Find the exact length of the polar curve. r=θ2,0≤θ≤5π4

Step 1 here the given polar curve. r=θ2,0≤θ≤5π4 Step 2 The length of curve is L=∫05π4r2+(drdθ)2dθ Thus L=∫05π4θ4+4θ2dθ L=∫05π4θθ2+4dθ Put θ2+4= ⇒2θdθ=d ⇒θdθ=dt2 Step 3 Also when θ⇒0 then t⇒4 And when θ⇒5π4,t⇒(5π4)2+4 Thus integration becomes L=12∫4(5π4)2+4t1/2dt L=[t32]4(5π4)2+4 L=[((5π4)2+4)32−432]

Solved: "Find the exact length of the polar curve...."

College
Calculus and Analysis
Integral Calculus
Parametric equations, polar coordinates, and vector-valued functions

jazzcutie0h

Answered question

2021-12-04

Find the exact length of the polar curve.

r = θ^{2}, 0 \leq θ \leq 5 \frac{π}{4}

Answer & Explanation

menerkupvd

Beginner2021-12-05Added 12 answers

Step 1
here the given polar curve.
$r = θ^{2}, 0 \leq θ \leq 5 \frac{π}{4}$
Step 2
The length of curve is
$L = \int_{0}^{\frac{5 π}{4}} \sqrt{r^{2} + {(\frac{d r}{d θ})}^{2}} d θ$
Thus
$L = \int_{0}^{\frac{5 π}{4}} \sqrt{θ^{4} + 4 θ^{2}} d θ$
$L = \int_{0}^{\frac{5 π}{4}} θ \sqrt{θ^{2} + 4} d θ$
Put
$θ^{2} + 4 =$
$\Rightarrow 2 θ d θ = d$
$\Rightarrow θ d θ = \frac{d t}{2}$
Step 3
Also
when $θ \Rightarrow 0$ then $t \Rightarrow 4$
And when $θ \Rightarrow \frac{5 π}{4}, t \Rightarrow {(\frac{5 π}{4})}^{2} + 4$
Thus integration becomes
$L = \frac{1}{2} \int_{4}^{(\frac{5 π}{4})^{2} + 4} t^{1 / 2} d t$
$L = {[t^{\frac{3}{2}}]}_{4}^{{(\frac{5 π}{4})}^{2} + 4}$
$L = [{({(\frac{5 π}{4})}^{2} + 4)}^{\frac{3}{2}} - 4^{\frac{3}{2}}]$

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