Problem 29 - Calculus Early Transcendentals: Pearson New International Edition

Question

Consider the curve \(\mathbf{r}(t)=2 t \mathbf{i}+t^{2} \mathbf{j}+\left(1-t^{2}\right) \mathbf{k}\) (a) Show that this curve lies on a plane and find the equation of this plane. (b) Where does the tangent line at \(t=2\) intersect the \(x y\) -plane?

Step-by-Step Solution

Answer

(a) Plane: \( y + z = 1 \), (b) Intersection: (2.5, 1, 0).

Step 1: Examine the curve components

The curve is given by \( \mathbf{r}(t) = 2t \mathbf{i} + t^2 \mathbf{j} + (1-t^2) \mathbf{k} \). Identify the components with respect to \( t \), which are \( x = 2t \), \( y = t^2 \), and \( z = 1 - t^2 \).

Step 2: Find relationships between the variables

From \( z = 1 - t^2 \), express \( t^2 = 1 - z \). Substitute back to get \( y = 1 - z \) and \( t = \frac{x}{2} \) from \( x = 2t \). The equation \( y = 1 - z \) shows the curve lies on the plane.

Step 3: Formulate the plane equation

The relationship \( y = 1 - z \) is indeed the equation of the plane on which the curve lies. Thus, the plane equation is \( y + z = 1 \).

Step 4: Find the tangent vector

To find the tangent line at \( t = 2 \), first find \( \mathbf{r}'(t) = \frac{d}{dt}(2t, t^2, 1-t^2) = (2, 2t, -2t) \). Evaluating at \( t = 2 \) gives the tangent vector \( \mathbf{r}'(2) = (2, 4, -4) \).

Step 5: Determine the point on the curve at \( t = 2 \)

Calculate \( \mathbf{r}(2) = 2(2) \mathbf{i} + 2^2 \mathbf{j} + (1-2^2) \mathbf{k} = 4 \mathbf{i} + 4 \mathbf{j} - 3 \mathbf{k} \), giving point (4, 4, -3) on the curve.

Step 6: Write the equation of the tangent line

The equation of the tangent line can be written as \( \mathbf{r}(t) = \mathbf{r}(2) + s \cdot \mathbf{r}'(2) \). This becomes \( (4, 4, -3) + s(2, 4, -4) \) which simplifies to \( x = 4 + 2s \), \( y = 4 + 4s \), \( z = -3 - 4s \).

Step 7: Find the intersection with the \( xy \)-plane

Setting \( z = 0 \) for intersection points on the \( xy \)-plane, solve \( -3 - 4s = 0 \) to find \( s = -\frac{3}{4} \). Substitute \( s \) back into the line equations: \( x = 4 + 2(-\frac{3}{4}) = 2.5 \), \( y = 4 + 4(-\frac{3}{4}) = 1 \).

Step 8: Conclusion for intersection

The tangent line intersects the \( xy \)-plane at the point (2.5, 1, 0).

Key Concepts

Parametric Equations

A parametric equation represents a curve or surface by defining coordinates as functions of a parameter, often denoted as \( t \). In the context of the exercise, the curve is expressed as \( \mathbf{r}(t) = 2t \mathbf{i} + t^2 \mathbf{j} + (1-t^2) \mathbf{k} \). Here, \( t \) is the parameter that defines points along the curve.

Parametric equations are versatile because they can describe complex curves and motions. For instance, the given curve has components:

• \( x = 2t \)
• \( y = t^2 \)
• \( z = 1 - t^2 \)

Understanding how each component varies with \( t \) helps in visualizing the trajectory or shape in three-dimensional space. These parametric forms allow for easy manipulation and understanding of the geometric figures, especially when calculating derivative vectors like tangents.
Employing parametric equations can significantly simplify the process of evaluating complex geometric problems.

Tangent Lines

Tangent lines are vital for understanding the direction and rate of change of a curve at a point. In vector calculus, finding a tangent line involves calculating the derivative of the parametric equation with respect to the parameter \( t \). In the problem, the curve \( \mathbf{r}(t) \) is differentiated to yield \( \mathbf{r}'(t) = (2, 2t, -2t) \). This derivative vector indicates the direction of the tangent line.

To determine the tangent at a specific point, say \( t = 2 \), substitute the value of \( t \) into the derivative:

• \( \mathbf{r}'(2) = (2, 4, -4) \)

This vector shows both the direction and the rates at which the coordinates change near the point. Further, the tangent line's equation can be parametrized using this directional vector and a point on the curve, yielding expressions like \( x = 4 + 2s \), \( y = 4 + 4s \), and \( z = -3 - 4s \).
This equips learners to explore where a curve is moving and how it behaves at specific intervals, a fundament of calculus that aids in deeper comprehension of graphical data.

Planes

To determine if a curve lies in a plane, we can identify relationships between its coordinates. In our exercise, the relationship \( y = 1 - z \) was established directly by manipulating the curve's parametric equations, ultimately leading to \( y + z = 1 \). This is the plane equation.

A plane in 3D space is essentially a flat, two-dimensional surface. It can be defined using:

• A point through which the plane passes.
• A normal vector perpendicular to the plane.

Having recognized that the curve lies on the plane \( y + z = 1 \), we understand that all points satisfy this equation, indicating coplanarity. Planes simplify analyzing orientation and alignment of curves or paths, such as intersections and distances, providing clarity in three-dimensional calculus problems.

Vector Calculus

Vector calculus extends calculus concepts to vector fields. It involves differentiation and integration of vectors, playing a significant role in physics and engineering.

In the given problem, vector calculus techniques are applied to:

• Differentiate the vector function to find tangent lines.
• Assess relationships for plane equations.

By leveraging vectors, we can capture directional information and construct linearly dependent components. This is crucial because it combines different calculus realms — differentiation and linearity — to analyze multidimensional systems.
The application of vector calculus here aids in understanding force and movement in space, — fundamental insights for tackling real-life problems. Mastery of vector calculus enhances one’s capability to handle complex systems by simplifying interactions into understandable vector expressions, aiding in the solution of multidimensional calculus inquiries.