Problem 25 - Calculus Early Transcendentals

Question

Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. $$ x=e^{-t} \cos t, \quad y=e^{-t} \sin t, \quad z=e^{-t} ; \quad(1,0,1) $$

Step-by-Step Solution

Answer

Parametric equations: \(x(t) = 1-t, y(t) = t, z(t) = 1-t\).

Step 1: Express Coordinates in Terms of Given Value

We need to find the value of \( t \) that corresponds to the point \((1, 0, 1)\). Given the parametric equations: \(x = e^{-t} \cos t\), \(y = e^{-t} \sin t\), and \(z = e^{-t}\), we focus on \(z\) first since the equation is \(z = e^{-t} = 1\). Solving for \(t\), we get \(t = 0\).

Step 2: Verify Parametric Equations with the Value of \(t\)

With \(t = 0\), substitute it into the parametric equation for \(x\) and \(y\) to verify the coordinates: \(x(0) = e^{0} \cos(0) = 1\) and \(y(0) = e^{0} \sin(0) = 0\). These values match the point \((1, 0, 1)\), confirming \(t = 0\).

Step 3: Find the Derivatives of x, y, and z

Compute the derivatives with respect to \(t\) for each coordinate. For \(x\), use the product and chain rules: \(\frac{dx}{dt} = -e^{-t}\cos t - e^{-t}\sin t\). For \(y\): \(\frac{dy}{dt} = -e^{-t}\sin t + e^{-t}\cos t\). For \(z\): \(\frac{dz}{dt} = -e^{-t}\).

Step 4: Evaluate Derivatives at \(t = 0\)

Evaluate the derivatives at \(t = 0\) to get the tangent direction. Substituting \(t = 0\) gives \(\frac{dx}{dt} = -1\), \(\frac{dy}{dt} = 1\), and \(\frac{dz}{dt} = -1\). The direction vector of the tangent line is thus \((-1, 1, -1)\).

Step 5: Write the Parametric Equations for the Tangent Line

Using the point \((1, 0, 1)\) and the direction vector \((-1, 1, -1)\), construct the parametric equations for the tangent line: \(x(t) = 1 - t\), \(y(t) = t\), \(z(t) = 1 - t\).

Key Concepts

Tangent Line

A tangent line is a straight line that touches a curve at a single point without crossing it. When dealing with parametric equations, finding a tangent line involves

• identifying the point of contact,
• calculating the direction in which the line is heading.

For a curve given by parametric equations, the tangent line at a specific point can be represented by new parametric equations derived from the original curve. These equations use the point of contact and a direction vector found through derivatives.
The key is to determine the precise direction in which the curve is moving as it passes through the specified point.
The result is a straight line that gives a good linear approximation of the curve near this point.

Derivatives

Derivatives are fundamental in understanding how functions change. For parametric equations, they reveal the rates of change of each coordinate with respect to the parameter \(t\).
Given a parametric curve defined by functions \(x(t)\), \(y(t)\), and \(z(t)\), the derivative with respect to \(t\) provides a snapshot of how each coordinate moves as \(t\) changes.

• The derivative \(\frac{dx}{dt}\) shows the rate of change of \(x\),
• \(\frac{dy}{dt}\) indicates how \(y\) varies,
• \(\frac{dz}{dt}\) expresses the change in \(z\).

By calculating these, you determine the direction vector that the tangent line will follow.
This direction vector is crucial for forming the equations of the tangent line.

Direction Vector

A direction vector provides a means of describing the orientation and direction of a line. In parametric equations, it's especially important for defining tangent lines.
The direction vector is derived from the derivatives of the parametric equations of the curve.
For the example curve:

• Using the derivatives, \((-1, 1, -1)\) is calculated as the direction vector for the tangent line at the point \((1, 0, 1)\).

The values \(-1, 1,\) and \(-1\) tell us in which directions and how quickly the coordinates are moving as seen from the specified point.
This vector becomes part of the new parametric equations for the tangent line, allowing us to describe its straight path.

Parametric Curve

A parametric curve is a way to represent a curve using a set of equations that depend on a parameter, usually \(t\).
Instead of describing \(y\) as a function of \(x\), both coordinates are expressed as functions of \(t\), giving more flexibility in representing complex shapes and paths. In three dimensions, this includes \(z(t)\) as well.

• For example, given \(x = e^{-t} \cos t\), \(y = e^{-t} \sin t\), and \(z = e^{-t}\), the curve is described entirely by \(t\).
  The coordinates change as \(t\) varies, tracing out a path in space.

Parametric curves make it easier to find tangent lines and points of interest. By increasing \(t\), the entire path of the curve can be explored, highlighting how the curve moves through space.