Problem 32 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 31-34, find the equation of the tangent line to the given curve at the given value of \(t\) without eliminating the parameter. Make a sketch. $$ x=3 t, y=8 t^{3} ; t=-\frac{1}{2} $$

Step-by-Step Solution

Answer

The tangent line equation is \(y = 2x + 2\).

Step 1: Find the Derivative

To find the equation of the tangent line, first, we need the derivative of the curve with respect to the parameter \(t\). For a parametric curve, we have \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\). Here, \(x = 3t\) and \(y = 8t^3\). Thus, \(\frac{dx}{dt} = 3\) and \(\frac{dy}{dt} = 24t^2\).

Step 2: Calculate the Slope of the Tangent

The slope \(m\) of the tangent line is given by \(m = \frac{dy/dt}{dx/dt}\). Substituting the derivatives, we get \(m = \frac{24t^2}{3} = 8t^2\). At \(t = -\frac{1}{2}\), the slope \(m\) becomes \(8\left(-\frac{1}{2}\right)^2 = 8 \times \frac{1}{4} = 2\).

Step 3: Find the Point on the Curve

Substitute \(t = -\frac{1}{2}\) into the parametric equations to find the coordinates of the point on the curve. For \(x\), we have \(x = 3\left(-\frac{1}{2}\right) = -\frac{3}{2}\). For \(y\), we have \(y = 8\left(-\frac{1}{2}\right)^3 = 8 \times \left(-\frac{1}{8}\right) = -1\). Therefore, the point is \((-\frac{3}{2}, -1)\).

Step 4: Write the Equation of the Tangent Line

Using the point-slope form of the line \(y - y_1 = m(x - x_1)\), where \((x_1, y_1) = (-\frac{3}{2}, -1)\) and \(m = 2\), the equation becomes \(y + 1 = 2\left(x + \frac{3}{2}\right)\). Simplifying, we get \(y = 2x + 2\).

Step 5: Sketch the Curve and Tangent Line

To sketch the curve and the tangent line, plot the parametric curve defined by \(x = 3t\) and \(y = 8t^3\). Mark the point \((-\frac{3}{2}, -1)\), and draw the line with equation \(y = 2x + 2\) through this point. The tangent line should just touch the curve at this point.

Key Concepts

Parametric Equations

Understanding parametric equations is crucial when dealing with curves in a plane. Unlike regular functions that express one variable in terms of another, parametric equations describe both variables using a third parameter, usually denoted as \( t \). This makes them very flexible and capable of modeling a wide range of curves.

In the given exercise, the equations \( x = 3t \) and \( y = 8t^3 \) express "x" and "y" coordinates as functions of \( t \). As \( t \) varies, it traces out a path on the plane. This representation is particularly helpful when the curve’s shape is too complex to be described easily with a single standard function.

Here are the key advantages of parametric equations:

• They can represent curves that loop or are multi-valued.
• They allow us to express motion, where \( t \) might represent time.
• They simplify calculus operations like differentiation and integration in certain cases.

Slope of Tangent

The slope of a tangent line gives us the direction and steepness of the line just touching the curve at a point. In parametric forms, this can still be calculated even though the curve isn't given in terms of \( y = f(x) \).

To find the slope of the tangent line for parametric equations, we first need to determine derivatives of each component with respect to \( t \): \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \). The slope \( m \) is given by the ratio \( \frac{dy/dt}{dx/dt} \).

In our problem, substituting \( t = -\frac{1}{2} \) yields the slope calculation: \( 8t^2 \) becomes 2, which gives us a precise direction of the tangent at the specified \( t \). This slope is what ensures the tangent line properly aligns with the curve at the point \((x, y) \).

• The tangent slope tells if the curve is increasing or decreasing at the point.
• A positive slope indicates an upward trend, whereas negative means downward.
• Zero slope suggests a flat, horizontal tangent.

Derivative

The concept of a derivative extends to parametric equations in calculus, where it represents rate of change. For a single variable function, a derivative at a point indicates its slope. This principle carries over to parametric equations too.

In the exercise, we took derivatives with respect to \( t \): \( \frac{dx}{dt} = 3 \) and \( \frac{dy}{dt} = 24t^2 \). Each derivative tells us how \( x \) and \( y \) change as \( t \) changes.

Derivatives in parametric equations allow us to compute not just the tangent slope but also other calculus operations, such as identifying points of inflection or maximum curvature.

• Helps in understanding the nature of the curve near points.
• Aids in sketching the curve's shape more accurately.
• Useful in various applications like physics for describing object motion.

Point-Slope Form

The point-slope form is a powerful way to express the equation of a line when you know a point on the line and the line's slope. It’s particularly helpful when constructing a tangent line.

For the given task, once we determined the slope \( m = 2 \) and identified the point on the curve as \((-\frac{3}{2}, -1) \), we could immediately use the point-slope formula: \( y - y_1 = m(x - x_1) \). Plugging in our values, we derive \( y + 1 = 2(x + \frac{3}{2}) \).

This formula is intuitive because it visually shows the slope behavior, starting from a known point. Thus, simplifying to \( y = 2x + 2 \) reveals the tangent line's equation.

• Quickly develops a linear equation when a point and slope are known.
• Can easily transition to slope-intercept form for graphing.
• Helps to visually connect analytic results with graphical interpretation.