Problem 25 - Essential Calculus Early Transcendentals

Question

At what points does the curve \(\mathbf{r}(t)=t \mathbf{i}+\left(2 t-t^{2}\right) \mathbf{k}\) inter- sect the paraboloid \(z=x^{2}+y^{2} ?\)

Step-by-Step Solution

Answer

The curve intersects the paraboloid at points \((0, 0, 0)\) and \((1, 0, 1)\).

Step 1: Parameterize the Curve

The given vector function is \( \mathbf{r}(t) = t \mathbf{i} + 0 \mathbf{j} + (2t - t^2) \mathbf{k} \). This means \( x = t \), \( y = 0 \), and \( z = 2t - t^2 \).

Step 2: Use the Paraboloid Equation

The equation of the paraboloid is \( z = x^2 + y^2 \). Substitute \( x = t \) and \( y = 0 \) into the paraboloid equation, giving \( z = t^2 \).

Step 3: Set the Two Expressions for z Equal

Since both expressions describe \( z \), set them equal to each other: \( 2t - t^2 = t^2 \).

Step 4: Simplify the Equation

Rearrange the equation from Step 3 to form a quadratic equation. \( 2t - t^2 = t^2 \) becomes \( 2t - 2t^2 = 0 \). Factor this equation to get \( 2t(1 - t) = 0 \).

Step 5: Solve the Quadratic Equation

Set each factor equal to zero: \( 2t = 0 \) gives \( t = 0 \), and \( 1 - t = 0 \) gives \( t = 1 \).

Step 6: Find Intersection Points

Substitute the values of \( t \) back into the expressions for \( x, y, \) and \( z \) to find the points of intersection: \( t = 0 \) gives \( (0, 0, 0) \) and \( t = 1 \) gives \( (1, 0, 1) \).

Key Concepts

Parametric Equations

A parametric equation is a way of representing a curve through parameters, often using a variable like \( t \).
In standard Cartesian coordinates, a point is usually defined by \( x \), \( y \), \( z \). In parametric equations, these coordinates are expressed as functions of one or more parameters. For example, consider the vector function \( \mathbf{r}(t) = t \mathbf{i} + 0 \mathbf{j} + (2t - t^2) \mathbf{k} \).

• Here, \( x = t \), \( y = 0 \), and \( z = 2t - t^2 \).
• This allows you to traverse through the points on a curve by varying \( t \).

Using parametric equations is beneficial because it simplifies problems, especially in vector calculus, by breaking down the components of complex figures into manageable parts.
It also helps in visualizing how curves and surfaces are formed in three-dimensional space.

Intersection of Surfaces

Finding the intersection points of surfaces is a common task in vector calculus and geometry.
It typically involves identifying points that satisfy both equations representing the surfaces involved.
In the given problem, we have a parametric curve defined by \( \mathbf{r}(t) = t \mathbf{i} + 0 \mathbf{j} + (2t - t^2) \mathbf{k} \), and a surface, the paraboloid described by \( z = x^2 + y^2 \).

• Realize that the equations of the curve give us the parameters for \( x \), \( y \), and \( z \).
• The paraboloid's equation represents a surface against which our curve must be compared to find intersections.
• Setting the curve's \( z \) value equal to the paraboloid's equation allows us to determine where they meet.

This concept taps into the powerful idea of mapping 3D space and understanding spatial relationships between lines, curves, and surfaces.

Quadratic Equations

Quadratic equations are fundamental in mathematics, appearing in various problems, including our exercise where they determine the intersection points.
A standard quadratic equation takes the form \( ax^2 + bx + c = 0 \). In this case, we derive it from comparing two expressions for \( z \):

• From the parametric curve: \( z = 2t - t^2 \)
• From the paraboloid: \( z = t^2 \)

When set equal, they form the quadratic equation \( 2t - t^2 = t^2 \). Simplifying gives us \( 2t - 2t^2 = 0 \).
This is solved by factoring, which in this problem:

• becomes \( 2t(1 - t) = 0 \)
• This yields solutions for \( t = 0 \) and \( t = 1 \).

Solving quadratics enables us to find specific parameter values, letting us determine precise intersections in spatial problems.
Thus, quadratic equations bridge equations into geometric solutions, revealing critical points of intersection.