Find the exact length of the polar curve. r=\theta^{2},\ 0\le\theta\le2

f480forever2rz 2021-11-20 Answered
Find the exact length of the polar curve.
r=θ2, 0θ2π
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

Opeance1951
Answered 2021-11-21 Author has 26 answers
Step 1
It is given that
r=θ2
Differentiate using the power rule
drdθ=2θ21
drdθ=2θ
Formula 6 from this section:
Length of a curve with polar equation r=f(θ), aθb is given by
L=abr2+[drdθ]2dθ
=02π(θ2)2+(2θ)2dθ
=02πθ4+4θ2dθ
Note that: θ4+4θ2=θ2(θ2+4)=θ2θ2+4=θθ2+4
Since we are integrating over [0, 2π], θ is positive and θ2=|θ|=θ
=02πθθ2+4dθ
=02πθ2+4[θ dθ]
We will now substitute
θ2+$=u
Step 2
On differentiation, we get
2θ,dθ=duθdθ=12du
The limits of integration will change from 02π to 02+44π2+4
=44π2+4u[12du]
=1244π2+4u12du
=12[u(12)+1(12)+1]44π2+4
=1
Not exactly what you’re looking for?
Ask My Question
huckelig75
Answered 2021-11-22 Author has 11 answers
Step 1
Given: r=θ2
l=abr2+(drdθ)2dθ - formula for length of a curve.
drdθ=2θ
L=02π(θ2)2+(2θ)2dθ
=02πθ4+4θ2dθ
=02πθ2(θ2+4)dθ
=02πθθ2+4dθ
Substitution:
t=θ2+4
dt=2θdθθdθ=dt2
04
2π4π2+4
L=1244π2+4tdt=12(23t32)44π2+4
=13((4π2+4)32432)92.89
L92.89
Not exactly what you’re looking for?
Ask My Question
user_27qwe
Answered 2021-11-29 Author has 208 answers

Step 1

The length of a polar function r=f(θ) is:

s=θ1θ2r2+(drdθ)2dθ sub in r=θ2 so drdθ=2θ

=02πθ4+4θ2dθ factor θ

=02πθθ2+4dθ let u=θ2+4 so du=2θdθ

=1244(π2+1)u12du

=12(23)(u32)44(π2+1)

=13(432(π2+1)32432)

=4332((π2+1)321)

The length of the curve is =4332((π2+1)321)

Not exactly what you’re looking for?
Ask My Question

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more