Problem 45 - Essential Calculus Early Transcendentals

Question

At what point on the curve \(x=t^{3}, y=3 t, z=t^{4}\) is the normal plane parallel to the plane \(6 x+6 y-8 z=1 ?\)

Step-by-Step Solution

Answer

The points are (1, 3, 1) and (-1, -3, 1).

Step 1: Understand the Problem

We need to find a point on the curve where the normal plane, which is the plane perpendicular to the tangent vector at that point, is parallel to the given plane.

Step 2: Find the Tangent Vector

The position vector of a point on the curve is given by \( \mathbf{r}(t) = (t^3, 3t, t^4) \). The tangent vector \( \mathbf{T}(t) \) is the derivative of the position vector. Compute the partial derivatives: \( \mathbf{T}(t) = \frac{d}{dt}(t^3, 3t, t^4) = (3t^2, 3, 4t^3) \).

Step 3: Calculate Normal Vector to the Tangent

The normal plane at a point is perpendicular to the tangent vector at that point. Therefore, its normal vector is the same as the tangent vector: \( \mathbf{n} = (3t^2, 3, 4t^3) \).

Step 4: Set the Plan Parallel Condition

For our normal vector \( (3t^2, 3, 4t^3) \) to align with the plane \( 6x + 6y - 8z = 1 \), it must be parallel to this plane's normal vector \( (6, 6, -8) \). Two vectors are parallel if they are scalar multiples of one another.

Step 5: Solve for t Using Scalar Multiple Condition

From the vector condition, set \( 3t^2 = 6 \mu \), \( 3 = 6 \mu \), and \( 4t^3 = -8 \mu \). Solving \( 3 = 6 \mu \) gives \( \mu = \frac{1}{2} \). Substitute \( \mu = \frac{1}{2} \) back into \( 3t^2 = 6 \mu \), to get \( t^2 = 1 \). Thus, \( t = \pm 1 \).

Step 6: Evaluate Curve Points for Each t

Substitute \( t = 1 \) into the curve: \( x = 1^3 = 1 \), \( y = 3(1) = 3 \), \( z = 1^4 = 1 \). For \( t = -1 \): \( x = (-1)^3 = -1 \), \( y = 3(-1) = -3 \), \( z = (-1)^4 = 1 \).

Step 7: Conclusion

The points on the curve are \( (1, 3, 1) \) and \( (-1, -3, 1) \). Since they both result from valid values of \( t \), either point makes the normal plane parallel to the given plane.

Key Concepts

Tangent Vector

In calculus and vector analysis, the tangent vector plays a fundamental role in understanding the direction of a curve at a particular point. Imagine a curve in three-dimensional space. The tangent vector at any given point on the curve gives the direction in which the curve progresses. Think of it as an indication of the curve's immediate path or trajectory.

To find the tangent vector, we take the derivative of the position vector that describes the point's coordinates as functions of a parameter, often denoted as \( t \). In the provided exercise, the curve is described by the position vector \( \mathbf{r}(t) = (t^3, 3t, t^4) \). By differentiating each component with respect to \( t \), we obtain the tangent vector \( \mathbf{T}(t) = (3t^2, 3, 4t^3) \).

• The first component, \( 3t^2 \), tells us how the x-coordinate changes.
• The second component, \( 3 \), indicates the y-direction remains constant.
• The third component, \( 4t^3 \), describes the z-direction adjustments.

This tangent vector is crucial for determining directions on the curve, and it is perpendicular to the normal plane at that point, providing insights into geometric relationships.

Normal Vector

The normal vector is a vector that is perpendicular to a surface or a plane at a given point. In the context of curves, the normal vector to a tangent represents a direction in three-dimensional space that is perpendicular to the direction of the tangent vector.

In the original exercise, the normal vector sought is the same as the tangent vector \( \mathbf{n} = (3t^2, 3, 4t^3) \) because the normal plane is perpendicular to this tangent vector. We are interested in identifying when this normal vector becomes parallel to a given plane, meaning it aligns with that plane's normal vector.

• Two vectors are parallel if they are scalar multiples of each other.
• The problem requires setting conditions such that the tangent vector, which also serves as the normal vector here, aligns with the normal vector of the plane in question.

Understanding the normal vector is essential as it indicates the plane's orientation in space, letting us analyze how the plane interacts with other geometrical entities such as curves and other planes.

Parallel Planes

Parallel planes are planes in three-dimensional space that never meet. They are always the same distance apart everywhere and have the same orientation. The concept of parallelism in terms of vectors is that two vectors are parallel if one is a scalar multiple of the other.

In our exercise, we deal with a condition where the normal plane of the curve must be parallel to the given plane defined by the equation \( 6x + 6y - 8z = 1 \). For parallelism, the normal vectors of these planes must be aligned. Therefore, the normal vector to the tangent vector, \( (3t^2, 3, 4t^3) \), must be a scalar multiple of the plane's normal vector, \( (6, 6, -8) \).

• By setting up the equality \( 3t^2 = 6 \mu \), \( 3 = 6 \mu \), and \( 4t^3 = -8 \mu \), we're equating components to find the scalar \( \mu \).
• Solving gives potential values for \( t \), which help identify points where the planes align.

This understanding of parallel planes is pivotal in geometry and calculus, as it helps us analyze spatial relationships and solve problems involving distances and alignments in space.