Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 10: Problem 45
Graph the curve and find its length. $$x=e^{t} \cos t, \quad y=e^{t} \sin t, \quad 0 \leqslant t \leqslant \pi$$
Short Answer
Expert verified
The curve length is \( \sqrt{2} (e^{\pi} - 1) \).
Step by step solution
01
Understanding the Parametric Equations
We have two parametric equations given for a curve: \[ x = e^t \cos(t) \] \[ y = e^t \sin(t) \] These describe the x and y coordinates in terms of the parameter \( t \) ranging from 0 to \( \pi \). To graph the curve, plot these points for \( t \) within the given interval.
02
Find the Derivatives of x and y
To find the length of the curve, we first need the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).Calculate \( \frac{dx}{dt} \) as follows:\[ \frac{dx}{dt} = \frac{d}{dt}(e^t \cos t) = e^t \cos t - e^t \sin t \]Calculate \( \frac{dy}{dt} \) as follows:\[ \frac{dy}{dt} = \frac{d}{dt}(e^t \sin t) = e^t \sin t + e^t \cos t \]
03
Calculate the Expression Under the Square Root for Length
Using the derivatives found, we need to calculate:\[ \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \]Substitute and simplify:\[ \left(\frac{dx}{dt}\right)^2 = (e^t \cos t - e^t \sin t)^2 = e^{2t} (\cos^2 t - 2 \cos t \sin t + \sin^2 t) \]\[ \left(\frac{dy}{dt}\right)^2 = (e^t \sin t + e^t \cos t)^2 = e^{2t} (\sin^2 t + 2 \cos t \sin t + \cos^2 t) \]Sum them:\[ e^{2t}[\cos^2 t - 2 \cos t \sin t + \sin^2 t] + e^{2t}[\sin^2 t + 2 \cos t \sin t + \cos^2 t] = 2e^{2t} \] So we get:\[ \sqrt{e^{2t} \, 2} = \sqrt{2} \, e^t \]
04
Set Up and Evaluate the Integral for Arc Length
To find the total arc length, integrate the expression from 0 to \( \pi \): \[ L = \int_{0}^{\pi} \sqrt{2} \, e^t \, dt \] Factor out \(\sqrt{2}\):\[ L = \sqrt{2} \int_{0}^{\pi} e^t \, dt \] Now evaluate \( \int e^t \, dt = e^t \): \[ L = \sqrt{2} \left[e^t \right]_{0}^{\pi} = \sqrt{2} (e^{\pi} - e^0) = \sqrt{2} (e^{\pi} - 1) \]
05
Conclusion
The length of the curve, described parametrically by the given equations over the interval \( 0 \leq t \leq \pi \), is \( \sqrt{2} (e^{\pi} - 1) \). Graphically, this curve corresponds to a spiral from \( t = 0 \) to \( t = \pi \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Arc Length
Finding the arc length of a curve defined by parametric equations involves a specific process. With parametric equations, the curve's x and y coordinates are expressed in terms of a third variable, which is often denoted as \( t \). To calculate the arc length, you'll need to follow these steps:
- First, determine the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).
- Next, you calculate the expression under the square root: \( \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } \).
- Finally, set up an integral of this expression over the given range of \( t \) to find the total arc length.
Derivatives
When working with parametric equations, derivatives are crucial for understanding how the curve behaves. The derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) represent the rates of change of \( x \) and \( y \) with respect to the parameter \( t \).
Calculating these derivatives involves basic rules of differentiation. For instance:
Calculating these derivatives involves basic rules of differentiation. For instance:
- The derivative of the product \( e^t \cos(t) \) is calculated using the product rule: \( \frac{d}{dt}(e^t \cos t) = e^t \cos t - e^t \sin t \).
- Similarly, \( \frac{d}{dt}(e^t \sin t) = e^t \sin t + e^t \cos t \).
Integration
Integration is a fundamental process in calculus used to find the accumulated quantity, such as area under a curve or, in this case, the arc length. When it comes to parametric equations, integrating helps calculate the curve's length over a specified range of \( t \).
For example, once you determine the expression for the arc length of a parametric curve: \( \sqrt{2} e^t \), the task is to integrate it from \( t = 0 \) to \( t = \pi \):
For example, once you determine the expression for the arc length of a parametric curve: \( \sqrt{2} e^t \), the task is to integrate it from \( t = 0 \) to \( t = \pi \):
- The integral \( \int_{0}^{\pi} e^t dt \) is solved as \( e^t \), evaluated from 0 to \( \pi \).
- Substituting the bounds gives \( e^{\pi} - e^0 \), simplifying to \( e^{\pi} - 1 \).
Curve Plotting
Plotting curves defined by parametric equations involves evaluating the equations at various intervals of \( t \) and then plotting the corresponding \( x \) and \( y \) values.
For the equations \( x = e^t \cos(t) \) and \( y = e^t \sin(t) \):
For the equations \( x = e^t \cos(t) \) and \( y = e^t \sin(t) \):
- Calculate sets of \( (x, y) \) by substituting different \( t \) values from 0 to \( \pi \).
- Plot these points on the coordinate plane to visualize the spiral-like curve.
