4.
x = a t 2 , y = 2 a t x=at^2,y=2at x = a t 2 , y = 2 a t
evolute of a curve is the locus of all its centers of curvature:
X ( t ) = x ( t ) − y ′ ( t ) ( x ′ ( t ) 2 + y ′ ( t ) 2 ) x ′ ( t ) y ′ ′ ( t ) − x ′ ′ ( t ) y ′ ( t ) X(t)=x(t)-\frac{y'(t)(x'(t)^2+y'(t)^2)}{x'(t)y''(t)-x''(t)y'(t)} X ( t ) = x ( t ) − x ′ ( t ) y ′′ ( t ) − x ′′ ( t ) y ′ ( t ) y ′ ( t ) ( x ′ ( t ) 2 + y ′ ( t ) 2 )
Y ( t ) = y ( t ) + x ′ ( t ) ( x ′ ( t ) 2 + y ′ ( t ) 2 ) x ′ ( t ) y ′ ′ ( t ) − x ′ ′ ( t ) y ′ ( t ) Y(t)=y(t)+\frac{x'(t)(x'(t)^2+y'(t)^2)}{x'(t)y''(t)-x''(t)y'(t)} Y ( t ) = y ( t ) + x ′ ( t ) y ′′ ( t ) − x ′′ ( t ) y ′ ( t ) x ′ ( t ) ( x ′ ( t ) 2 + y ′ ( t ) 2 )
x ′ = 2 a t , x ′ ′ = 2 a x'=2at,x''=2a x ′ = 2 a t , x ′′ = 2 a
y ′ = 2 a , y ′ ′ = 0 y'=2a,y''=0 y ′ = 2 a , y ′′ = 0
X ( t ) = a t 2 − 2 a ( 4 a 2 t 2 + 4 a 2 ) − 4 a 2 = 3 a t 2 + 2 a X(t)=at^2-\frac{2a(4a^2t^2+4a^2)}{-4a^2}=3at^2+2a X ( t ) = a t 2 − − 4 a 2 2 a ( 4 a 2 t 2 + 4 a 2 ) = 3 a t 2 + 2 a
Y ( t ) = 2 a t + 2 a t ( 4 a 2 t 2 + 4 a 2 ) − 4 a 2 = − 2 a t 3 Y(t)=2at+\frac{2at(4a^2t^2+4a^2)}{-4a^2}=-2at^3 Y ( t ) = 2 a t + − 4 a 2 2 a t ( 4 a 2 t 2 + 4 a 2 ) = − 2 a t 3
5.
The envelope of the family of curves is a curve such that at each point it touches tangentially one of the curves of the family.
y = m x ± a 2 m 2 − b y = mx±\sqrt{a^2m^2 − b} y = m x ± a 2 m 2 − b
f ( x , y , m ) = y − m x ± a 2 m 2 − b = 0 f(x,y,m)= y - mx±\sqrt{a^2m^2 − b}=0 f ( x , y , m ) = y − m x ± a 2 m 2 − b = 0
f m ′ ( x , y , m ) = − x ± m / a 2 m 2 − b = 0 f'_m(x,y,m)=-x±m/\sqrt{a^2m^2 − b}=0 f m ′ ( x , y , m ) = − x ± m / a 2 m 2 − b = 0
± a 2 m 2 − b = m x − y ±\sqrt{a^2m^2 − b}=mx-y ± a 2 m 2 − b = m x − y
− x + m m x − y = 0 -x+\frac{m}{mx-y}=0 − x + m x − y m = 0
x y − m x 2 + m = 0 xy-mx^2+m=0 x y − m x 2 + m = 0
m = x y x 2 − 1 m=\frac{xy}{x^2-1} m = x 2 − 1 x y
( y − x 2 y x 2 − 1 ) 2 = a 2 x 2 y 2 ( x 2 − 1 ) 2 − b (y-\frac{x^2y}{x^2-1})^2=\frac{a^2x^2y^2}{(x^2-1)^2}-b ( y − x 2 − 1 x 2 y ) 2 = ( x 2 − 1 ) 2 a 2 x 2 y 2 − b
6.
