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Question Number 12185 by prakash jain last updated on 15/Apr/17

$$\mathrm{A}\mathrm{particle}\mathrm{is}\mathrm{moving}\mathrm{with}\mathrm{velocity}$$$$v=K(y\hat{i}+x\hat{j})$$$$provethat$$$$y^2=x^2+constant$$

Answered by ajfour last updated on 16/Apr/17

$$\frac{dx}{dt}\hat{i}+\frac{dy}{dt}\hat{j}=K(y\hat{i}+x\hat{j})$$$$Kdt=\frac{dx}{y}=\frac{dy}{x}$$$$\Rightarrow ydy=xdx$$$$y^2=x^2+C$$$$$$

Commented by prakash jain last updated on 16/Apr/17

$$\mathrm{Thanks}.$$

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