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UNO [17]
3 years ago
9

Select the curve generated by the parametric equations. Indicate with an arrow the direction in which the curve is traced as t i

ncreases. x = t + sin(t), y = cos(t), −π ≤ t ≤ π
Mathematics
1 answer:
bixtya [17]3 years ago
7 0

Answer:

length of the curve = 8

Step-by-step explanation:

Given parametric equations are x = t + sin(t) and y = cos(t) and given interval is

−π ≤ t ≤ π

Given data the arrow the direction in which the curve is traces means

the length of the curve of the given parametric equations.

The formula of length of the curve is

\int\limits^a_b {\sqrt{\frac{(dx}{dt}) ^{2}+(\frac{dy}{dt}) ^2 } } \, dx

Given limits values are −π ≤ t ≤ π

x = t + sin(t) ...….. (1)

y = cos(t).......(2)

differentiating equation (1)  with respective to 'x'

\frac{dx}{dt} = 1+cost

differentiating equation (2)  with respective to 'y'

\frac{dy}{dt} = -sint

The length of curve is

\int\limits^\pi_\pi  {\sqrt{(1+cost)^{2}+(-sint)^2 } } \, dt

\int\limits^\pi_\pi  \,   {\sqrt{(1+cost)^{2}+2cost+(sint)^2 } } \, dt

on simplification , we get

here using sin^2(t) +cos^2(t) =1 and after simplification , we get

\int\limits^\pi_\pi  \,   {\sqrt{(2+2cost } } \, dt

\sqrt{2} \int\limits^\pi_\pi  \,   {\sqrt{(1+1cost } } \, dt

again using formula, 1+cost = 2cos^2(t/2)

\sqrt{2} \int\limits^\pi _\pi  {\sqrt{2cos^2\frac{t}{2} } } \, dt

Taking common \sqrt{2} we get ,

\sqrt{2}\sqrt{2}  \int\limits^\pi _\pi ( {\sqrt{cos^2\frac{t}{2} } } \, dt

2(\int\limits^\pi _\pi  {cos\frac{t}{2} } \, dt

2(\frac{sin(\frac{t}{2} }{\frac{t}{2} } )^{\pi } _{-\pi }

length of curve = 4(sin(\frac{\pi }{2} )- sin(\frac{-\pi }{2} ))

length of the curve is = 4(1+1) = 8

<u>conclusion</u>:-

The arrow of the direction or the length of curve = 8

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