We can find the length of the curve by the next formula:
l = ∫ θ 1 θ 2 ( r ( θ ) ) 2 + r ′ ( θ ) ) 2 d θ . l=\int_{ \theta_1}^{ \theta_2} \sqrt{(r(\theta))^2+r '(\theta))^2}d \theta. l = ∫ θ 1 θ 2 ( r ( θ ) ) 2 + r ′ ( θ ) ) 2 d θ .
r ( θ ) = 4 ( 1 − s i n θ ) , r(\theta)=4(1-sin\theta), r ( θ ) = 4 ( 1 − s in θ ) ,
r ′ ( θ ) = − 4 c o s θ , r'(\theta)=-4cos\theta, r ′ ( θ ) = − 4 cos θ ,
so we have:
l = ∫ 90 ° 270 ° ( 4 ( 1 − s i n θ ) ) 2 + ( − 4 c o s θ ) 2 d θ = l=\int_{ 90\degree}^{ 270\degree} \sqrt{(4(1-sin\theta))^2+(-4cos\theta)^2}d \theta= l = ∫ 90° 270° ( 4 ( 1 − s in θ ) ) 2 + ( − 4 cos θ ) 2 d θ =
= ∫ 90 ° 270 ° 16 − 32 s i n θ + 16 s i n 2 θ + 16 c o s 2 θ d θ = =\int_{ 90\degree}^{ 270\degree} \sqrt{16-32sin\theta+16sin^2\theta+16cos^2\theta}d \theta= = ∫ 90° 270° 16 − 32 s in θ + 16 s i n 2 θ + 16 co s 2 θ d θ =
= ∫ 90 ° 270 ° 16 − 32 s i n θ + 16 d θ = =\int_{ 90\degree}^{ 270\degree} \sqrt{16-32sin\theta+16}d \theta= = ∫ 90° 270° 16 − 32 s in θ + 16 d θ =
= ∫ 90 ° 270 ° 32 ( 1 − s i n θ ) d θ = =\int_{ 90\degree}^{ 270\degree} \sqrt{32(1-sin\theta)}d \theta= = ∫ 90° 270° 32 ( 1 − s in θ ) d θ =
= 4 2 ∫ 90 ° 270 ° 1 − s i n θ d θ = =4\sqrt{2}\int_{ 90\degree}^{ 270\degree} \sqrt{1-sin\theta}d \theta= = 4 2 ∫ 90° 270° 1 − s in θ d θ =
= 4 2 ∫ 90 ° 270 ° ( 1 − s i n θ ) 1 + s i n θ 1 + s i n θ d θ = =4\sqrt{2}\int_{ 90\degree}^{ 270\degree} \sqrt{(1-sin\theta) \frac{1+sin\theta}{1+sin\theta}}d \theta= = 4 2 ∫ 90° 270° ( 1 − s in θ ) 1 + s in θ 1 + s in θ d θ =
= 4 2 ∫ 90 ° 270 ° 1 − s i n 2 θ 1 + s i n θ d θ = =4\sqrt{2}\int_{ 90\degree}^{ 270\degree} \sqrt{ \frac{1-sin^2\theta}{1+sin\theta}}d \theta= = 4 2 ∫ 90° 270° 1 + s in θ 1 − s i n 2 θ d θ =
= 4 2 ∫ 90 ° 270 ° c o s 2 θ 1 + s i n θ d θ = =4\sqrt{2}\int_{ 90\degree}^{ 270\degree} \sqrt{ \frac{cos^2\theta}{1+sin\theta}}d \theta= = 4 2 ∫ 90° 270° 1 + s in θ co s 2 θ d θ =
= 4 2 ∫ 90 ° 270 ° − c o s θ 1 + s i n θ d θ = =4\sqrt{2}\int_{ 90\degree}^{ 270\degree} \frac{-cos\theta}{\sqrt{{1+sin\theta}}}d \theta= = 4 2 ∫ 90° 270° 1 + s in θ − cos θ d θ =
= − 4 2 ∫ 90 ° 270 ° c o s θ 1 + s i n θ d θ . =-4\sqrt{2}\int_{ 90\degree}^{ 270\degree} \frac{cos\theta}{\sqrt{{1+sin\theta}}}d \theta. = − 4 2 ∫ 90° 270° 1 + s in θ cos θ d θ .
Substitution: 1 + s i n θ = u , 1+sin\theta=u, 1 + s in θ = u , d u = c o s θ d θ , du=cos\theta d\theta, d u = cos θ d θ ,
1 + s i n ( 90 ° ) = 1 + 1 = 2 , 1+sin(90\degree)=1+1=2, 1 + s in ( 90° ) = 1 + 1 = 2 ,
1 + s i n ( 270 ° ) = 1 + ( − 1 ) = 0. 1+sin(270\degree)=1+(-1)=0. 1 + s in ( 270° ) = 1 + ( − 1 ) = 0.
We have:
l = − 4 2 ∫ 2 0 d u u = 4 2 ⋅ 2 u ∣ 0 2 = 8 2 ( 2 − 0 ) = 16. l=-4\sqrt{2}\int_{ 2}^{0} \frac{du}{\sqrt{u}}=4\sqrt{2} \cdot 2\sqrt{u}|_0^2=8\sqrt{2}(\sqrt{2}-0)=16. l = − 4 2 ∫ 2 0 u d u = 4 2 ⋅ 2 u ∣ 0 2 = 8 2 ( 2 − 0 ) = 16.
Answer: 16.
#339914 on Dec 2023
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