the chain rule for terms involving \(y\), writing the derivative as \(\frac{dy}{dx}\) whenever you differentiate \(y\), since \(y\) is considered as a function of \(x\).
• Simplify and solve for \(\frac{dy}{dx}\), which gives the slope of the tangent line.

Normal Lines in Calculus

Normal lines are lines that are perpendicular to tangent lines at a given point on a curve. If you understand the concept of perpendicularity, finding normal lines in calculus becomes straightforward. When dealing with a curve defined by an implicit equation like \(x^3 + y^3 = 3xy\), normal lines can help you understand the geometry and behavior of the curve.

To find the normal line, first find the slope of the tangent line at the point of interest using implicit differentiation. Next, determine the slope of the normal line, which will be the negative reciprocal of the tangent line's slope. This step emphasizes the perpendicular nature, as the product of slopes of two perpendicular lines in a plane is \(-1\).

• Tangent line: The line that "just touches" the curve at a particular point.
• Normal line: The line perpendicular to the tangent line.
• Perpendicularity: \(m_1 \times m_2 = -1\), where \(m_1\) and \(m_2\) are the slopes of perpendicular lines.

Slope of Tangent and Normal Lines

The slope of tangent and normal lines is critical in analyzing curves. The tangent line's slope at a specific point gives an instantaneous rate of change of the function at that point. This is crucial in understanding how the curve behaves locally around that point.

For our function, \(x^3 + y^3 = 3xy\), the tangent slope \(m_t\) at \(\left(\frac{3}{2}, \frac{3}{2}\right)\) was found using implicit differentiation to be \(-1\). Consequently, the normal line's slope \(m_n\) is the negative reciprocal, \(1\).

• The slope of a tangent line is derived from the derivative \(\frac{dy}{dx}\).
• The slope of a normal line can be computed as \(m_n = -\frac{1}{m_t}\).

These slopes are vital as they provide insights into directions and angles at which the curves change or tilt.

Point-Slope Form

The point-slope form is a linear equation format used to easy write the equation of a line when you know the slope and a point on the line. It's particularly handy in calculus when dealing with tangent and normal lines. The formula is given by \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is the point.

Using the point-slope form simplifies finding the equation of the normal line for implicit curves like \(x^3 + y^3 = 3xy\). With the normal line slope \(1\) and the point \(\left(\frac{3}{2}, \frac{3}{2}\right)\), the normal line can be expressed as \(y - \frac{3}{2} = 1\times(x - \frac{3}{2})\), which simplifies to \(y = x\).

• Helps simplify the process of finding specific line equations.
• Useful for describing the relationship between a curve and tangent or normal lines at specific points.

Understanding and applying the point-slope form correctly is crucial for writing accurate equations of lines in calculus.