Problem 51 - Essential Calculus Early Transcendentals

Question

\(49-52=\) Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. $$x=e^{-t} \cos t, \quad y=e^{-t} \sin t, \quad z=e^{-t_{ ;}},(1,0,1)$$

Step-by-Step Solution

Answer

The parametric equations are \(x = 1 - t\), \(y = t\), and \(z = 1 - t\).

Step 1: Identify the Given Equations and Point

We are provided with the parametric equations for a curve: \(x = e^{-t} \cos t\), \(y = e^{-t} \sin t\), and \(z = e^{-t}\). The specified point at which we need to find the tangent line is \((1, 0, 1)\).

Step 2: Find the Parameter Value

First, we determine the value of \(t\) that corresponds to the given point. For \(z = e^{-t} = 1\), we solve \(1 = e^{-t}\), which implies \(-t = 0\) or \(t = 0\).

Step 3: Calculate the Derivatives

Next, we compute the derivatives \(\frac{dx}{dt}\), \(\frac{dy}{dt}\), and \(\frac{dz}{dt}\).- \(\frac{dx}{dt} = \frac{d}{dt}(e^{-t} \cos t) = -e^{-t} \cos t - e^{-t} \sin t\)- \(\frac{dy}{dt} = \frac{d}{dt}(e^{-t} \sin t) = -e^{-t} \sin t + e^{-t} \cos t\)- \(\frac{dz}{dt} = \frac{d}{dt}(e^{-t}) = -e^{-t}\)

Step 4: Evaluate at the Parameter Value

Evaluate the derivatives at \(t = 0\).- \(\frac{dx}{dt}\bigg|_{t=0} = -e^{0} \cos 0 - e^{0} \sin 0 = -1\)- \(\frac{dy}{dt}\bigg|_{t=0} = -e^{0} \sin 0 + e^{0} \cos 0 = 1\)- \(\frac{dz}{dt}\bigg|_{t=0} = -e^{0} = -1\)

Step 5: Construct the Parametric Equations for the Tangent Line

The parametric equations for the tangent line are derived using the point and the evaluated derivatives.- \(x = 1 - t\)- \(y = 1\cdot t = t\)- \(z = 1 - t\)These represent the tangent line at the point \((1, 0, 1)\).

Key Concepts

Parametric Equations

Parametric equations are a way of expressing mathematical functions where each coordinate is represented as a function of a separate independent variable, typically denoted as a parameter. In our exercise, the functions provided are shown as follows:
\[x = e^{-t} \cos t, \quad y = e^{-t} \sin t, \quad z = e^{-t}\]
These equations describe a curve in three-dimensional space. The parameter \(t\) varies over a specific range, allowing us to plot the trajectory of the curve as \(t\) increments. One great advantage of parametric equations is their flexibility in representing complex curves, which are hard to model using only one equation for each coordinate. Using this representation, we can easily analyze and find tangent lines or other geometric properties.

Derivatives

Derivatives are fundamental in calculus and are used to determine the rate of change of one quantity with respect to another.
In the context of parametric equations, derivatives help us understand how each coordinate of a curve changes with respect to the parameter \(t\). For instance, to find the velocity (change in position) of an object following the parametric path, we calculate the derivatives:

• \(\frac{dx}{dt}\) represents the rate of change of the \(x\)-coordinate with respect to \(t\)
• \(\frac{dy}{dt}\) represents the rate of change of the \(y\)-coordinate with respect to \(t\)
• \(\frac{dz}{dt}\) represents the rate of change of the \(z\)-coordinate with respect to \(t\)

In our problem, calculating these derivatives for:
- \(x = e^{-t} \cos t\) gives \(-e^{-t} \cos t - e^{-t} \sin t\)
- \(y = e^{-t} \sin t\) gives \(-e^{-t} \sin t + e^{-t} \cos t\)
- \(z = e^{-t}\) gives \(-e^{-t}\)
These results provide vital insights into the curve's orientation and movement characteristics, crucial for finding tangent lines.

Evaluation of Derivatives

After obtaining the derivatives, the next step is evaluating them at a specific point, which in this case is when \(t = 0\). Evaluating the derivatives gives us the tangent vector, which points in the direction of the curve's immediate movement at the specified point.
For the point \((1, 0, 1)\), we resolve \(t = 0\) which simplifies the derivatives to:
- \(\frac{dx}{dt}\bigg|_{t=0} = -1\)
- \(\frac{dy}{dt}\bigg|_{t=0} = 1\)
- \(\frac{dz}{dt}\bigg|_{t=0} = -1\)
The resulting values of \(-1\), \(1\), and \(-1\) represent the components of the tangent vector at the specified point on the curve. With this tangent vector in hand, constructing the parametric equations for the tangent line becomes straightforward:
- \(x = 1 - t\)
- \(y = 0 + t\) which simplifies to \(y = t\)
- \(z = 1 - t\)
This method demonstrates how derivatives and their evaluations inform us about the behavior and geometric properties of parametric curves.