Problem 12 - Calculus III

Question

Find the arc length of the curve on the given interval.\(\mathbf{r}(t)=t^{2} \mathbf{i}+\left(2 t^{2}+1\right) \mathbf{j}, 1 \leq t \leq 3\)

Step-by-Step Solution

Answer

The arc length is \( 8\sqrt{5} \).

Step 1: Understand the Arc Length Formula for a Curve

The arc length \( L \) of a parametric curve \( \mathbf{r}(t) = (x(t), y(t)) \) from \( t = a \) to \( t = b \) is given by \[ L = \int_{a}^{b} \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } \, dt. \] Here, \( \mathbf{r}(t) = (t^2, 2t^2 + 1) \).

Step 2: Find the Derivatives of x(t) and y(t)

Find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). For \( x(t) = t^2 \), we get \( \frac{dx}{dt} = 2t \). For \( y(t) = 2t^2 + 1 \), we have \( \frac{dy}{dt} = 4t \).

Step 3: Express the Arc Length Integral

Substitute \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) into the arc length formula: \[ L = \int_{1}^{3} \sqrt{(2t)^2 + (4t)^2} \, dt = \int_{1}^{3} \sqrt{4t^2 + 16t^2} \, dt. \] Simplify the expression under the square root.

Step 4: Simplify the Expression Under the Square Root

Combine the terms to get \( 4t^2 + 16t^2 = 20t^2 \). Therefore, the integral becomes \[ \int_{1}^{3} \sqrt{20t^2} \, dt = \int_{1}^{3} \sqrt{20}t \, dt. \] As \( \sqrt{20} = 2\sqrt{5} \), the expression is \( 2\sqrt{5}t \).

Step 5: Compute the Integral

Integrate \( 2\sqrt{5}t \): \[ \int 2\sqrt{5}t \, dt = 2\sqrt{5} \cdot \frac{t^2}{2} = \sqrt{5}t^2. \] Evaluate from \( t = 1 \) to \( t = 3 \).

Step 6: Evaluate the Definite Integral

Calculate the arc length: \[ \sqrt{5}[(3)^2 - (1)^2] = \sqrt{5}[9 - 1] = \sqrt{5} \times 8 = 8\sqrt{5}. \] Thus, the arc length of the curve from \( t = 1 \) to \( t = 3 \) is \( 8 \sqrt{5} \).

Key Concepts

Understanding Parametric Curves

Parametric curves provide a powerful and flexible method to describe complicated shapes and paths using parameters. Instead of using a single equation like we do in Cartesian coordinates, a parametric curve is defined by separate functions for each coordinate.
For instance, in this exercise, we have the parametric curve \[ \mathbf{r}(t) = (t^2, 2t^2 + 1). \] Here, both the x-coordinate and y-coordinate are expressed in terms of the parameter \( t \). This allows us to describe a wide variety of curves, including those that loop or have multiple sections, by simply varying the values that \( t \) can take.

• It gives great flexibility for modeling paths or curves that can't be easily described by traditional functions.
• Parametric equations can describe both closed loops and open paths clearly.
• They are useful in computer graphics and animations where movement along a path is defined by a parameter.

By analyzing the parametric equations, we can gain information about the curve's shape, direction, and other properties that might be hidden in a simple Cartesian setup.

Integral Calculus and Its Role

Integral calculus is essential for computations involving areas and lengths. It allows us to "sum up" infinitely small pieces to get a total measure. When it comes to finding arc length of a parametric curve, like in this exercise, integral calculus is used to add up tiny distances along the curve.
We utilize the formula\[ L = \int_{a}^{b} \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } \, dt, \] which essentially measures the curve's length piece by piece over a specified interval.

• The integral sums infinitesimal lengths along the curve to give the total arc length.
• Integral calculus provides the foundation for precise calculations in physics and engineering, where measurements over a range must be summed.
• Without integral calculus, calculating precise lengths of curves would be very challenging, or even impossible for complex shapes.

In this specific exercise, we found the arc length of the curve described by \( \mathbf{r}(t) = (t^2, 2t^2 + 1) \) by evaluating the integral over the interval from \( t = 1 \) to \( t = 3 \).

The Importance of Derivatives

Derivatives describe how a function changes as its input changes. In the context of parametric curves, derivatives are fundamental because they help us understand the rate of change of the curve's coordinates with respect to the parameter.
For the parametric equations given:\[ x(t) = t^2 \quad \text{and} \quad y(t) = 2t^2 + 1, \] the derivatives \( \frac{dx}{dt} = 2t \) and \( \frac{dy}{dt} = 4t \) tell us how quickly each coordinate changes as \( t \) changes.

• Derivatives highlight the behavior of a curve at different points, revealing where it speeds up, slows down, or maintains constant motion.
• They are crucial in determining tangents, normals, and the overall geometry of curves.
• In practical applications, understanding derivatives allows us to predict a curve's path, making them invaluable in fields like physics, engineering, and computer graphics.

By finding these derivatives, we effectively laid the groundwork for integrating the arc length, providing insight into the geometry of the curve over the specified domain.