Problem 29 - Thomas Calculus

Question

In Exercises \(27-30\) , find the value(s) of \(t\) so that the tangent line to the given curve contains the given point. $$ \mathbf{r}(t)=2 t \mathbf{i}+t^{2} \mathbf{j}-t^{2} \mathbf{k} ; \quad(0,-4,4) $$

Step-by-Step Solution

Answer

The values of \( t \) are \( t = 2 \) and \( t = -2 \).

Step 1: Understand the Given Vector Function

The vector function given is \( \mathbf{r}(t) = 2t \mathbf{i} + t^{2} \mathbf{j} - t^{2} \mathbf{k} \). It describes the position of a point in 3D space as a function of \( t \).

Step 2: Find the Derivative to Determine the Tangent Vector

To find the tangent line at \( t \), compute the derivative of the vector function: \[ \mathbf{r}'(t) = \frac{d}{dt}\left(2t\mathbf{i} + t^2\mathbf{j} - t^2\mathbf{k}\right) = 2\mathbf{i} + 2t\mathbf{j} - 2t\mathbf{k}. \] This vector \( \mathbf{r}'(t) \) gives the direction of the tangent line at any point \( t \).

Step 3: Set Up the Equation for the Tangent Line

A point on the tangent line can be expressed as \( \mathbf{r}(t) + \, s\mathbf{r}'(t) \), where \( s \) is a scalar. Therefore, the line is \( (2t + 2s)\mathbf{i} + (t^2 + 2ts)\mathbf{j} - (t^2 + 2ts)\mathbf{k} \).

Step 4: Incorporate the Given Point into the Tangent Line Equation

We want the line to pass through the point \((0, -4, 4)\). Thus, we set equations: \( 2t + 2s = 0 \), \( t^2 + 2ts = -4 \), \( -t^2 - 2ts = 4 \).

Step 5: Solve the System of Equations

From \( 2t + 2s = 0 \), solve for \( s \): \( s = -t \). Substitute \( s = -t \) into the other two equations. This results in: \( t^2 - 2t^2 = -4 \) which simplifies to \( -t^2 = -4 \), giving \( t^2 = 4 \). Thus, \( t = 2 \) or \( t = -2 \).

Step 6: Verify Solutions with Remaining Equations

Substituting \( t = 2 \) into \( -2t^2 = 4 \) verifies it as a valid solution. Checking \( t = -2 \) gives the same satisfaction in the equations. Thus both \( t=2 \) and \( t = -2 \) are correct.

Key Concepts

Understanding Vector Functions in 3D Space

A vector function, like the one shown in this exercise, is a function that assigns a vector to each value of the input variable, typically denoted as \( t \). In this particular case, the vector function is \( \mathbf{r}(t) = 2t \mathbf{i} + t^{2} \mathbf{j} - t^{2} \mathbf{k} \).
This function describes how a point moves through 3-dimensional space.

• Each component of the vector \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) denotes a direction along the x, y, and z axes respectively.
• The coefficients \( 2t \), \( t^2 \), and \( -t^2 \) tell us how the position along each axis changes as \( t \) varies.

For instance, as \( t \) increases, the position along the x-axis moves linearly due to \( 2t \). In contrast, the positions along the y and z axes experience quadratic changes due to \( t^2 \) and \( -t^2 \), respectively.Understanding vector functions and how they behave is crucial when analyzing motion or curves in 3D space.

The Importance of Derivatives in Finding Tangents

The derivative of a vector function provides vital information about the curve's behavior, specifically its tangent.
The derivative of the given vector function, \( \mathbf{r}'(t) = 2\mathbf{i} + 2t\mathbf{j} - 2t\mathbf{k} \), represents the tangent vector at any point \( t \).

• In simpler terms, this tangent vector points in the direction that the curve is headed at that particular point.
• This is essential for determining the tangent line's orientation.

By finding the derivative, we unlock information about the speed and direction of the curve at a particular instant. This process transforms the abstract art of calculus into a concrete tool for understanding motion and geometry.

Navigating Systems of Equations in 3D Space

When dealing with problems in 3D space, systems of equations often arise. In this exercise, three separate equations are formed to ensure the derived tangent line passes through a particular point \((0, -4, 4)\).

• The first equation, \( 2t + 2s = 0 \), relates to the position in the x-direction.
• The second, \( t^2 + 2ts = -4 \), ensures the y-position meets the condition.
• The third equation, \( -t^2 - 2ts = 4 \), ensures the z-position matches as well.

These equations form a system that must be solved simultaneously. BrkSolving such systems usually involves substitution or elimination methods. For this problem, solving \( s = -t \) allows substitution into the remaining equations, leading to the discovery that potential solutions for \( t \) are \( t = 2 \) or \( t = -2 \).
Systems of equations are a cornerstone of mathematical problem-solving, allowing us to reconcile multiple conditions at once.

3D Space and Its Complexities

Exploring mathematical problems in 3D space adds a layer of complexity that mirrors the real world. In this exercise, the vector function determines a curve in three-dimensional space, characterized by three axes \( x \), \( y \), and \( z \).

• The representation \( \mathbf{i} \), \( \mathbf{j} \), \( \mathbf{k} \) depicts directional components in this space.
• Each axis interacts in a manner that can lead to diverse geometric forms when mapped by vector functions.

Understanding how these three dimensions interact is crucial for disciplines like physics, engineering, and computer graphics. Working in 3D involves visualizing how changes in one dimension can impact others, offering a more holistic understanding of spatial problems.
Mastering these concepts unlocks a broader perspective on the intricacies of our world.