Question

Use calculus to find the length of the line segment \(\mathbf{r}(t)=\langle t,-8 t, 4 t\rangle,\) for \(0 \leq t \leq 2 .\) Verify your answer without using calculus.

Short answer

Answer: The length of the line segment is 18 units.

Step-by-step solution

Find the derivatives of the parametric equations

We are given that \(\mathbf{r}(t)=\langle t, -8 t, 4 t \rangle\). We will now find the derivatives with respect to t: $$\frac{dx}{dt} = \frac{d}{dt} (t) = 1 \\ \frac{dy}{dt} = \frac{d}{dt} (-8t) = -8 \\ \frac{dz}{dt} = \frac{d}{dt} (4t) = 4$$

Find the arc length

Using the formula for the arc length, which is \(L = \int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 + (\frac{dz}{dt})^2} dt\), we will now integrate: $$L = \int_0^2 \sqrt{(1)^2 + (-8)^2 + (4)^2} dt$$ The expression inside the square root simplifies as follows: $$L = \int_0^2 \sqrt{1 + 64 + 16} dt = \int_0^2 \sqrt{81} dt$$ As \(\sqrt{81} = 9\), the integral becomes: $$L = \int_0^2 9 dt = 9\int_0^2 dt$$ Now we can integrate and find the arc length: $$L = 9(t\Big|_0^2) = 9(2-0) = 9(2) = 18$$

Verify the answer without using calculus

To verify our answer, we have to find the length of the line segment using the non-calculus method. We know that \(\mathbf{r}(t)=\langle t, -8 t, 4 t \rangle\). Using this equation, we can find the coordinates of the two points of the line segment at the bounds of t: $$\mathbf{r}(0)=\langle 0, 0, 0 \rangle \,\,\text{and}\,\, \mathbf{r}(2)=\langle 2, -16, 8 \rangle$$ Now we can use the Pythagorean theorem in three dimensions to find the distance between these two points: $$L = \sqrt{(2-0)^2 + (-16-0)^2 + (8-0)^2} = \sqrt{4 + 256 + 64} = \sqrt{324} = 18$$ This confirms that our answer is correct, as the length of the line segment found using both methods is 18 units.

Key concepts

Parametric Equations

Parametric equations represent the coordinates of points on a curve as functions of a parameter. For instance, in our exercise, the parameter 't' was used to express the x, y, and z coordinates of a 3-dimensional curve as \( \mathbf{r}(t)=\langle t, -8t, 4t \rangle \). This equation effectively describes how the position of a point on the curve changes as 't' varies.

Understanding parametric equations is crucial because they provide a method to trace the path of an object along a curve in a coordinated system. This is especially helpful in fields like physics and engineering, where the motion of objects needs to be described precisely. Parametric functions are advantageous as they can describe lines, curves, and surfaces where standard forms would fail or be overly complex.

Arc Length Formula

The arc length formula is a fundamental concept in calculus used to determine the distance along a curved line (or an arc). The formula is based on the idea of approximating a curve with several tiny line segments and then finding the limit as these segments become infinitesimally small. The arc length for a curve described by parametric equations \( x(t), y(t), z(t) \) from \( t=a \) to \( t=b \) is given by:

\[ L = \int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 + (\frac{dz}{dt})^2} \, dt \]

This formula integrates the square root of the sum of the squares of the derivatives of the parametric equations with respect to 't'. In essence, it captures the element of distance for each infinitesimal piece of the curve and adds them up to find the total length. In the exercise, we applied this formula to find the length of a specific line segment.

Pythagorean Theorem in Three Dimensions

The Pythagorean theorem is well-known for finding the length of the hypotenuse of a right-angled triangle in two dimensions. However, this theorem can be extended to three dimensions to measure the distance between two points in space. The three-dimensional version is given by:

\[ L = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]

Here, \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \) are the coordinates of the two points. This equation is derived from the application of the Pythagorean theorem to the three perpendicular sides of a rectangular prism. It's highly beneficial for confirming the results obtained using calculus-based methods, as done in the exercise, where we found the length of the line segment both by integration and by direct application of the three-dimensional Pythagorean theorem.

Parametric Derivatives

Parametric derivatives are the rates at which the coordinates of a point on a parametric curve change with respect to the parameter. In the context of our problem, we found how fast the x, y, and z coordinates change as 't' varies. The derivatives \( \frac{dx}{dt} \) for x, \( \frac{dy}{dt} \) for y, and \( \frac{dz}{dt} \) for z was calculated as constants in this particular example:

\[ \frac{dx}{dt} = 1, \quad \frac{dy}{dt} = -8, \quad \frac{dz}{dt} = 4 \]

These derivatives are the building blocks for finding the arc length of a curve. They give us information about the slope of the curve at any point 't' and are used under the square root in the arc length formula, which integrates these rates of change to find the total length of the curve.