Problem 92 - Calculus Volume 3

Question

Find points on the curve at which tangent line is horizontal or vertical. $$ x=\frac{3 t}{1+t^{3}}, \quad y=\frac{3 t^{2}}{1+t^{3}} $$

Step-by-Step Solution

Answer

The tangent is horizontal at (0,0) and (3√2/3,2√2/3) and vertical at (2√1/2,3(1/2)^(2/3)).

Step 1: Understand Horizontal Tangent Line Condition

A horizontal tangent line occurs when the derivative of the curve with respect to \(x\), \(\frac{dy}{dx}\), is zero. We first need expressions for \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) to find \(\frac{dy}{dx}\).

Step 2: Compute Derivatives dx/dt and dy/dt

Differentiate \(x = \frac{3t}{1+t^3}\) to get \(\frac{dx}{dt}\). Using the quotient rule, we have:\[\frac{dx}{dt} = \frac{(1+t^3)(3) - 3t(3t^2)}{(1+t^3)^2} = \frac{3 - 6t^3}{(1+t^3)^2}.\]Similarly, differentiate \(y = \frac{3t^2}{1+t^3}\) to get \(\frac{dy}{dt}\):\[\frac{dy}{dt} = \frac{(1+t^3)(6t) - 3t^2(3t^2)}{(1+t^3)^2} = \frac{6t - 9t^4}{(1+t^3)^2}.\]

Step 3: Calculate dy/dx

Use the chain rule to find \(\frac{dy}{dx}\):\[\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{6t - 9t^4}{3 - 6t^3}.\]

Step 4: Set dy/dx to Zero for Horizontal Tangent

To find \(t\) for which the tangent line is horizontal, set \(\frac{dy}{dx} = 0\):\[6t - 9t^4 = 0.\]Factor out \(3t\):\[3t(2 - 3t^3) = 0.\]Solutions are \(t = 0\) or \(t = \sqrt[3]{\frac{2}{3}}\).

Step 5: Find Corresponding Points for Horizontal Tangents

For \(t = 0\), substitute into \(x\) and \(y\):\[x = \frac{3(0)}{1+0^3} = 0, \quad y = \frac{3(0)^2}{1+0^3} = 0.\]Thus, point \((0, 0)\) has a horizontal tangent.For \(t = \sqrt[3]{\frac{2}{3}}\), substitute into \(x\) and \(y\):\[x = \frac{3 \cdot \sqrt[3]{\frac{2}{3}}}{1 + \left(\sqrt[3]{\frac{2}{3}}\right)^3} = 3\cdot\sqrt[3]{\frac{2}{3}}, \quad y = \frac{3 \cdot \left(\sqrt[3]{\frac{2}{3}}\right)^2}{1 + \left(\sqrt[3]{\frac{2}{3}}\right)^3} = 2\cdot\sqrt[3]{\frac{2}{3}}.\]

Step 6: Understand Vertical Tangent Line Condition

A vertical tangent line occurs when \(\frac{dx}{dt} = 0\) and \(\frac{dy}{dt} eq 0\). We have an expression for \(\frac{dx}{dt}\) from Step 2.

Step 7: Set dx/dt to Zero for Vertical Tangent

Set \(\frac{dx}{dt} = 0\):\[3 - 6t^3 = 0.\]Solving for \(t\) gives:\[t = \sqrt[3]{\frac{1}{2}}.\]

Step 8: Find Corresponding Points for Vertical Tangents

Plug \(t = \sqrt[3]{\frac{1}{2}}\) back into \(x\) and \(y\):\[x = \frac{3 \cdot \sqrt[3]{\frac{1}{2}}}{1 + \left(\sqrt[3]{\frac{1}{2}}\right)^3} = 2 \cdot \sqrt[3]{\frac{1}{2}}, \quad y = \frac{3 \cdot \left(\sqrt[3]{\frac{1}{2}}\right)^2}{1 + \left(\sqrt[3]{\frac{1}{2}}\right)^3} = 3 \cdot \left(\sqrt[3]{\frac{1}{2}}\right)^2.\]

Key Concepts

Horizontal Tangent

When studying tangent lines, a horizontal tangent is where the slope of the line is zero. This happens when the change in y with respect to x, or the derivative \( \frac{dy}{dx} \), is zero. To find a horizontal tangent for a curve defined by parametric equations, we calculate the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \), and then find \( \frac{dy}{dx} \) using the formula \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \). The challenge is to determine \( t \) where \( \frac{dy}{dx} = 0 \). This involves solving the equation \( 6t - 9t^4 = 0 \), leading to specific solutions for \( t \) that correspond to points where the tangent is horizontal, such as \( (0,0) \) and another point found through substitution.

Vertical Tangent

A vertical tangent occurs when the curve's rate of change in x terms is zero but not in y terms. Mathematically, this means \( \frac{dx}{dt} = 0 \) but \( \frac{dy}{dt} eq 0 \). In our exercise, we use the expression for \( \frac{dx}{dt} \) to find such a point. We set the equation \( 3 - 6t^3 = 0 \) and solve for \( t \). This value of \( t \) is used to substitute back into the parametric equations of x and y to find the exact coordinates where the vertical tangent line occurs. This provides a clear understanding of how the vertical tangent is defined independently of the horizontal.

Derivatives

Derivatives are fundamental when working with tangent lines. They help us understand how a curve behaves and changes. For parametric equations, you must calculate \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) separately. Using the derivative rules, especially the quotient rule, is necessary when the equations are fractions. For instance, differentiating \( x = \frac{3t}{1+t^3} \) and \( y = \frac{3t^2}{1+t^3} \) requires careful application of these rules to obtain \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). From these, \( \frac{dy}{dx} \) is found by dividing \( \frac{dy}{dt} \) by \( \frac{dx}{dt} \), crucial for determining the slope of tangent lines.

Parametric Equations

Parametric equations express a set of related quantities as functions of a parameter, usually \( t \). They are incredibly useful for describing complex curves that can't be represented with a single function. In this context, \( x = \frac{3t}{1+t^3} \) and \( y = \frac{3t^2}{1+t^3} \) are parametric equations with \( t \) as an independent parameter. Understanding these allows us to trace out the curve as \( t \) changes. Each value of \( t \) provides coordinates \( (x,y) \) on the plane. This understanding is key when we need to analyze features like tangent lines, because it reveals how changes in \( t \) affect the entire curve.