English, asked by vaibhavkrkandhway, 2 months ago

orthogonal trajectories of the family of curves x⅔+y⅔=a⅔ is
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Answers

Answered by pulakmath007
1

SOLUTION

TO DETERMINE

The orthogonal trajectories of the family of curves

\displaystyle\sf{ {x}^{ \frac{2}{3} } +  {y}^{ \frac{2}{3} } =  {a}^{ \frac{2}{ 3} }   }

EVALUATION

Here the given equation of family of curves is

\displaystyle\sf{ {x}^{ \frac{2}{3} } +  {y}^{ \frac{2}{3} } =  {a}^{ \frac{2}{ 3} }   }

Differentiating both sides with respect to x we get

\displaystyle\sf{ \frac{2}{3}  {x}^{ \frac{2}{3} - 1 } +  {y}^{ \frac{2}{3} - 1 } \frac{dy}{dx}  =  0 }

\displaystyle\sf{ \implies \:    {x}^{ -  \frac{1}{3}  } +  {y}^{  - \frac{1}{3} } \frac{dy}{dx}  =  0 }

\displaystyle\sf{ \implies \:    \frac{dy}{dx}  =   -  \frac{ {x}^{ -  \frac{1}{3}  } }{ {y}^{  - \frac{1}{3} }}  }

\displaystyle\sf{ \implies \:    \frac{dy}{dx}  =   -  \frac{ {y}^{   \frac{1}{3}  } }{ {x}^{  \frac{1}{3} }}  }

For orthogonal trajectories we replace

\displaystyle\sf{  \frac{dy}{dx}   \:  by \:  -  \frac{dx}{dy}  }

Thus we get

\displaystyle\sf{    -  \frac{dx}{dy}  =   -  \frac{ {y}^{   \frac{1}{3}  } }{ {x}^{  \frac{1}{3} }}  }

\displaystyle\sf{ \implies \:     \frac{dx}{dy}  =    \frac{ {y}^{   \frac{1}{3}  } }{ {x}^{  \frac{1}{3} }}  }

\displaystyle\sf{ \implies \:    {x}^{  \frac{1}{3} } \: dx \:   =     {y}^{   \frac{1}{3}  }dy}

On integration we get

\displaystyle\sf{ \int  {x}^{  \frac{1}{3} } \: dx \:   =  \int    {y}^{   \frac{1}{3}  }dy}

\displaystyle\sf{ \implies  \:  \frac{{x}^{  \frac{1}{3} + 1 }}{ \frac{1}{3} + 1 }     =   \frac{  {y}^{   \frac{1}{3}  + 1 }}{ \frac{1}{3}  + 1}  + c}

Where C is integration constant

\displaystyle\sf{ \implies  \:  \frac{{x}^{  \frac{4}{3} }}{ \frac{4}{3} }     =   \frac{  {y}^{   \frac{4}{3}  }}{ \frac{4}{3} }  + c}

\displaystyle\sf{ \implies  \:  {x}^{  \frac{4}{3} }     =     {y}^{  \frac{4}{3}   }  +  \frac{4c}{3} }

\displaystyle\sf{ \implies  \:  {x}^{  \frac{4}{3} }     =     {y}^{  \frac{4}{3}   }  + k \:  \:  \: where \:  \: k =  \frac{4c}{3}  }

\displaystyle\sf{ \implies  \:  {x}^{  \frac{4}{3} }    -      {y}^{  \frac{4}{3}   }   =  k \:  \:  }

FINAL ANSWER

Hence the required orthogonal trajectories of the given family of curves is

 \boxed{  \: \displaystyle\sf{ {x}^{  \frac{4}{3} }    -      {y}^{  \frac{4}{3}   }   =  k \:  \:  } }

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Answered by barani79530
0

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