Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the parametric equations of the tangent line to the curve \(x=2 t^{2}, y=4 t, z=t^{3}\) at \(t=1\).

Step-by-Step Solution

Answer

The parametric equations are \(x(t) = 2 + 4t\), \(y(t) = 4 + 4t\), \(z(t) = 1 + 3t\).

Step 1: Finding the Point on the Curve

First, substitute the parameter value \(t = 1\) into the parametric equations to find the point on the curve. Thus we have \(x = 2(1)^2 = 2\), \(y = 4(1) = 4\), and \(z = (1)^3 = 1\). Therefore, the point is \((2, 4, 1)\).

Step 2: Finding the Derivatives

To find the direction vector of the tangent line, we need the derivatives of \(x\), \(y\), and \(z\) with respect to \(t\). Compute \(\frac{dx}{dt} = 4t\), \(\frac{dy}{dt} = 4\), and \(\frac{dz}{dt} = 3t^2\).

Step 3: Evaluating the Derivatives at t=1

Substitute \(t = 1\) into the derivatives: \(\frac{dx}{dt} = 4(1) = 4\), \(\frac{dy}{dt} = 4\), and \(\frac{dz}{dt} = 3(1)^2 = 3\). So, the direction vector is \((4, 4, 3)\).

Step 4: Writing the Parametric Equations

The parametric equations of the tangent line can be written using the point \((2, 4, 1)\) and the direction vector \((4, 4, 3)\). Therefore, the equations are \(x(t) = 2 + 4t\), \(y(t) = 4 + 4t\), \(z(t) = 1 + 3t\).

Key Concepts

Parametric Equations

Parametric equations are a powerful way to describe curves in mathematics. Instead of defining a curve with a single equation, such as a line or a parabola, parametric equations use a parameter, often denoted by \(t\), to express the coordinates of points on the curve. This approach allows us to handle more complex shapes and movements. For example, in our exercise, the curve is given by the equations:

• \(x = 2t^2\)
• \(y = 4t\)
• \(z = t^3\)

Each coordinate \(x, y,\) and \(z\) is expressed as a function of \(t\). These equations describe the position of the points along the curve as \(t\) changes. For every value of \(t\), there is a corresponding point \((x, y, z)\) on the curve. This method is particularly useful for finding tangent lines, since the changes in \(t\) can easily be used to find derivatives.

Derivatives

In calculus, derivatives help us understand how functions change. When dealing with parametric equations, taking derivatives with respect to the parameter \(t\) allows us to analyze the behavior of each coordinate. In our exercise, we found the derivatives of \(x\), \(y\), and \(z\) with respect to \(t\):

• \(\frac{dx}{dt} = 4t\)
• \(\frac{dy}{dt} = 4\)
• \(\frac{dz}{dt} = 3t^2\)

These derivatives tell us the rate of change of each coordinate with respect to \(t\). By substituting a specific value of \(t\), such as \(t = 1\), these derivatives reveal the slope of the tangent line to the curve at that point. Thus, they help in constructing the direction vector for the tangent line. Understanding this concept is key to tackling many 3D motion problems and geometric computations.

Direction Vector

The direction vector is an essential component for determining the tangent line to a curve at a specific point. It provides the direction in which the tangent line extends. Once we have the derivatives of the coordinates, as in our exercise, we can evaluate them at a specific \(t\) to find the components of the direction vector. At \(t = 1\), the derivatives calculated were \(\frac{dx}{dt} = 4\), \(\frac{dy}{dt} = 4\), and \(\frac{dz}{dt} = 3\). Therefore, the direction vector is \((4, 4, 3)\). The direction vector gives us a sense of how sharp the curve is at that point and dictates the precise slope of the tangent line. It's like an arrow pointing to where the curve "heads next" at that particular spot, capturing the instantaneous rate of change in all coordinate directions.

Point on the Curve

Determining a specific point on the curve is crucial for writing the equation of a tangent line. In parametric equations, we do this by plugging a specific parameter value into the equations. In our exercise, we used \(t = 1\). By substituting \(t\) in the parametric equations:

• \(x = 2(1)^2 = 2\)
• \(y = 4(1) = 4\)
• \(z = (1)^3 = 1\)

We find the point \((2, 4, 1)\) on the curve. This point acts as the starting position for the tangent line. Knowing the exact location on the curve is essential because it specifies the point from which the tangent extends. Together with the direction vector, it allows us to form the full parametric equations of the tangent line, effectively describing its path in space.