find the length of each curve: x=1+3t, y=4t-1, 1<= t<=7, and

[atul]

Im stuck in this 2 problems help is much appriciated:

1. x=1+3t y=4t-1 1<= t <=7

2 y= (1/10)X^5+(1/6)X^-3 1<=X<=2

thanks if someone can help

 

[atul]

Im stuck in this 2 problems help is much appriciated:

1. x=1+3t y=4t-1 1<= t <=7

2 y= (1/10)X^5+(1/6)X^-3 1<=X<=2

thanks if someone can help

 

[daon]

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,

$\displaystyle \int _{x_0} ^{x_1} \sqrt{1+(y')^2}dx$

 

[soroban]

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?

$\displaystyle 1)\;\;x\:=\:1+3t,\quad y\:=\:4t-1,\quad 1\leq t \leq 7$

$\displaystyle \text{Formula (for parametric functions): }\;L \;=\;\int^b_a\sqrt{\left(\tfrac{dx}{dt}\right)^2 + \left(\tfrac{dy}{dt}\right)^2}\,dt$


\(\displaystyle \text{We have: }\;\begin{Bmatrix}x &=& 1+3t & \Rightarrow & \frac{dt}{dt} &=& 3 \\ \\[-3mm] y &=& 4t-1 & \Rightarrow & \frac{dy}{dt} &=& 4 \end{Bmatrix}\)

\(\displaystyle \text{Then: }\;\sqrt{\left(\tfrac{dx}{dt}\right)^2 + \left(\tfrac{dy}{dt}\right)^2}} \;=\;\sqrt{3^2+4^2} \;=\;\sqrt{25} \;=\;5\)

\(\displaystyle \text{Therefore: }\;L \;=\;\int^7_1 5\,dt\quad\hdots\text{ etc.}\)

$\displaystyle 2)\;\;y\:=\: \frac{x^5}{10} + \frac{x^{-3}}{6},\quad 1\leq x \leq 2$

$\displaystyle \text{Formiula: }\;L \;=\;\int^b_a\sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx$

$\displaystyle \text{We have: }\;y \:=\:\frac{x^5}{10} + \frac{x^{\:\text{-}3}}{6}$

$\displaystyle \text{Then: }\;\frac{dy}{dx} \;=\;\frac{x^4}{2} - \frac{x^{\:\text{-}4}}{2}$

. . $\displaystyle \left(\frac{dy}{dx}\right)^2 \;\;=\;\;\left(\frac{x^4}{2} - \frac{x^{\:\text{-}4}}{2}\right)^2 \;\;=\;\;\frac{x^8}{4} - \frac{1}{2} + \frac{x^{\:\text{-}8}}{4}$

. . $\displaystyle 1 + \left(\frac{dy}{dx}\right)^2 \;\;=\;\;1 + \frac{x^8}{4} - \frac{1}{2} + \frac{x^{\:\text{-}4}}{4} \;\;=\;\;\frac{x^8}{4} + \frac{1}{2} + \frac{x^{\:\text{-}8}}{4} \;\;=\;\;\left(\frac{x^4}{2} + \frac{x^{\:\text{-}4}}{2}\right)^2$

. . \(\displaystyle \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \;\;=\;\;\sqrt{\left(\frac{x^4}{2} + \frac{x^{\:\text{-}4}}{2}\right)^2} \;\;=\;\;\frac{x^4}{2}+\frac{x^{\:\text{-}4}}{2} \;\;=\;\;\frac{1}{2}\left(x^4} + x^{\:\text{-}4}\right)\)


\(\displaystyle \text{Therefore: }\;L \;=\;\frac{1}{2}\int^2_1\left(x^4 - x^{\:\text{-}4}\right)\,dx \quad\hdots\:\text{ etc.}\)

 

[stapel]

Duplicate threads merged.

 

[atul]

thank you i was struggling to get through the power part and didnt knew how to go ahead with it. It was really helpful.