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If-x-asin2t-bcos2t-prove-that-dx-dt-2-a-2-b-2-x-2-Solution-

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Question Number 141785 by 7770 last updated on 23/May/21
If x=asin2t+bcos2t,prove that (dx/dt)=2(√(a^2 +b^2 −x^2 )) Solution....
$$\boldsymbol{\mathrm{If}}\:\:\boldsymbol{\mathrm{x}}=\boldsymbol{\mathrm{asin}}\mathrm{2}\boldsymbol{\mathrm{t}}+\boldsymbol{\mathrm{bcos}}\mathrm{2}\boldsymbol{\mathrm{t}},\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{that}}\: \\ $$$$\frac{\boldsymbol{\mathrm{dx}}}{\boldsymbol{\mathrm{dt}}}=\mathrm{2}\sqrt{\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{b}}^{\mathrm{2}} −\boldsymbol{\mathrm{x}}^{\mathrm{2}} } \\ $$$$\boldsymbol{\mathrm{Solution}}…. \\ $$
Answered by iloveisrael last updated on 23/May/21
(dx/dt) = 2a cos 2t−2b sin 2t (dx/dt) = 2 (√((acos 2t−bsin 2t)^2 )) (dx/dt) =2(√(a^2 cos^2 2t+b^2 sin^2 2t−ab sin 4t)) (dx/dt) =2(√(a^2 (1−sin^2 2t)+b^2 (1−cos^2 2t)−ab sin 4t)) (dx/dt) =2(√(a^2 +b^2 −(a^2 sin^2 2t+b^2 cos^2 2t+ab sin 4t)) (dx/dt) = 2(√(a^2 +b^2 −x^2 )) (1)x^2 = a^2 sin^2 2t + b^2 cos^2 2t +ab sin 4t
$$\:\frac{{dx}}{{dt}}\:=\:\mathrm{2}{a}\:\mathrm{cos}\:\mathrm{2}{t}−\mathrm{2}{b}\:\mathrm{sin}\:\mathrm{2}{t} \\ $$$$\:\frac{{dx}}{{dt}}\:=\:\mathrm{2}\:\sqrt{\left({a}\mathrm{cos}\:\mathrm{2}{t}−{b}\mathrm{sin}\:\mathrm{2}{t}\right)^{\mathrm{2}} } \\ $$$$\:\frac{{dx}}{{dt}}\:=\mathrm{2}\sqrt{{a}^{\mathrm{2}} \:\mathrm{cos}\:^{\mathrm{2}} \mathrm{2}{t}+{b}^{\mathrm{2}} \:\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}{t}−{ab}\:\mathrm{sin}\:\mathrm{4}{t}} \\ $$$$\:\frac{{dx}}{{dt}}\:=\mathrm{2}\sqrt{{a}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}{t}\right)+{b}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{2}{t}\right)−{ab}\:\mathrm{sin}\:\mathrm{4}{t}} \\ $$$$\:\frac{{dx}}{{dt}}\:=\mathrm{2}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\left({a}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \mathrm{2}{t}+{b}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \mathrm{2}{t}+{ab}\:\mathrm{sin}\:\mathrm{4}{t}\right.} \\ $$$$\:\frac{{dx}}{{dt}}\:=\:\mathrm{2}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{x}^{\mathrm{2}} } \\ $$$$\left(\mathrm{1}\right){x}^{\mathrm{2}} \:=\:{a}^{\mathrm{2}} \:\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}{t}\:+\:{b}^{\mathrm{2}} \:\mathrm{cos}\:^{\mathrm{2}} \mathrm{2}{t}\:+{ab}\:\mathrm{sin}\:\mathrm{4}{t} \\ $$$$ \\ $$

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