Question

The orthogonal trajectories of the family of curves x2/3and y2/3

Answer 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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Answer 2

Step-by-step explanation:

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