Problem 29 - Calculus Early Transcendentals

Question

Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. Illustrate by graphing both the curve and the tangent line on a common screen. $$ x=t \cos t, y=t, z=t \sin t ; \quad(-\pi, \pi, 0) $$

Step-by-Step Solution

Answer

The tangent line at \((-\pi, \pi, 0)\) is given by \((x, y, z) = (-\pi - t, \pi + t, -\pi t)\).

Step 1: Differentiate the parametric equations

To find the tangent line, we first need to determine the direction vector of the tangent by differentiating the given parametric equations with respect to the parameter \(t\). Given the curve:- \(x = t \cos t\)- \(y = t\)- \(z = t \sin t\)Differentiate each:- \(\frac{dx}{dt} = \cos t - t \sin t\)- \(\frac{dy}{dt} = 1\)- \(\frac{dz}{dt} = \sin t + t \cos t\)

Step 2: Evaluate derivatives at the point

Evaluate these derivatives at the given point \((-\pi, \pi, 0)\), which corresponds to \(t = \pi\).- \(\frac{dx}{dt}\bigg|_{t=\pi} = \cos\pi - \pi\sin\pi = -1\)- \(\frac{dy}{dt}\bigg|_{t=\pi} = 1\)- \(\frac{dz}{dt}\bigg|_{t=\pi} = \sin\pi + \pi\cos\pi = \pi(-1) = -\pi\)

Step 3: Formulate the parametric equations for the tangent line

The parametric equations for the tangent line use the point and the direction vector: \((-\pi, \pi, 0)\) and \((-1, 1, -\pi)\).The parametric equations are:- \(x = -\pi - t\)- \(y = \pi + t\)- \(z = 0 - \pi t = -\pi t\)

Step 4: Graph the curve and the tangent line

To illustrate this, graph the original parametric curve and the tangent line on the same plot. Plot the curve defined by:- \(x = t \cos t\)- \(y = t\)- \(z = t \sin t\).Then plot the tangent line with the equations:- \(x = -\pi - t\)- \(y = \pi + t\)- \(z = -\pi t\).Verify visually that the tangent line appropriately touches the curve at the point \((-\pi, \pi, 0)\).

Key Concepts

Tangent Line

A tangent line to a curve at a given point is a straight line that just "touches" the curve without crossing it at that point. It's like imagining the road that takes the slightest turn to match the curve, just for an instant. When the point is well defined in a parametric context, the tangent line helps us describe the behavior of the curve at that instant. For the curve defined by parametric equations, we find the tangent line by first identifying the specific point where the curve meets the line. Then, using a direction vector, which we will discuss below, we derive the formula for this line. This results in new parametric equations that represent the tangent. For example, if a curve passes through the point (-π, π, 0), the tangent line at this point can be expressed with similar parameter equations, as found in the original exercise. The tangent line provides a linear approximation of the curve near the specified point, often simplifying complex calculations.

Differentiation

Differentiation is a core element in finding tangent lines for parametric equations. By carrying out differentiation, we gain insight into how each coordinate changes as the parameter, usually time or some other variable, varies. When a curve is given in parametric form, we can differentiate each equation with respect to the parameter to find the velocity vector components:

• For x = t cos t, the derivative \(\frac{dx}{dt} = \cos t - t \sin t\)
• For y = t, the derivative \(\frac{dy}{dt} = 1\)
• For z = t sin t, the derivative \(\frac{dz}{dt} = \sin t + t \cos t\)

The differentiation provides a directional insight that is critical in characterizing the tangent line. Evaluating these derivatives at a specific point helps us determine the slope of the tangent line exactly at that point. It tells us how steeply the curve is rising or falling.

Direction Vector

The direction vector is crucial in defining the tangent line to a curve at a specific point. It provides the direction in which the tangent line points. Think of it as an arrow that shows how straight the line is pulling away from a touched curve.In our exercise, after differentiating the parametric equations and computing the derivative values at the specific point, we form a direction vector. For the point (-π, π, 0), this results in (-1, 1, -π). These components directly translate into the parametric equations of the tangent line:

• \(x = -\pi - t\)
• \(y = \pi + t\)
• \(z = -\pi \cdot t\)

The direction vector not only impacts how we set these equations but also confirms that the line and curve align correctly at the designated tangent point. Hence, it acts as the cornerstone for validating the equation formulation.