Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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

$$\:\frac{{dx}}{{dt}}\:=\:\mathrm{2}{a}\:\mathrm{cos}\:\mathrm{2}{t}−\mathrm{2}{b}\:\mathrm{sin}\:\mathrm{2}{t} \\ $$$$\:\frac{{dx}}{{dt}}\:=\:\mathrm{2}\:\sqrt{\left({a}\mathrm{cos}\:\mathrm{2}{t}−{b}\mathrm{sin}\:\mathrm{2}{t}\right)^{\mathrm{2}} } \\ $$$$\:\frac{{dx}}{{dt}}\:=\mathrm{2}\sqrt{{a}^{\mathrm{2}} \:\mathrm{cos}\:^{\mathrm{2}} \mathrm{2}{t}+{b}^{\mathrm{2}} \:\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}{t}−{ab}\:\mathrm{sin}\:\mathrm{4}{t}} \\ $$$$\:\frac{{dx}}{{dt}}\:=\mathrm{2}\sqrt{{a}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}{t}\right)+{b}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{2}{t}\right)−{ab}\:\mathrm{sin}\:\mathrm{4}{t}} \\ $$$$\:\frac{{dx}}{{dt}}\:=\mathrm{2}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\left({a}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \mathrm{2}{t}+{b}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \mathrm{2}{t}+{ab}\:\mathrm{sin}\:\mathrm{4}{t}\right.} \\ $$$$\:\frac{{dx}}{{dt}}\:=\:\mathrm{2}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{x}^{\mathrm{2}} } \\ $$$$\left(\mathrm{1}\right){x}^{\mathrm{2}} \:=\:{a}^{\mathrm{2}} \:\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}{t}\:+\:{b}^{\mathrm{2}} \:\mathrm{cos}\:^{\mathrm{2}} \mathrm{2}{t}\:+{ab}\:\mathrm{sin}\:\mathrm{4}{t} \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com