Problem 35 - Calculus Early Transcendentals

Question

Show that the curve with parametric equations \(x=t^{2}\) , \(y=1-3 t, z=1+t^{3}\) passes through the points \((1,4,0)\) and \((9,-8,28)\) but not through the point \((4,7,-6)\)

Step-by-Step Solution

Answer

The curve passes through (1,4,0) and (9,-8,28), but not through (4,7,-6).

Step 1: Determine t for Point (1,4,0)

To check if the curve passes through the point \((1,4,0)\), equate the parametric equations to the point. Start with \(x = t^2 = 1\). Solving for \(t\) gives us \(t = \pm 1\). Next, we check these values in the other two equations: \(y = 1 - 3t\) and \(z = 1 + t^3\).

Step 2: Verify Point (1,4,0) for t=1

Substitute \(t = 1\) into the equations. \(y = 1 - 3(1) = -2\) and \(z = 1 + (1)^3 = 2\). Since \( (1, -2, 2)\) does not match \((1, 4, 0)\), we discard \(t = 1\).

Step 3: Verify Point (1,4,0) for t=-1

Substitute \(t = -1\) into the equations. \(y = 1 - 3(-1) = 4\) and \(z = 1 + (-1)^3 = 0\). Thus, \((1, 4, 0)\) matches the expected coordinates for \(t = -1\). This point is on the curve.

Step 4: Determine t for Point (9,-8,28)

For the point \((9, -8, 28)\), set \(x = t^2 = 9\). Solving gives \(t = \pm 3\). Check these values in the equations for \(y\) and \(z\).

Step 5: Verify Point (9,-8,28) for t=3

For \(t = 3\), substitute into the equations: \(y = 1 - 3(3) = -8\) and \(z = 1 + (3)^3 = 28\). The coordinates match \((9, -8, 28)\), so the point is on the curve.

Step 6: Verify Point (9,-8,28) for t=-3

For \(t = -3\), substitute into the equations: \(y = 1 - 3(-3) = 10\) and \(z = 1 + (-3)^3 = -26\). Since \((9, 10, -26)\) does not match the desired point, discard \(t = -3\).

Step 7: Determine t for Point (4,7,-6)

For the point \((4, 7, -6)\), start with \(x = t^2 = 4\), giving \(t = \pm 2\). Check these in the equations for \(y\) and \(z\).

Step 8: Verify Point (4,7,-6) for t=2

For \(t = 2\), substitute into the equations: \(y = 1 - 3(2) = -5\) and \(z = 1 + (2)^3 = 9\). The coordinates \((4, -5, 9)\) do not match \((4, 7, -6)\).

Step 9: Verify Point (4,7,-6) for t=-2

For \(t = -2\), substitute into the equations: \(y = 1 - 3(-2) = 7\) and \(z = 1 + (-2)^3 = -7\). While the value for \(y\) matches, \(z = -7 eq -6\), so the point is not on the curve.

Key Concepts

Vector Calculus

Vector calculus is a mathematical tool used to analyze and describe vectors and scalar fields. In this exercise, we deal with a curve in three-dimensional space defined by parametric equations. These equations use a parameter, usually denoted as \(t\), to express the \(x\), \(y\), and \(z\) coordinates of points on a curve. This method is powerful because:

• It unambiguously defines each point along the curve.
• Allows easy calculations of derivatives and integrals, important for understanding the behavior of the curve.

To find whether a point lies on this curve, we equate the parametric equations to the corresponding coordinates of the point and solve for \(t\). If a consistent \(t\) value satisfies all three equations, the point lies on the curve. Using vector calculus simplifies complex problems, leading to solutions that reveal fundamental properties about the curve's geometry and behavior.

Curve

A curve, in mathematical terms, is a continuous mapping from a parameter space (usually a time-like interval) to geometric space. Here, the curve is described by the parametric equations:

• \(x(t) = t^2\)
• \(y(t) = 1 - 3t\)
• \(z(t) = 1 + t^3\)

Each value of \(t\) represents a point \((x, y, z)\) in 3D space. This representation allows us to trace the path of a curve by varying \(t\). Parametric equations are particularly useful when the relationship between coordinates is not easily described by a single functional form. They allow:

• Describing complex shapes that cannot be expressed as the graph of a function of \(x\) or \(y\) alone.
• Studying the orientation and direction of the curve as \(t\) varies.

In practical applications, these curves can represent trajectories of objects in physics or paths traced by particles.

Coordinate Geometry

Coordinate geometry, or analytic geometry, involves the study of geometry using a coordinate system. In the current exercise, coordinate geometry integrates with vector calculus to determine if given points lie on a specified curve. Using coordinates allows us to:

• Plot points in a space with respect to \(x, y, z\) axes.
• Measure distances and angles between points conveniently.
• Apply algebraic methods to solve geometric problems.

To verify that a point \((x, y, z)\) exists on a curve defined by the parametric equations, each coordinate is checked individually by substituting and solving for the parameter \(t\). If all conditions are met for the same \(t\), the point belongs to the curve. This method connects algebraic functions with geometric visualization, improving our understanding of spatial relationships.

3D Space

Three-dimensional space is a geometric setting in which three values (or coordinates) are required to determine the position of an element (i.e., a point). Our exercise considers points defined in 3D with coordinates \((x, y, z)\). Key points about 3D space and parametric equations include:

• 3D space models real-world phenomena accurately since it accommodates width, height, and depth.
• Lines or curves can indicate motion and trajectory, essential for fields like physics and engineering.
• The use of parametric equations in 3D allows seamless calculation of intersections or points of contact like those in this exercise.

Understanding these equations and their representation in 3D helps in visualizing complex spatial concepts and comprehending how different dimensions interact, making it a fundamental concept in many scientific and mathematical applications.