Problem 118 - Calculus Volume 3

Question

Find the length of the curve \(x=e^{t}-t, y=4 e^{t / 2},-8 \leq t \leq 3 .\)

Step-by-Step Solution

Answer

The length of the curve is \( L = e^3 - e^{-8} + 11 \).

Step 1: Understand the Curve Equation

The curve is parametrized by the equations \( x = e^t - t \) and \( y = 4e^{t/2} \) with the parameter \( t \) ranging from \(-8\) to \(3\). Our task is to find the length of this curve over the given interval of \( t \).

Step 2: Derivatives of Parametric Equations

Find the derivatives of the parametric equations with respect to \( t \). \( \frac{dx}{dt} = \frac{d}{dt}(e^t - t) = e^t - 1 \) and \( \frac{dy}{dt} = \frac{d}{dt}(4e^{t/2}) = 2e^{t/2} \).

Step 3: Determine the Length Formula

The length \( L \) of a parametric curve \((x(t), y(t))\) is given by the integral \( L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \). Here, \( a = -8 \) and \( b = 3 \).

Step 4: Substitute Derivatives into Length Formula

Substitute the derivatives into the length formula: \( L = \int_{-8}^{3} \sqrt{(e^t - 1)^2 + (2e^{t/2})^2} \, dt \).

Step 5: Simplify the Integrand

Simplify the expression under the square root: \( (e^t - 1)^2 = e^{2t} - 2e^t + 1 \) and \( (2e^{t/2})^2 = 4e^t \). Thus, the integrand simplifies to \( \sqrt{e^{2t} + 2e^t + 1} = \sqrt{(e^t + 1)^2} = |e^t + 1| \).

Step 6: Evaluate the Integral

Since \( e^t + 1 > 0 \) for all \( t \), \( |e^t + 1| = e^t + 1 \). Therefore, the integral is \( L = \int_{-8}^{3} (e^t + 1) \, dt \). Evaluate the integral: \( \int e^t \, dt = e^t \), and \( \int 1 \, dt = t \). Thus, \( L = [e^t + t]_{-8}^{3} \).

Step 7: Compute the Definite Integral

Calculate \([e^t + t]_{-8}^{3} = (e^3 + 3) - (e^{-8} - 8) = e^3 + 3 - e^{-8} + 8 \). Simplify to get \( L = e^3 - e^{-8} + 11 \).

Step 8: Final Length

The final length of the curve is \( L = e^3 - e^{-8} + 11 \), which is an exact expression involving exponential functions.

Key Concepts

Parametric Equations

Parametric equations are a way to define a curve by expressing both the x and y coordinates as functions of a third variable, usually denoted as t, known as the parameter. In this exercise, the curve is defined by the equations \( x = e^t - t \) and \( y = 4e^{t/2} \). These equations describe the path traced by a point as t changes.
When you use parametric equations, you can represent more complex curves and shapes that might be difficult to express with a single function, especially when the graph may loop or intersect itself. This makes them especially powerful for describing paths and trajectories in physics and engineering.

• Parametrization allows separate control of the curve's path (trajectory) by varying the parameter \( t \).
• Is more flexible compared to regular Cartesian equations, enabling the modeling of dynamic systems.

Understanding parametric equations requires interpreting how the x and y values relate to changes in the parameter t over a given domain, such as the range from \(-8 \) to \(3 \) used in this specific problem.

Integral Calculus

Integral calculus is concerned with the accumulation of quantities and the area under and between curves. In the context of finding the length of a curve, integral calculus is used to sum infinitesimal line segments along the curve. The length of a parametric curve \((x(t), y(t))\) over an interval \([a, b]\) is determined using the integral:\[ L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]
This formula adds up small distances along the curve, which are calculated from the derivatives of the parametric equations. The formula is derived by considering the differential arc length \( ds \) of the curve and applying the Pythagorean theorem in a very small segment.

• The integrand \( \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \) represents the infinitesimal distance along the curve.
• The limits of integration \( a \) and \( b \) specify the interval of the parameter \( t \) over which we evaluate the curve.

This method is crucial in applications where the total distance along a path needs to be calculated, such as in navigation and material science.

Derivatives

The derivative of a function represents the rate of change of the function's output with respect to changes in its input. In a parametric context, these derivatives give us the instantaneous rate of change along the curve for both the x and y parameters. For the given curve, we calculate:

• The derivative \( \frac{dx}{dt} = e^t - 1 \), which captures how x changes relative to t.
• The derivative \( \frac{dy}{dt} = 2e^{t/2} \), which captures how y changes relative to t.

Derivatives are fundamental in determining the shape and behavior of a curve. They help us understand where vertices, maximums, minimums, or inflection points might occur.
When finding the length of the curve, these derivatives are plugged into the length formula to account for how sharp or gentle the curves are in both dimensions. The computation uses the derivatives to assess changes along the path implicitly traced by the parametric equation.

Exponential Functions

Exponential functions are mathematical functions of the form \( f(t) = a^t \), where \( a \) is a positive constant. These functions are important in modeling situations where growth or decay happens at a constant rate, as seen in compound interest, populations, and radioactive decay.
In the exercise, both x and y parametric equations involve the exponential function \( e^t \):

• The function \( e^t \) is raised to a power that is directly affected by the parameter \( t \).
• The function naturally lends to growth and decay patterns, shaping the curve as \( t \) increases or decreases.

Exponential functions are incredibly useful due to their mathematical properties:

• The derivative of \( e^t \) is \( e^t \), simplifying calculations, especially in calculus.
• They have a constant proportional rate of change, which is unique among functions.

In the solution, the exponential function is seen impacting the length calculation through its persistent presence in both \( x \) and \( y \) derivatives. It ensures that the function \( e \) shapes the overall path in a significant, yet predictable manner across the interval given for \( t \).