Problem 25 - Calculus Early Transcendentals

Question

Find the curvature of \(\mathbf{r}(t)=\left\langle t, t^{2}, t^{3}\right\rangle\) at the point \((1,1,1)\)

Step-by-Step Solution

Answer

Curvature at (1,1,1) is \( \frac{\sqrt{76}}{196} \).

Step 1: Determine the Value of t at the Given Point

First, we need to find the value of \( t \) corresponding to the point \((1, 1, 1)\). Since \( \mathbf{r}(t) = \langle t, t^2, t^3 \rangle \), we set \[ t = 1, \] \[ t^2 = 1, \] \[ t^3 = 1. \]These conditions are satisfied when \( t = 1 \).

Step 2: Compute the First Derivative of r(t)

The first derivative of \( \mathbf{r}(t) \) is the velocity vector \( \mathbf{r}'(t) \):\[\mathbf{r}'(t) = \left\langle \frac{d}{dt}(t), \frac{d}{dt}(t^2), \frac{d}{dt}(t^3) \right\rangle = \langle 1, 2t, 3t^2 \rangle.\]At \( t = 1 \), this becomes \( \mathbf{r}'(1) = \langle 1, 2, 3 \rangle \).

Step 3: Compute the Second Derivative of r(t)

The second derivative of \( \mathbf{r}(t) \) is the acceleration vector \( \mathbf{r}''(t) \):\[\mathbf{r}''(t) = \left\langle \frac{d}{dt}(1), \frac{d}{dt}(2t), \frac{d}{dt}(3t^2) \right\rangle = \langle 0, 2, 6t \rangle.\]At \( t = 1 \), this becomes \( \mathbf{r}''(1) = \langle 0, 2, 6 \rangle \).

Step 4: Compute the Curvature Formula

The curvature \( \kappa \) of the vector function \( \mathbf{r}(t) \) is given by:\[\kappa = \frac{\| \mathbf{r}'(t) \times \mathbf{r}''(t) \|}{\| \mathbf{r}'(t) \|^3}.\]

Step 5: Calculate the Cross Product of r'(t) and r''(t)

Compute \( \mathbf{r}'(1) \times \mathbf{r}''(1) \):\[\mathbf{r}'(1) = \langle 1, 2, 3 \rangle, \quad \mathbf{r}''(1) = \langle 0, 2, 6 \rangle\]The cross product:\[\mathbf{r}'(1) \times \mathbf{r}''(1) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 2 & 3 \ 0 & 2 & 6 \end{vmatrix} = \mathbf{i}(12 - 6) - \mathbf{j}(6 - 0) + \mathbf{k}(2 - 0) = \langle 6, -6, 2 \rangle\]

Step 6: Calculate the Magnitudes Needed for the Curvature Formula

First, find \( \| \mathbf{r}'(1) \times \mathbf{r}''(1) \| \):\[\| \langle 6, -6, 2 \rangle \| = \sqrt{6^2 + (-6)^2 + 2^2} = \sqrt{76}.\]Next, calculate \( \| \mathbf{r}'(1) \| \):\[\| \langle 1, 2, 3 \rangle \| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}.\]

Step 7: Calculate the Curvature

Using the values calculated, substitute into the curvature formula:\[\kappa = \frac{\sqrt{76}}{(\sqrt{14})^3} = \frac{\sqrt{76}}{14\sqrt{14}} = \frac{\sqrt{76}}{196}.\]Therefore, the curvature at the point \((1, 1, 1)\) is \( \frac{\sqrt{76}}{196} \).

Key Concepts

Parametric Equations

Parametric equations are a way of defining a mathematical relationship by using parameters. Instead of using functions like \( y = f(x) \) where one variable is dependent on another, parametric equations express sets of related quantities as explicit functions of an independent parameter.

For instance, in our problem, we have \( \mathbf{r}(t) = \langle t, t^2, t^3 \rangle \). Here, \( t \) is the parameter, and the output is a vector dependent on \( t \):

• The first component, \( t \), is simply \( t \) itself.
• The second component, \( t^2 \), means the square of \( t \).
• The third component, \( t^3 \), represents the cube of \( t \).

In three-dimensional space, parametric equations help you to plot curves that may represent paths traveled over time. Understanding how to manipulate these equations enables solving complex problems in calculus and physics.

Derivatives

Derivatives are a crucial concept in calculus that denote the rate of change. In the context of parametric equations, derivatives help us understand how each component of the curve changes as the parameter \( t \) changes.

Given a vector-valued function \( \mathbf{r}(t) = \langle t, t^2, t^3 \rangle \), its first derivative, \( \mathbf{r}'(t) \), known as the velocity vector, measures how the position vector changes with respect to time. This is computed as:
\[\mathbf{r}'(t) = \left\langle 1, 2t, 3t^2 \right\rangle\]
At the specific point \( t = 1 \), this derivative gives the velocity at that point, \( \mathbf{r}'(1) = \langle 1, 2, 3 \rangle \).

The second derivative, known as the acceleration vector, \( \mathbf{r}''(t) \), is computed similarly:
\[\mathbf{r}''(t) = \left\langle 0, 2, 6t \right\rangle\]
At \( t = 1 \), it gives us \( \mathbf{r}''(1) = \langle 0, 2, 6 \rangle \). These derivatives are vital for determining the curvature, providing an understanding of how sharply a path curves at any given point.

Cross Product

The cross product is a vector multiplication of two vectors in three-dimensional space, resulting in a third vector that is perpendicular to the plane formed by the initial two vectors.

In our exercise, we calculate the cross product of the velocity vector \( \mathbf{r}'(1) = \langle 1, 2, 3 \rangle \) and the acceleration vector \( \mathbf{r}''(1) = \langle 0, 2, 6 \rangle \). This cross product helps in finding the curvature of the curve as it measures how the vectors span a parallelogram. The formula is given by:

• \[ \mathbf{r}'(1) \times \mathbf{r}''(1) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 2 & 3 \ 0 & 2 & 6 \end{vmatrix} \]
• Simplified to: \( \langle 6, -6, 2 \rangle \)

The magnitude of this resulting vector: \( \sqrt{6^2 + (-6)^2 + 2^2} = \sqrt{76} \), is used to obtain the curvature. Understanding cross products is key for problems involving rotational and directional change.

Curvature Formula

Curvature measures how sharply a curve bends at a given point. For vector functions, the curvature \( \kappa \) at a point is calculated using the derivatives and their cross product.

The formula for the curvature of a parametric curve given by a vector-valued function \( \mathbf{r}(t) \) is:
\[ \kappa = \frac{\| \mathbf{r}'(t) \times \mathbf{r}''(t) \|}{\| \mathbf{r}'(t) \|^3} \]
In our problem, we calculated:

• The cross product \( \| \langle 6, -6, 2 \rangle \| = \sqrt{76} \) from the vectors \( \mathbf{r}'(1) \) and \( \mathbf{r}''(1) \).
• The magnitude of \( \mathbf{r}'(1) \), which is \( \| \langle 1, 2, 3 \rangle \| = \sqrt{14} \).

Substituting these into the curvature formula gives:
\[ \kappa = \frac{\sqrt{76}}{(\sqrt{14})^3} = \frac{\sqrt{76}}{196} \]
This calculation provides the curvature at the specified point \((1, 1, 1)\), depicting the degree of the path's bend at that position. Mastery of the curvature formula allows the analysis of paths in physics and geometry, giving insights into motion and structure.