Home > Calculus > Determining the Length of a Parametric Curve (Parametric Form)

What is the arclength of #f(t) = (sin^2t-cos2t,t/pi)# on #t in [-pi/4,pi/4]#?

Answer 1

#approx 3.06967#

we have #x(t)=sin^2(t)-cos(2t)# so #x'(t)=2sin(t)cos(t)+2sin(t)# #y(t)=t/pi# #y'(t)=1/pi# and we have the integral #int_(-pi/4)^(pi/4)sqrt(1/pi^2+(2sin(t)cos(t)+2sin(2t))^2)dt# This leads to an elliptic integral.

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Answer 2

Charles Arocha

To find the arc length of (f(t) = (\sin^2 t - \cos 2t, \frac{t}{\pi})) on (t) in ([- \frac{\pi}{4}, \frac{\pi}{4}]), you would use the formula for arc length:

[L = \int_a^b \sqrt{(f'(t))^2 + (g'(t))^2} , dt]

where (f(t)) and (g(t)) are the component functions of the curve. In this case, (f(t) = \sin^2 t - \cos 2t) and (g(t) = \frac{t}{\pi}). Then, compute (f'(t)) and (g'(t)), plug them into the formula, and evaluate the integral over the given interval.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.