Find the exact length of the polar curve. r=θ2, 0≤θ≤2π

Question

Find the exact length of the polar curve.  r=θ2, 0≤θ≤2π

Opeance1951 · Accepted Answer

Step 1  It is given that  r=θ2  Differentiate using the power rule  drdθ=2⋅θ2−1  drdθ=2θ  Formula 6 from this section:  Length of a curve with polar equation r=f(θ), a≤θ≤b is given by   L=∫abr2+[drdθ]2dθ  =∫02π(θ2)2+(2θ)2dθ  =∫02πθ4+4θ2dθ  Note that: θ4+4θ2=θ2(θ2+4)=θ2⋅θ2+4=θ⋅θ2+4  Since we are integrating over [0, 2π], θ is positive and θ2=|θ|=θ  =∫02πθ⋅θ2+4dθ  =∫02πθ2+4[θ dθ]  We will now substitute  θ2+$=u  Step 2  On differentiation, we get  2θ,dθ=du⇒θdθ=12du  The limits of integration will change from ∫02π to ∫02+44π2+4  =∫44π2+4u[12du]  =12∫44π2+4u12du  =12[u(12)+1(12)+1]44π2+4  =1

huckelig75 · Accepted Answer

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=2θdθ→θdθ=dt2  0→4  2π→4π2+4  L=12∫44π2+4tdt=12(23t32)∣44π2+4  =13((4π2+4)32−432)≈92.89  L≈92.89

user_27qwe · Accepted Answer

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  =12∫44(π2+1)u12du  =12(23)(u32)44(π2+1)  =13(432(π2+1)32−432)  =4332((π2+1)32−1)  The length of the curve is =4332((π2+1)32−1)

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

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