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Magazine Chapter 10: Problem 47
Graph the curve and find its length. $$x=e^{t}-t, \quad y=4 e^{t / 2}, \quad-8 \leq t \leqslant 3$$
Short Answer
Expert verified
The length of the curve is \(e^3 - e^{-8} + 11\).
Step by step solution
01
Parametric Equations
The given problem involves parametric equations where \(x\) and \(y\) are functions of the parameter \(t\). We have \(x = e^t - t\) and \(y = 4e^{t/2}\). The task is to graph this curve and find its length over the interval \(-8 \leq t \leq 3\).
02
Graphing the Curve
To graph the curve, evaluate the parametric equations at a series of values within the interval \(-8 \leq t \leq 3\). For each \(t\), calculate \(x\) and \(y\) and plot the resulting points on the coordinate plane. Connect these points smoothly to visualize the curve. A graphing calculator or software can facilitate this step.
03
Arc Length Formula
To find the length of the curve, we use the parametric arc length formula: \[L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\]. This requires finding \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).
04
Derivatives of x and y
Calculate the derivatives: \( \frac{dx}{dt} = e^t - 1 \) and \( \frac{dy}{dt} = 2e^{t/2} \). These are found by differentiating \(x = e^t - t\) and \(y = 4e^{t/2}\) with respect to \(t\).
05
Integrate the Arc Length Formula
Substitute \( \frac{dx}{dt} = e^t - 1 \) and \( \frac{dy}{dt} = 2e^{t/2} \) into the arc length formula to get \[L = \int_{-8}^{3} \sqrt{(e^t - 1)^2 + (2e^{t/2})^2} \, dt\]. Simplifying, this becomes:\[L = \int_{-8}^{3} \sqrt{e^{2t} - 2e^t + 1 + 4e^t} \, dt = \int_{-8}^{3} \sqrt{e^{2t} + 2e^t + 1} \, dt\].
06
Simplify the Integrand
Recognize that the integrand \(\sqrt{e^{2t} + 2e^t + 1}\) simplifies to \(\sqrt{(e^t + 1)^2}\). This is simply \(e^t + 1\) as it is the positive root, thus:\[L = \int_{-8}^{3} (e^t + 1) \, dt\].
07
Compute the Integral
Break the integral into simpler parts:\[L = \int_{-8}^{3} e^t \, dt + \int_{-8}^{3} 1 \, dt\]. These integrals evaluate to \[L = \left[e^t\right]_{-8}^{3} + \left[t\right]_{-8}^{3}\].
08
Evaluate the Integral
First evaluate \( \left[e^t\right]_{-8}^{3} = e^3 - e^{-8} \) and \( \left[t\right]_{-8}^{3} = 3 - (-8) = 11 \). Combine these results to find the length of the curve: \[L = (e^3 - e^{-8}) + 11\]. Compute these for the final length.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Arc Length Formula
In calculus, the arc length formula is a powerful tool for finding the length of a curve defined by parametric equations. It allows you to calculate the distance along a curve from one point to another. The general formula for the arc length, \( L \), of a parametric curve defined by \( x(t) \) and \( y(t) \) from \(t=a\) to \(t=b\) is given by:
\[L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\]
This formula combines the derivatives of the curve鈥檚 parametric equations. Here's how to use it:
\[L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\]
This formula combines the derivatives of the curve鈥檚 parametric equations. Here's how to use it:
- Start by differentiating the given parametric equations to find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).
- Plug these derivatives into the formula under the square root sign, sum their squares, and take the square root.
- Integrate this result from the beginning parameter \(a\) to the endpoint \(b\).
Parametric Curve Graphing
Graphing a parametric curve involves plotting points calculated from the parametric equations over a specific range of the parameter, \( t \). This process allows us to visualize the trajectory of a moving point, which is particularly useful for curves that are not functions but still represent paths through the plane.
To graph these curves:
To graph these curves:
- Select a range for \( t \), such as \(-8 \leq t \leq 3\) for our example.
- Calculate the corresponding \(x\) and \(y\) values for several points within this range.
- Plot these \((x, y)\) coordinate points on the graph.
- Connect the plotted points smoothly to form the curve.
Derivatives of Parametric Functions
Derivatives of parametric functions play a crucial role in various calculations such as finding tangents, rates of change, and lengths of curves. When dealing with parametric equations, \(x(t)\) and \(y(t)\), we compute the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) with respect to the parameter \(t\).
This process involves differentiating each of the parametric equations individually:
This process involves differentiating each of the parametric equations individually:
- Differentiate \(x(t)\) to find \( \frac{dx}{dt} \).
- Differentiate \(y(t)\) to find \( \frac{dy}{dt} \).
- They are used in calculating the arc length of the parametric curve.
- They help in analyzing the curve's slope and understanding changes in direction.
- Moreover, combined derivatives can offer insight into the curve's trajectory and speed.
