Find the arc length of the curve
π₯(π‘)=cosπ‘+π‘sinπ‘, 0β€π‘β€π/2
π¦(π‘) = sin π‘ β π‘ cos π‘ ^2
(ds)^2 = (dx/dt)^2 + (dy/dt)^2
= (-sint + sint + tcost)^2 + (cost - cost + tsint)^2
= t^2cos^2(t) + t^2sin^2(t)
= t^2
so,
ds = dt
that's pretty easy to integrate, right?
Hmmm. I missed that ^2 hanging out there. Is
y(t) = sint - t(cost)^2
or
y(t) = sint - tcos(t^2)?
In either case, adjust the expression for ds above.
To find the arc length of the curve given by π₯(π‘) = cosπ‘ + π‘sinπ‘ and π¦(π‘) = sinπ‘ β π‘cosπ‘^2, 0 β€ π‘ β€ π/2, we can use the arc length formula:
L = β« [β(π₯'(π‘))^2 + (π¦'(π‘))^2] dt
First, let's find the derivatives π₯'(π‘) and π¦'(π‘):
π₯'(π‘) = -sinπ‘ + sinπ‘ + π‘(cosπ‘) + sinπ‘ = π‘cosπ‘ + sinπ‘
π¦'(π‘) = cosπ‘ - π inπ‘ - 2π‘cosπ‘(-sinπ‘) = cosπ‘ - sinπ‘ + 2π‘(cosπ‘)(sinπ‘)
Now, substitute these derivatives back into the arc length formula and evaluate the integral:
L = β« [β(π‘cosπ‘ + sinπ‘)^2 + (cosπ‘ - sinπ‘ + 2π‘(cosπ‘)(sinπ‘))^2] dt
Simplifying, we get:
L = β« β(π‘^2cos^2π‘ + 2π‘cosπ‘sinπ‘ + sin^2π‘ + cos^2π‘ - 2cosπ‘sinπ‘ + 4π‘^2(cos^2π‘)(sin^2π‘)) dt
L = β« β(π‘^2 + 4π‘^2(cos^2π‘)(sin^2π‘)) dt
L = β« β(π‘^2(1 + 4cos^2π‘sin^2π‘)) dt
L = β« β(π‘^2(1 + 4(sin^2π‘)(1 - cos^2π‘))) dt
L = β« β(π‘^2(1 + 4sin^2π‘ - 4sin^2π‘cos^2π‘)) dt
L = β« β(π‘^2(1 + 4sin^2π‘ - 4sin^2π‘(1 - sin^2π‘))) dt
L = β« β(π‘^2(1 + 4sin^2π‘ - 4sin^2π‘ + 4sin^4π‘)) dt
L = β« β(π‘^2(4sin^4π‘ + 4sin^2π‘ + 1)) dt
Now, integrate with respect to t:
L = β« π‘β(4sin^4π‘ + 4sin^2π‘ + 1) dt
At this point, we can simplify the integrand to an expression involving only sinπ‘:
L = β« π‘β(4(sin^2π‘)^2 + 4sin^2π‘ + 1) dt
After that, we can try to substitute for sinπ‘:
Let u = sinπ‘,
du = cosπ‘ dt,
dt = du / cosπ‘
Substituting, we get:
L = β« π‘β(4(u^2)^2 + 4u^2 + 1) (du / cosπ‘)
Simplifying the integrand and changing the limits of integration, we have:
L = β« (π’β(4u^4 + 4u^2 + 1)) du from 0 to sin(π/2)
Now, we can solve this definite integral to find the arc length of the curve.
To find the arc length of a curve, we use the formula:
L = β«β[dx/dt]^2 + [dy/dt]^2 dt
where dx/dt and dy/dt represent the derivative of x and y with respect to t, respectively.
Let's calculate dx/dt and dy/dt first.
Given:
x(t) = cos(t) + t*sin(t)
y(t) = sin(t) - t*cos(t)^2
To find dx/dt, we take the derivative of x(t) with respect to t:
dx/dt = -sin(t) + sin(t) + t*cos(t) = t*cos(t)
To find dy/dt, we take the derivative of y(t) with respect to t:
dy/dt = cos(t) - 1*cos(t)^2 + 2t*cos(t)*sin(t) = cos(t) - cos(t)^2 + 2t*cos(t)*sin(t)
Now, we can substitute dx/dt and dy/dt into the formula for arc length:
L = β«β[(t*cos(t)]^2 + [cos(t) - cos(t)^2 + 2t*cos(t)*sin(t)]^2 dt, where t ranges from 0 to Ο/2.
Simplifying the expression under the square root:
L = β«β[t^2*cos(t)^2 + cos(t)^2 - 2t*cos(t)^3 + 2t*cos(t)*sin(t) - 2t*cos(t)^2*sin(t) + 4t^2*cos(t)^2*sin(t)^2] dt
Now, we can calculate the integral from 0 to Ο/2 numerically or using an appropriate software or calculator.