x / a + y / b = 1 x/
a
+
y/
b
= 1 x / a + y / b = 1
f ( x , y , a ) = x / a + y / b − 1 = 0 f(x,y,a)=x/a+y/b-1=0 f ( x , y , a ) = x / a + y / b − 1 = 0
b = 1 − a b=1-a b = 1 − a
f ( x , y , a ) = x / a + y / ( 1 − a ) − 1 = 0 f(x,y,a)=x/a+y/(1-a)-1=0 f ( x , y , a ) = x / a + y / ( 1 − a ) − 1 = 0
f a ′ ( x , y , a ) = − x / a 2 + y / ( 1 − a ) 2 = 0 f'_a(x,y,a)=-x/a^2+y/(1-a)^2=0 f a ′ ( x , y , a ) = − x / a 2 + y / ( 1 − a ) 2 = 0
x ( 1 − a ) 2 = y a 2 x(1-a)^2=ya^2 x ( 1 − a ) 2 = y a 2
x / a + x ( 1 + a ) / a 2 − 1 = 0 x/a+x(1+a)/a^2-1=0 x / a + x ( 1 + a ) / a 2 − 1 = 0
x / ( 1 − b ) + x ( 2 − b ) / ( b − 1 ) 2 − 1 = 0 x/(1-b)+x(2-b)/(b-1)^2-1=0 x / ( 1 − b ) + x ( 2 − b ) / ( b − 1 ) 2 − 1 = 0
− x ( b − 1 ) + x ( 2 − b ) − ( b − 1 ) 2 = 0 -x(b-1)+x(2-b)-(b-1)^2=0 − x ( b − 1 ) + x ( 2 − b ) − ( b − 1 ) 2 = 0
x ( 3 − 2 b ) = ( b − 1 ) 2 x(3-2b)=(b-1)^2 x ( 3 − 2 b ) = ( b − 1 ) 2
7.
x 2 / a 2 + y 2 / b 2 = 1 x^
2/
a^2
+
y^
2/
b^
2
= 1 x 2 / a 2 + y 2 / b 2 = 1
x = a c o s t , y = b s i n t x=acost,y=bsint x = a cos t , y = b s in t
evolute of a curve:
X ( t ) = x ( t ) − y ′ ( t ) ( x ′ ( t ) 2 + y ′ ( t ) 2 ) x ′ ( t ) y ′ ′ ( t ) − x ′ ′ ( t ) y ′ ( t ) X(t)=x(t)-\frac{y'(t)(x'(t)^2+y'(t)^2)}{x'(t)y''(t)-x''(t)y'(t)} X ( t ) = x ( t ) − x ′ ( t ) y ′′ ( t ) − x ′′ ( t ) y ′ ( t ) y ′ ( t ) ( x ′ ( t ) 2 + y ′ ( t ) 2 )
Y ( t ) = y ( t ) + x ′ ( t ) ( x ′ ( t ) 2 + y ′ ( t ) 2 ) x ′ ( t ) y ′ ′ ( t ) − x ′ ′ ( t ) y ′ ( t ) Y(t)=y(t)+\frac{x'(t)(x'(t)^2+y'(t)^2)}{x'(t)y''(t)-x''(t)y'(t)} Y ( t ) = y ( t ) + x ′ ( t ) y ′′ ( t ) − x ′′ ( t ) y ′ ( t ) x ′ ( t ) ( x ′ ( t ) 2 + y ′ ( t ) 2 )
X ( t ) = a c o s t − b c o s t ( b 2 c o s 2 t + a 2 s i n 2 t ) a b s i n 2 t + a b c o s 2 t = a c o s t − c o s t ( b 2 c o s 2 t + a 2 s i n 2 t ) a X(t)=acost-\frac{bcost(b^2cos^2t+a^2sin^2t)}{absin^2t+abcos^2t}=acost-\frac{cost(b^2cos^2t+a^2sin^2t)}{a} X ( t ) = a cos t − ab s i n 2 t + ab co s 2 t b cos t ( b 2 co s 2 t + a 2 s i n 2 t ) = a cos t − a cos t ( b 2 co s 2 t + a 2 s i n 2 t )
Y ( t ) = b s i n t − a s i n t ( b 2 c o s 2 t + a 2 s i n 2 t ) a b s i n 2 t + a b c o s 2 t = b s i n t − s i n t ( b 2 c o s 2 t + a 2 s i n 2 t ) b Y(t)=bsint-\frac{asint(b^2cos^2t+a^2sin^2t)}{absin^2t+abcos^2t}=bsint-\frac{sint(b^2cos^2t+a^2sin^2t)}{b} Y ( t ) = b s in t − ab s i n 2 t + ab co s 2 t a s in t ( b 2 co s 2 t + a 2 s i n 2 t ) = b s in t − b s in t ( b 2 co s 2 t + a 2 s i n 2 t )
equation of normals ar point (x1 ,y1 ):
( y − y 1 ) = − y ′ ( x 1 ) ( x − x 1 ) (y-y_1)=-y'(x_1)(x-x_1) ( y − y 1 ) = − y ′ ( x 1 ) ( x − x 1 )
a 2 x / x 1 − b 2 y / y 1 = a 2 − b 2 a^2x/x_1 - b^2y/y_1 = a^2 - b^2 a 2 x / x 1 − b 2 y / y 1 = a 2 − b 2
#339914 on Dec 2023
Comments