find the radius of curvature at any point on the cartesian curve x2/3+y2/3=a2/3
Answers
Step-by-step explanation:
2/3
+y
2/3
=a
2/3
let x=a
3
cos
3
α, y=a
3
sin
3
α
Differentiate wrt α
dα
dx
=a
3
3cos
2
α(−sin α)
dα
dy
=a
3
3sin
2
α(cos α)
dx
dy
=−tan α
dx
2
d
2
y
=−sec
2
α ∗
dx
dα
=
−a
3
3cos
2
α∗sin α
−sec
2
α
\begin{gathered}\frac{1}{3a^3\ cos^4\ \alpha\ * \ sin\ \alpha} \\ \\ NOW,\ \ \ R\ =\ Radius\ of\ curvature\ at\ (x,y)\ =\ \ \frac{(1+tan^2\ \alpha)^3/2}{\frac{1}{3a^3\ cos^4\ \alpha\ * \ sin\ \alpha}} \\ \\ \ \ \ \ =\ sec^3\ \alpha\ *\ 3a^3\ *\ cos^4\ \alpha\ *\ sin\ \alpha \\ \\ \ \ \ \ \ = 3a^3\ cos\ \alpha\ sin\ \alpha \\ \\ 3a^3\ (xy)^{\frac{1}{3}} / a^2\\ \\ 3a (xy)^{\frac{1}{3}} \\ \\ \end{gathered}
3a
3
cos
4
α ∗ sin α
1
NOW, R = Radius of curvature at (x,y) =
3a
3
cos
4
α ∗ sin α
1
(1+tan
2
α)
3
/2
= sec
3
α ∗ 3a
3
∗ cos
4
α ∗ sin α
=3a
3
cos α sin α
3a
3
(xy)
3
1
/a
2
3a(xy)
3
1
Step-by-step explanation:
x
2/3
+y
2/3
=a
2/3
let x=a
3
cos
3
α, y=a
3
sin
3
α
Differentiate wrt α
dα
dx
=a
3
3cos
2
α(−sin α)
dα
dy
=a
3
3sin
2
α(cos α)
dx
dy
=−tan α
dx
2
d
2
y
=−sec
2
α ∗
dx
dα
=
−a
3
3cos
2
α∗sin α
−sec
2
α
\begin{gathered}\frac{1}{3a^3\ cos^4\ \alpha\ * \ sin\ \alpha} \\ \\ NOW,\ \ \ R\ =\ Radius\ of\ curvature\ at\ (x,y)\ =\ \ \frac{(1+tan^2\ \alpha)^3/2}{\frac{1}{3a^3\ cos^4\ \alpha\ * \ sin\ \alpha}} \\ \\ \ \ \ \ =\ sec^3\ \alpha\ *\ 3a^3\ *\ cos^4\ \alpha\ *\ sin\ \alpha \\ \\ \ \ \ \ \ = 3a^3\ cos\ \alpha\ sin\ \alpha \\ \\ 3a^3\ (xy)^{\frac{1}{3}} / a^2\\ \\ 3a (xy)^{\frac{1}{3}} \\ \\ \end{gathered}
3a
3
cos
4
α ∗ sin α
1
NOW, R = Radius of curvature at (x,y) =
3a
3
cos
4
α ∗ sin α
1
(1+tan
2
α)
3
/2
= sec
3
α ∗ 3a
3
∗ cos
4
α ∗ sin α
=3a
3
cos α sin α
3a
3
(xy)
3
1
/a
2
3a(xy)
3
1