calc practice help

NameisJames said:
find the arclength for 0<= t <= 1/2pi of the curve (cos 2t + 3,7 - sin 2t)
Click to expand...
Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:


Please share your work/thoughts about this problem
 
This is the equation of a circle.
Where is the center of this circle?
What is the radius of this circle?
What part of the circle is the requested arclength?
What is the arclength?
 
x=cos⁡2t+3\displaystyle x = \cos 2t + 3
y=7−sin⁡2t\displaystyle y = 7 - \sin 2t

Arc length =∫0π/2(dxdt)2+(dydt)2 dt\displaystyle = \int_{0}^{\pi/2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \ dt
 
nasi112 said:
x=cos⁡2t+3\displaystyle x = \cos 2t + 3
y=7−sin⁡2t\displaystyle y = 7 - \sin 2t

Arc length =∫0π/2(dxdt)2+(dydt)2 dt\displaystyle = \int_{0}^{\pi/2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \ dt
Click to expand...
Tutors please note that the OP has not posted any work, neither did s/he answer the questions posed by StevenG.
 
nasi112 said:
x=cos⁡2t+3\displaystyle x = \cos 2t + 3
y=7−sin⁡2t\displaystyle y = 7 - \sin 2t

Arc length =∫0π/2(dxdt)2+(dydt)2 dt\displaystyle = \int_{0}^{\pi/2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \ dt
Click to expand...
I personally like doing a problem at the lowest level possible. Right or wrong this is my style.
You should immediately see that this is a circle. I actually drew the circle and the the answer was then obvious after considering the domain.
 
Steven G said:
I personally like doing a problem at the lowest level possible. Right or wrong this is my style.
You should immediately see that this is a circle. I actually drew the circle and the the answer was then obvious after considering the domain.
Click to expand...
Your approach works fine. I just used the calculus way to solve the problem.
 
You must log in or register to reply here.
Top