find the length of each curve: x=1+3t, y=4t-1, 1<= t<=7, and
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Re: finding the length of a curve
For 1, dx/dt = 3, dy/dt = 4
\(\displaystyle \int _{t_o} ^{t_f} \sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}dt = \int _1 ^7 \sqrt{(3)^2+(4)^2}dt = ?\)
For 2,
\(\displaystyle \int _{x_0} ^{x_1} \sqrt{1+(y')^2}dx\)
For 1, dx/dt = 3, dy/dt = 4
\(\displaystyle \int _{t_o} ^{t_f} \sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}dt = \int _1 ^7 \sqrt{(3)^2+(4)^2}dt = ?\)
For 2,
\(\displaystyle \int _{x_0} ^{x_1} \sqrt{1+(y')^2}dx\)
Re: find the length of each curve
Hello, atul!
Exactly where are you stuck?
You don't know the Arc Length formulas?
You have them but don't know how to use them?
You really suck at algebra?
You don't know how to integrate?
You can't evaluate definite integrals?
\(\displaystyle \text{Formula (for parametric functions): }\;L \;=\;\int^b_a\sqrt{\left(\tfrac{dx}{dt}\right)^2 + \left(\tfrac{dy}{dt}\right)^2}\,dt\)
\(\displaystyle \text{We have: }\;\begin{Bmatrix}x &=& 1+3t & \Rightarrow & \frac{dt}{dt} &=& 3 \\ \\[-3mm] y &=& 4t-1 & \Rightarrow & \frac{dy}{dt} &=& 4 \end{Bmatrix}\)
\(\displaystyle \text{Then: }\;\sqrt{\left(\tfrac{dx}{dt}\right)^2 + \left(\tfrac{dy}{dt}\right)^2}} \;=\;\sqrt{3^2+4^2} \;=\;\sqrt{25} \;=\;5\)
\(\displaystyle \text{Therefore: }\;L \;=\;\int^7_1 5\,dt\quad\hdots\text{ etc.}\)
\(\displaystyle \text{Formiula: }\;L \;=\;\int^b_a\sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx\)
\(\displaystyle \text{We have: }\;y \:=\:\frac{x^5}{10} + \frac{x^{\:\text{-}3}}{6}\)
\(\displaystyle \text{Then: }\;\frac{dy}{dx} \;=\;\frac{x^4}{2} - \frac{x^{\:\text{-}4}}{2}\)
. . \(\displaystyle \left(\frac{dy}{dx}\right)^2 \;\;=\;\;\left(\frac{x^4}{2} - \frac{x^{\:\text{-}4}}{2}\right)^2 \;\;=\;\;\frac{x^8}{4} - \frac{1}{2} + \frac{x^{\:\text{-}8}}{4}\)
. . \(\displaystyle 1 + \left(\frac{dy}{dx}\right)^2 \;\;=\;\;1 + \frac{x^8}{4} - \frac{1}{2} + \frac{x^{\:\text{-}4}}{4} \;\;=\;\;\frac{x^8}{4} + \frac{1}{2} + \frac{x^{\:\text{-}8}}{4} \;\;=\;\;\left(\frac{x^4}{2} + \frac{x^{\:\text{-}4}}{2}\right)^2\)
. . \(\displaystyle \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \;\;=\;\;\sqrt{\left(\frac{x^4}{2} + \frac{x^{\:\text{-}4}}{2}\right)^2} \;\;=\;\;\frac{x^4}{2}+\frac{x^{\:\text{-}4}}{2} \;\;=\;\;\frac{1}{2}\left(x^4} + x^{\:\text{-}4}\right)\)
\(\displaystyle \text{Therefore: }\;L \;=\;\frac{1}{2}\int^2_1\left(x^4 - x^{\:\text{-}4}\right)\,dx \quad\hdots\:\text{ etc.}\)
Hello, atul!
Exactly where are you stuck?
You don't know the Arc Length formulas?
You have them but don't know how to use them?
You really suck at algebra?
You don't know how to integrate?
You can't evaluate definite integrals?
\(\displaystyle \text{Formula (for parametric functions): }\;L \;=\;\int^b_a\sqrt{\left(\tfrac{dx}{dt}\right)^2 + \left(\tfrac{dy}{dt}\right)^2}\,dt\)
\(\displaystyle \text{We have: }\;\begin{Bmatrix}x &=& 1+3t & \Rightarrow & \frac{dt}{dt} &=& 3 \\ \\[-3mm] y &=& 4t-1 & \Rightarrow & \frac{dy}{dt} &=& 4 \end{Bmatrix}\)
\(\displaystyle \text{Then: }\;\sqrt{\left(\tfrac{dx}{dt}\right)^2 + \left(\tfrac{dy}{dt}\right)^2}} \;=\;\sqrt{3^2+4^2} \;=\;\sqrt{25} \;=\;5\)
\(\displaystyle \text{Therefore: }\;L \;=\;\int^7_1 5\,dt\quad\hdots\text{ etc.}\)
\(\displaystyle \text{Formiula: }\;L \;=\;\int^b_a\sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx\)
\(\displaystyle \text{We have: }\;y \:=\:\frac{x^5}{10} + \frac{x^{\:\text{-}3}}{6}\)
\(\displaystyle \text{Then: }\;\frac{dy}{dx} \;=\;\frac{x^4}{2} - \frac{x^{\:\text{-}4}}{2}\)
. . \(\displaystyle \left(\frac{dy}{dx}\right)^2 \;\;=\;\;\left(\frac{x^4}{2} - \frac{x^{\:\text{-}4}}{2}\right)^2 \;\;=\;\;\frac{x^8}{4} - \frac{1}{2} + \frac{x^{\:\text{-}8}}{4}\)
. . \(\displaystyle 1 + \left(\frac{dy}{dx}\right)^2 \;\;=\;\;1 + \frac{x^8}{4} - \frac{1}{2} + \frac{x^{\:\text{-}4}}{4} \;\;=\;\;\frac{x^8}{4} + \frac{1}{2} + \frac{x^{\:\text{-}8}}{4} \;\;=\;\;\left(\frac{x^4}{2} + \frac{x^{\:\text{-}4}}{2}\right)^2\)
. . \(\displaystyle \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \;\;=\;\;\sqrt{\left(\frac{x^4}{2} + \frac{x^{\:\text{-}4}}{2}\right)^2} \;\;=\;\;\frac{x^4}{2}+\frac{x^{\:\text{-}4}}{2} \;\;=\;\;\frac{1}{2}\left(x^4} + x^{\:\text{-}4}\right)\)
\(\displaystyle \text{Therefore: }\;L \;=\;\frac{1}{2}\int^2_1\left(x^4 - x^{\:\text{-}4}\right)\,dx \quad\hdots\:\text{ etc.}\)
