Thread: length of parametric curve

consider the parametric equation

x=10(cos∅+∅sin∅)
y=10(sin∅-∅cos∅)

what is the length of the curve for ∅=0 to ∅=1/6∏?

Originally Posted by waite3

consider the parametric equation

x=10(cos∅+∅sin∅)
y=10(sin∅-∅cos∅)

what is the length of the curve for ∅=0 to ∅=1/6∏?

do you have the right integral?

the arc length is given by: $\displaystyle L = \int_0^{1/6 \pi} \sqrt{\left( \frac {dx}{d \theta} \right)^2 + \left( \frac {dy}{d \theta} \right)^2 }~d \theta$

ya i have that but everytime i take the derivative of x and y then plug them into the equation i come out with the wrong answer. what do you come out with for an answer. thanks for your help

Originally Posted by waite3

ya i have that but everytime i take the derivative of x and y then plug them into the equation i come out with the wrong answer. what do you come out with for an answer. thanks for your help

for the integral, i got $\displaystyle 5 \theta^2 + C$, now just evaluate that between the given limits

i come out with 0.239245 and its not right.

Originally Posted by waite3

i come out with 0.239245 and its not right.

of course that is wrong. that is not what you get when you evaluate.

you have $\displaystyle 5 \theta^2 \big|_0^{1 / 6 \pi} = 5 \left( \frac 1{6 \pi} \right)^2 {\color{red}\ne} ~0.239245$