Consider the parametric equation #x = 9(cost+tsint)# and #y = 9(sint-tcost)#, What is the length of the curve for #t= 0# to #t=3pi/10#?

Question

Aurora Alvarado · Accepted Answer

#9/2((3pi)/10)^2# We have #{(dx/dt=9tcost),(dy/dt=9tsint):}# and #(ds)/dt=sqrt((dx/dt)^2+(dy/dt)^2) = 9t# then #s=int_(t=0)^(t=(3pi)/10)9tdt = 9/2((3pi)/10)^2#

Christopher Agresta · Answer

undefined To find the length of the curve defined by the parametric equations $ x = 9(\cos t + t \sin t) $ and $ y = 9(\sin t - t \cos t) $ for $ t = 0 $ to $ t = \frac{3\pi}{10} $, we can use the arc length formula for parametric curves:

$$ L = \int_{t_1}^{t_2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt $$

where $ t_1 = 0 $ and $ t_2 = \frac{3\pi}{10} $.

$$ \frac{dx}{dt} = -9t \sin t $$
$$ \frac{dy}{dt} = 9t \cos t $$

Substituting these derivatives into the arc length formula:

$$ L = \int_{0}^{\frac{3\pi}{10}} \sqrt{\left(-9t \sin t\right)^2 + \left(9t \cos t\right)^2} dt $$
$$ L = \int_{0}^{\frac{3\pi}{10}} \sqrt{81t^2 \sin^2 t + 81t^2 \cos^2 t} dt $$
$$ L = \int_{0}^{\frac{3\pi}{10}} \sqrt{81t^2(\sin^2 t + \cos^2 t)} dt $$
$$ L = \int_{0}^{\frac{3\pi}{10}} 9t \, dt $$
$$ L = \left[\frac{9t^2}{2}\right]_{0}^{\frac{3\pi}{10}} $$
$$ L = \frac{81\pi^2}{200} $$

Therefore, the length of the curve for $ t = 0 $ to $ t = \frac{3\pi}{10} $ is $ \frac{81\pi^2}{200} $.

Aurora Alvarado · Answer

undefined To find the length of the curve defined by the parametric equations $x = 9(\cos t + t \sin t)$ and $y = 9(\sin t - t \cos t)$ for $t = 0$ to $t = \frac{3\pi}{10}$, we can use the formula for the arc length of a parametric curve:

$$ L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt $$

First, we need to find the derivatives of $x$ and $y$ with respect to $t$:

$$ \frac{dx}{dt} = 9(-\sin t + t \cos t + \cos t - t \sin t) $$

$$ \frac{dy}{dt} = 9(\cos t - t \sin t + \sin t + t \cos t) $$

Next, we plug these derivatives into the formula:

$$ L = \int_{0}^{\frac{3\pi}{10}} \sqrt{\left(9(-\sin t + t \cos t + \cos t - t \sin t)\right)^2 + \left(9(\cos t - t \sin t + \sin t + t \cos t)\right)^2} \, dt $$

$$ L = \int_{0}^{\frac{3\pi}{10}} \sqrt{81(1 + t^2)} \, dt $$

$$ L = 9 \int_{0}^{\frac{3\pi}{10}} \sqrt{1 + t^2} \, dt $$

We recognize this as the integral form of the arc length of the curve $y = \sqrt{1 + x^2}$, which is $L = \frac{1}{2}(e^{2\pi}-1)$. Thus, the length of the curve for $t = 0$ to $t = \frac{3\pi}{10}$ is approximately $9(\frac{1}{2}(e^{2\pi}-1))$.