17Calculus - Path and Line Integrals
Your experience with the power of integrals has been expanding since you first learned about them. You started with evaluating the area under a curve and then quickly went to evaluating area between curves. Then you were able to calculate volumes of revolution. Double and triple integrals expanded your ability to solve all kinds of problems in two and three dimensions. Now you are about to embark on another integration adventure involving line integrals.
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If you want complete lectures on this topic, here are some videos.
Prof Leonard - How to Compute Line Integrals (Over Non-Conservative V.Fields) [2hr-17min-23secs]
video by Prof Leonard |
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Prof Leonard - Line Integrals on CONSERVATIVE V. Fields (Independence of Path) [1hr-53min-57secs]
video by Prof Leonard |
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Basically, the idea is that you can evaluate an integral along a specific path. This allows you to solve all kinds of problems involving vector fields in physics and engineering. Before you jump into this technique, you need an important skill that is usually not covered very well as a separate topic. Many instructors just assume you can pick it up as you go. However, it does help to have a bit of explanation and some examples. This skill is the ability to parameterize a smooth line or curve, i.e. to be able to describe a line or curve using parametric equations. Make sure you understand the material on this other page before going on.
Line integrals and path integrals are two different things, even though some instructors don't separate the two ideas.
A path integral, also called a scalar line integral, is an integral along a certain path and applies to scalar functions. |
A line integral is also an integral along a certain path but we require the integral to be evaluated in a specific direction, i.e. we need to take into account the direction in which the curve is traced, and applies to vector fields. |
We start by discussing the more basic idea of path integrals and then discuss how line integrals are different.
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Topics You Need To Understand For This Page
Related Topics and Links
external links you may find helpful |
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Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
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Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
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