Math, asked by kadapakareem1423, 3 months ago


find radius of curvatue of the curve x^2/3+y^2/3=a^2/3​

Answers

Answered by AdityaYeole
0

Answer:

\begin{gathered} x^{2/3} + y^{2/3} = a^{2/3} \\ \\ let\ x = a^3 cos^3\ \alpha,\ \ \ y = a^3\ sin^3\ \alpha \\ \\ Differentiate\ wrt\ \alpha\ \\ \\ \frac{dx}{d\alpha} = a^3\ 3cos^2\ \alpha (-sin\ \alpha) \\ \\ \frac{dy}{d\alpha} = a^3\ 3 sin^2\ \alpha (cos\ \alpha) \\ \\ \frac{dy}{dx} = - tan\ \alpha \\ \\ \frac{d^2y}{dx^2} = - sec^2\ \alpha\ *\ \frac{d\alpha}{dx} \\ \\ \ \ \ = \frac{-sec^2\ \alpha}{-a^3\ 3 cos^2\ \alpha * sin\ \alpha} \\ \\ \end{gathered}

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

dx

=a

3

3cos

2

α(−sin α)

dy

=a

3

3sin

2

α(cos α)

dx

dy

=−tan α

dx

2

d

2

y

=−sec

2

α ∗

dx

=

−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

Answered by trrisha2609
0

Answer:

2/3

+y

2/3

=a

2/3

let x=a

3

cos

3

α, y=a

3

sin

3

α

Differentiate wrt α

dx

=a

3

3cos

2

α(−sin α)

dy

=a

3

3sin

2

α(cos α)

dx

dy

=−tan α

dx

2

d

2

y

=−sec

2

α ∗

dx

=

−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

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