What is the arclength of #f(t) = (cos2t-sin2t,tan^2t)# on #t in [pi/12,(5pi)/12]#?

Question

Christopher Allers · Accepted Answer

≈14.69 (no closed form)

#f(t) = (cos2t-sin2t,tan^2t)# To find the arc length #Deltas# over a time #Deltat#, you can try to approximate it with a straight line, using the hypotenuse of the right-angled triangle with sides #Deltax(t), Deltay(t)#, which is the change in the x- and y- coordinates over the time #t#. #Deltas ~~ sqrt(Deltax^2+Deltay^2)# Of course, this will often be wildly inaccurate since most curves aren't exactly a straight line. However, for a continuous function, the smaller time interval you choose, the more accurate the above approximation is (you can imagine zooming in on a function's graph: the more you zoom, the closer the function looks like a straight line) Thus, at the limit when you shorten the time interval, you get #ds = sqrt(dx^2+dy^2)# To manipulate it to suit the given parametric, we can adjust the equation such that #ds# is in terms of #dt#. #ds = sqrt((dx/dt*dt)^2+(dy/dt*dt)^2)# #=sqrt(((dx/dt)^2+(dy/dt)^2)dt^2)# #=sqrt((dx/dt)^2+(dy/dt)^2)dt# Since #dx/dt=-2sin2t-2cos2t, dy/dt=2tan(t)sec^2(t)#, #ds=sqrt((-2sin2t-2cos2t)^2+(2tan(t)sec^2(t))^2)dt# Factorising a 2 out of the entire expression and simplifying, we have #ds=2sqrt((cos2t+sin2t)^2+(tan(t)sec^2(t))^2)dt# To find the total arc length from #pi/12# to #5/12pi#, we can add up tiny segments of #ds# from #t=pi/12# to #t=5/12pi# Mathematically, this is equivalent to integrating from #t=pi/12# to #t=5/12pi# #s=int_(pi/12)^(5/12pi)2sqrt((cos2t+sin2t)^2+(tan(t)sec^2(t))^2)dt# Now this expression has no closed form and is approximately 14.69

Elijah Abram · Answer

undefined To find the arc length of the curve $ f(t) = (\cos(2t) - \sin(2t), \tan^2(t)) $ on the interval $ t $ in $ \left[ \frac{\pi}{12}, \frac{5\pi}{12} \right] $, you can use the formula for arc length:

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

where $ a $ and $ b $ are the limits of integration, and $ \frac{dx}{dt} $ and $ \frac{dy}{dt} $ are the derivatives of $ x $ and $ y $ with respect to $ t $, respectively.

Therefore, the arc length can be calculated as:

$$ L = \int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} \sqrt{\left(\frac{d(\cos(2t) - \sin(2t))}{dt}\right)^2 + \left(\frac{d(\tan^2(t))}{dt}\right)^2} \, dt $$