Question

For the following exercises, find points on the curve at which tangent line is horizontal or vertical.\(x=\frac{3 t}{1+t^{3}}, \quad y=\frac{3 t^{2}}{1+t^{2}}\)

Short answer

Tangent is horizontal at (0, 0), (\(\frac{3}{2}, \frac{3}{2}\)), (-3, 3); vertical at (\(\frac{3}{2}, \frac{3}{2}\)).

Step-by-step solution

Find Derivatives

To find the points where the tangent line is horizontal or vertical, we need to compute the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\). First, determine \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\):For \(x(t)=\frac{3t}{1+t^{3}}\):Using the quotient rule, \(\frac{dx}{dt} = \frac{(1+t^3)(3) - 3t(3t^2)}{(1+t^3)^2} = \frac{3(1-t^3)}{(1+t^3)^2}\).For \(y(t)=\frac{3t^2}{1+t^2}\):Again using the quotient rule, \(\frac{dy}{dt} = \frac{(1+t^2)(6t) - 3t^2(2t)}{(1+t^2)^2} = \frac{6t - 6t^3}{(1+t^2)^2}\).

Horizontal Tangent Condition

The tangent line is horizontal when \(\frac{dy}{dt} = 0\). So we set \[\frac{dy}{dt} = \frac{6t - 6t^3}{(1+t^2)^2} = 0\]This simplifies to solving \(6t(1-t^2) = 0\). Thus, the solutions are: \(t = 0\) or \(t = 1\) or \(t = -1\).

Vertical Tangent Condition

The tangent line is vertical when \(\frac{dx}{dt} = 0\). Set \[\frac{dx}{dt} = \frac{3(1-t^3)}{(1+t^3)^2} = 0\]This simplifies to solving \(1-t^3 = 0\), which gives: \(t = 1\).

Calculate Points for Horizontal Tangent

Substitute \(t = 0, 1, -1\) into the original equations for \(x\) and \(y\) to find the points:- For \(t = 0\): \(x = \frac{3\times 0}{1+0^3} = 0\), \(y = \frac{3\times 0^2}{1+0^2} = 0\); Point is (0, 0).- For \(t = 1\): \(x = \frac{3\times 1}{1+1^3} = \frac{3}{2}\), \(y = \frac{3\times 1^2}{1+1^2} = \frac{3}{2}\); Point is (\(\frac{3}{2}\), \(\frac{3}{2}\)).- For \(t = -1\): \(x = \frac{3\times (-1)}{1+(-1)^3} = -3\), \(y = \frac{3\times (-1)^2}{1+(-1)^2} = 3\); Point is (-3, 3).

Calculate Points for Vertical Tangent

Substitute \(t = 1\) (since this was found in Step 3) into the original equations for \(x\) and \(y\) to verify: Point at \(t = 1\) is already calculated as (\(\frac{3}{2}, \frac{3}{2}\)).

Key concepts

Derivatives

In calculus, derivatives represent how a function changes as its input changes. They are crucial for understanding the behavior of curves defined by equations. Here, we focus on parametric equations, which define points on a curve using a parameter, often denoted as \(t\). Each point on the curve can be described by a set of equations for \(x\) and \(y\) as functions of \(t\).To analyze the curve, we find the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\). This task requires the use of the **Quotient Rule** because our equations for \(x(t)\) and \(y(t)\) are rational functions, involving both polynomials and division. The quotient rule states \(\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}\), which allows us to compute the rate of change of these functions.- For \(x(t) = \frac{3t}{1+t^3}\), this means applying...- For \(y(t) = \frac{3t^2}{1+t^2}\), similarly, the quotient rule is applied...These derivatives give us insight into how \(x\) and \(y\) change with \(t\), foundational for finding the conditions for horizontal and vertical tangents.

Tangent Line Conditions

Understanding tangent line conditions is key to analyzing curve behavior in calculus. A tangent line is a straight line that just "touches" a curve at a given point, showing the curve's direction at that point. We focus on two types of tangents: horizontal and vertical.### Horizontal Tangent ConditionA tangent line is horizontal if its slope is zero. For parametric curves, this condition is realized when \(\frac{dy}{dt} = 0\) because the line does not rise or fall as \(x\) changes. By setting the derivative \(\frac{dy}{dt} = 0\), we solve for the specific values of \(t\) that make it zero. This involves managing the equation \(6t(1-t^2) = 0\), simplifying it, and solving.- This results in solutions \(t = 0\), \(t = 1\), and \(t = -1\).### Vertical Tangent ConditionA vertical tangent occurs when the line's slope is undefined, notably when \(\frac{dx}{dt} = 0\). Here, we attempt to find when the change in \(x\) per change in \(t\) is zero, determined by the equation \(1-t^3 = 0\).- Simplifying gives us the solution \(t = 1\). These conditions let us pinpoint exactly when a curve might turn horizontally or rise/fall steeply with an undefined slope.

Parametric Equations

Parametric equations are a method of defining geometrical shapes by using parameters, such as \(t\). Different values of \(t\) correspond to different points on the curve, and collectively, these points represent the entire path of the curve.The given parametric forms \(x = \frac{3t}{1+t^3}\) and \(y = \frac{3t^2}{1+t^2}\) allow every point on the curve to be uniquely described. These forms are particularly helpful when a curve can't be easily expressed as a single function in the form \(y = f(x)\).### Advantages of Parametric Representation:- Simplifies complex shapes and multi-valued functions.- Separates motion or shape into horizontal and vertical components, allowing for more versatile or cyclic motion, like loops or spirals.By analyzing the system using derivatives and tangent line conditions, we can make precise conclusions about the curve, such as where it flattens out or becomes vertical. These tools aid in deeper calculus problems, not just visualizing but quantitatively describing tangent behavior. Parametric equations thus serve as a powerful means to describe complex curves straightforwardly and comprehensively, offering insight into the hidden behavior of the curve.