3 a y 2 = x 2 ( a − x ) , 3ay^2=x^2(a-x), 3 a y 2 = x 2 ( a − x ) ,
y 2 = x 2 ( a − x ) 3 a , y^2=\frac{x^2(a-x)}{3a}, y 2 = 3 a x 2 ( a − x ) ,
y = ± x a − x 3 a . y=\pm \frac{x\sqrt{a-x}}{\sqrt{3a}}. y = ± 3 a x a − x .
x=0: y=0;
x=a: y=0.
The curve is symmetrical about the x-axis and meets it in points (0, 0) and (a, 0), so the loop lies between this points:
To find the perimeter of the loop, we can double its part above the x-axis.
P = 2 ∫ 0 a 1 + ( d y d x ) 2 d x . P=2 \int_0^a \sqrt{1+(\frac{dy}{dx})^2}dx. P = 2 ∫ 0 a 1 + ( d x d y ) 2 d x .
d y d x = d d x ( x a − x 3 a ) = 1 3 a ( a − x − x 2 a − x ) = \frac{dy}{dx}=\frac{d}{dx}(\frac{x\sqrt{a-x}}{\sqrt{3a}})=\frac{1}{\sqrt{3a}}(\sqrt{a-x}-\frac{x}{2\sqrt{a-x}})= d x d y = d x d ( 3 a x a − x ) = 3 a 1 ( a − x − 2 a − x x ) =
= 1 3 a 2 ( a − x ) − x 2 a − x = 2 a − 3 x 2 3 a a − x . =\frac{1}{\sqrt{3a}}\frac{2(a-x)-x}{2\sqrt{a-x}}=\frac{2a-3x}{2\sqrt{3a}\sqrt{a-x}}. = 3 a 1 2 a − x 2 ( a − x ) − x = 2 3 a a − x 2 a − 3 x .
1 + ( d y d x ) 2 = 1 + ( 2 a − 3 x 2 3 a a − x ) 2 = 1+(\frac{dy}{dx})^2=1+(\frac{2a-3x}{2\sqrt{3a}\sqrt{a-x}})^2= 1 + ( d x d y ) 2 = 1 + ( 2 3 a a − x 2 a − 3 x ) 2 =
= 1 + ( 2 a − 3 x ) 2 12 a ( a − x ) = 12 a 2 − 12 a x + 4 a 2 − 12 a + 9 x 2 12 a ( a − x ) = =1+\frac{(2a-3x)^2}{12a(a-x)}=\frac{12a^2-12ax+4a^2-12a+9x^2}{12a(a-x)}= = 1 + 12 a ( a − x ) ( 2 a − 3 x ) 2 = 12 a ( a − x ) 12 a 2 − 12 a x + 4 a 2 − 12 a + 9 x 2 =
= 16 a 2 − 24 a x + 9 x 2 12 a ( a − x ) = ( 4 a − 3 x ) 2 12 a ( a − x ) . =\frac{16a^2-24ax+9x^2}{12a(a-x)}=\frac{(4a-3x)^2}{12a(a-x)}. = 12 a ( a − x ) 16 a 2 − 24 a x + 9 x 2 = 12 a ( a − x ) ( 4 a − 3 x ) 2 .
So we have
P = 2 ∫ 0 a ( 4 a − 3 x ) 2 12 a ( a − x ) d x = 2 2 3 a ∫ 0 a 4 a − 3 x a − x d x = 1 3 a ∫ 0 a 4 a − 3 x a − x d x P=2 \int_0^a \sqrt{\frac{(4a-3x)^2}{12a(a-x)}}dx=\frac{2}{2\sqrt{3a}} \int_0^a \frac{4a-3x}{\sqrt{a-x}}dx=
\frac{1}{\sqrt{3a}} \int_0^a \frac{4a-3x}{\sqrt{a-x}}dx P = 2 ∫ 0 a 12 a ( a − x ) ( 4 a − 3 x ) 2 d x = 2 3 a 2 ∫ 0 a a − x 4 a − 3 x d x = 3 a 1 ∫ 0 a a − x 4 a − 3 x d x
Substitution: a − x = u a-x=u a − x = u d u = − d x , du=-dx, d u = − d x , x = a − u . x=a-u. x = a − u .
x = 0 : u = a − x = a − 0 = a ; x=0: u=a-x=a-0=a; x = 0 : u = a − x = a − 0 = a ;
x = a : u = a − x = a − a = 0. x=a: u=a-x=a-a=0. x = a : u = a − x = a − a = 0.
P = 1 3 a ∫ 0 a 4 a − 3 x a − x d x = − 1 3 a ∫ a 0 4 a − 3 ( a − u ) u d u = P=\frac{1}{\sqrt{3a}} \int_0^a \frac{4a-3x}{\sqrt{a-x}}dx=-\frac{1}{\sqrt{3a}} \int_a^0\frac{4a-3(a-u)}{\sqrt{u}}du= P = 3 a 1 ∫ 0 a a − x 4 a − 3 x d x = − 3 a 1 ∫ a 0 u 4 a − 3 ( a − u ) d u =
= − 1 3 a ∫ a 0 a + 3 u u d u = − 1 3 a ∫ a 0 ( a u + 3 u ) d u = =-\frac{1}{\sqrt{3a}} \int_a^0\frac{a+3u}{\sqrt{u}}du=-\frac{1}{\sqrt{3a}} \int_a^0 (\frac{a}{\sqrt{u}}+3\sqrt{u})du= = − 3 a 1 ∫ a 0 u a + 3 u d u = − 3 a 1 ∫ a 0 ( u a + 3 u ) d u =
= − 1 3 a ( 2 a u + 2 u u ) ∣ a 0 = =-\frac{1}{\sqrt{3a}} (2a\sqrt{u}+2u\sqrt{u})|_a^0= = − 3 a 1 ( 2 a u + 2 u u ) ∣ a 0 =
− 1 3 a [ ( 2 a 0 + 2 ⋅ 0 ⋅ 0 ) − ( 2 a a + 2 a a ) ] = -\frac{1}{\sqrt{3a}}[ (2a\sqrt{0}+2\cdot 0 \cdot \sqrt{0})- (2a\sqrt{a}+2a\sqrt{a})]= − 3 a 1 [( 2 a 0 + 2 ⋅ 0 ⋅ 0 ) − ( 2 a a + 2 a a )] =
= 4 a a 3 a = 4 a 3 . =\frac{4a\sqrt{a}}{\sqrt{3a}}=\frac{4a}{\sqrt{3}}. = 3 a 4 a a = 3 4 a .
Answer: 4 a 3 . \frac{4a}{\sqrt{3}}. 3 4 a .
#340153 on Dec 2023
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