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Question Number 59753 by Aditya789 last updated on 14/May/19
x=(a+b)cosx−bcos(((a+b)/b))x y=(a+b)sinx−bsin(((a+b)/b))x find (dy/dx)=tan((a/(2b))+1)x
$$\mathrm{x}=\left(\mathrm{a}+\mathrm{b}\right)\mathrm{cosx}−\mathrm{bcos}\left(\frac{\mathrm{a}+\mathrm{b}}{\mathrm{b}}\right)\mathrm{x} \\ $$$$\mathrm{y}=\left(\mathrm{a}+\mathrm{b}\right)\mathrm{sinx}−\mathrm{bsin}\left(\frac{\mathrm{a}+\mathrm{b}}{\mathrm{b}}\right)\mathrm{x} \\ $$$$\mathrm{find}\:\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{tan}\left(\frac{\mathrm{a}}{\mathrm{2b}}+\mathrm{1}\right)\mathrm{x} \\ $$
Answered by tanmay last updated on 14/May/19
x+y=(a+b)(sinx+cosx)−b[sin(((a+b)/b))x+cos(((a+b)/b))x] 1+(dy/dx)=(a+b)(cosx−sinx)−b[cos(((a+b)/b))x−sin(((a+b)/b))x]×((a+b)/b) 1+(dy/dx)=(a+b)(cosx−sinx)−(a+b)[cos(((a+b)/b))x−sin(((a+b)/b))x] 1+(dy/dx)=(a+b)[cosx−cos(((a+b)/b))x]+(a+b)[sin(((a+b)/b))x−sinx] 1+(dy/dx)=(a+b)[2sin{((x+(((a+b)/b))x)/2)}sin{(((((a+b)/b))x−x)/2)}]+(a+b)[2cos{(((((a+b)/b))x+x)/2)}sin{(((((a+b)/b))x−x)/2)}] 1+(dy/dx)=(a+b)[2sin(((ax+2bx)/(2b)))sin(((ax)/(2b)))]+(a+b)[2cos(((ax+2bx)/(2b)))sin(((ax)/(2b)))] (dy/dx)+1=2(a+b)sin(((ax)/(2b)))[sin(((ax+2bx)/(2b)))+cos(((ax+2bx)/(2b)))] 1+(dy/dx)=2(a+b)sin(((ax)/(2b)))[sin(((ax)/(2b))+x)+cos(((ax)/(2b))+x)] 1+(dy/dx)=((2(a+b)sin(((ax)/(2b)))cos(((ax)/(2b))+x)[1+tan(((ax)/(2b))+x)])/) pls check the question...
$${x}+{y}=\left({a}+{b}\right)\left({sinx}+{cosx}\right)−{b}\left[{sin}\left(\frac{{a}+{b}}{{b}}\right){x}+{cos}\left(\frac{{a}+{b}}{{b}}\right){x}\right] \\ $$$$\mathrm{1}+\frac{{dy}}{{dx}}=\left({a}+{b}\right)\left({cosx}−{sinx}\right)−{b}\left[{cos}\left(\frac{{a}+{b}}{{b}}\right){x}−{sin}\left(\frac{{a}+{b}}{{b}}\right){x}\right]×\frac{{a}+{b}}{{b}} \\ $$$$\mathrm{1}+\frac{{dy}}{{dx}}=\left({a}+{b}\right)\left({cosx}−{sinx}\right)−\left({a}+{b}\right)\left[{cos}\left(\frac{{a}+{b}}{{b}}\right){x}−{sin}\left(\frac{{a}+{b}}{{b}}\right){x}\right] \\ $$$$\mathrm{1}+\frac{{dy}}{{dx}}=\left({a}+{b}\right)\left[{cosx}−{cos}\left(\frac{{a}+{b}}{{b}}\right){x}\right]+\left({a}+{b}\right)\left[{sin}\left(\frac{{a}+{b}}{{b}}\right){x}−{sinx}\right] \\ $$$$\mathrm{1}+\frac{{dy}}{{dx}}=\left({a}+{b}\right)\left[\mathrm{2}{sin}\left\{\frac{{x}+\left(\frac{{a}+{b}}{{b}}\right){x}}{\mathrm{2}}\right\}{sin}\left\{\frac{\left(\frac{{a}+{b}}{{b}}\right){x}−{x}}{\mathrm{2}}\right\}\right]+\left({a}+{b}\right)\left[\mathrm{2}{cos}\left\{\frac{\left(\frac{{a}+{b}}{{b}}\right){x}+{x}}{\mathrm{2}}\right\}{sin}\left\{\frac{\left(\frac{{a}+{b}}{{b}}\right){x}−{x}}{\mathrm{2}}\right\}\right] \\ $$$$\mathrm{1}+\frac{{dy}}{{dx}}=\left({a}+{b}\right)\left[\mathrm{2}{sin}\left(\frac{{ax}+\mathrm{2}{bx}}{\mathrm{2}{b}}\right){sin}\left(\frac{{ax}}{\mathrm{2}{b}}\right)\right]+\left({a}+{b}\right)\left[\mathrm{2}{cos}\left(\frac{{ax}+\mathrm{2}{bx}}{\mathrm{2}{b}}\right){sin}\left(\frac{{ax}}{\mathrm{2}{b}}\right)\right] \\ $$$$\frac{{dy}}{{dx}}+\mathrm{1}=\mathrm{2}\left({a}+{b}\right){sin}\left(\frac{{ax}}{\mathrm{2}{b}}\right)\left[{sin}\left(\frac{{ax}+\mathrm{2}{bx}}{\mathrm{2}{b}}\right)+{cos}\left(\frac{{ax}+\mathrm{2}{bx}}{\mathrm{2}{b}}\right)\right] \\ $$$$\mathrm{1}+\frac{{dy}}{{dx}}=\mathrm{2}\left({a}+{b}\right){sin}\left(\frac{{ax}}{\mathrm{2}{b}}\right)\left[{sin}\left(\frac{{ax}}{\mathrm{2}{b}}+{x}\right)+{cos}\left(\frac{{ax}}{\mathrm{2}{b}}+{x}\right)\right] \\ $$$$\mathrm{1}+\frac{{dy}}{{dx}}=\frac{\mathrm{2}\left({a}+{b}\right){sin}\left(\frac{{ax}}{\mathrm{2}{b}}\right){cos}\left(\frac{{ax}}{\mathrm{2}{b}}+{x}\right)\left[\mathrm{1}+{tan}\left(\frac{{ax}}{\mathrm{2}{b}}+{x}\right)\right]}{} \\ $$$${pls}\:{check}\:{the}\:{question}… \\ $$
Answered by $@ty@m last updated on 14/May/19
In my opinion, the question should be:z x=(a+b)cosα−bcos(((a+b)/b))α ...(1) y=(a+b)sinα−bsin(((a+b)/b))α ....(2) find (dy/dx)=tan((a/(2b))+1)α Solution: Differentiating (1) w.r.t. α (dx/dα)=−(a+b)sin α+b(((a+b)/b))sin (((a+b)/b))α =(a+b){sin (((a+b)/b))α−sin α} =(a+b){2cos (((a+2b)/b))α.sin (a/b)α} ...(3) Differentiating (2) w.r.t. α (dy/dα)=(a+b)cos α−b(((a+b)/b))cos (((a+b)/b))α =(a+b){cos α−cos (((a+b)/b))α} =(a+b).2sin (((a+2b)/b))αsin (a/b)α...(4) ∴ (dy/dx)=((dy/dα)/(dx/dα)) =tan ((a+2b)/b)α
$${In}\:{my}\:{opinion},\:{the}\:{question}\:{should}\:{be}:{z} \\ $$$$\mathrm{x}=\left(\mathrm{a}+\mathrm{b}\right)\mathrm{cos}\alpha−\mathrm{bcos}\left(\frac{\mathrm{a}+\mathrm{b}}{\mathrm{b}}\right)\alpha\:…\left(\mathrm{1}\right) \\ $$$$\mathrm{y}=\left(\mathrm{a}+\mathrm{b}\right)\mathrm{sin}\alpha−\mathrm{bsin}\left(\frac{\mathrm{a}+\mathrm{b}}{\mathrm{b}}\right)\alpha\:….\left(\mathrm{2}\right) \\ $$$$\mathrm{find}\:\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{tan}\left(\frac{\mathrm{a}}{\mathrm{2b}}+\mathrm{1}\right)\alpha \\ $$$${Solution}: \\ $$$${Differentiating}\:\left(\mathrm{1}\right)\:{w}.{r}.{t}.\:\alpha \\ $$$$\frac{{dx}}{{d}\alpha}=−\left({a}+{b}\right)\mathrm{sin}\:\alpha+{b}\left(\frac{{a}+{b}}{{b}}\right)\mathrm{sin}\:\left(\frac{{a}+{b}}{{b}}\right)\alpha \\ $$$$=\left({a}+{b}\right)\left\{\mathrm{sin}\:\left(\frac{{a}+{b}}{{b}}\right)\alpha−\mathrm{sin}\:\alpha\right\} \\ $$$$=\left({a}+{b}\right)\left\{\mathrm{2cos}\:\left(\frac{{a}+\mathrm{2}{b}}{{b}}\right)\alpha.\mathrm{sin}\:\frac{{a}}{{b}}\alpha\right\}\:…\left(\mathrm{3}\right) \\ $$$${Differentiating}\:\left(\mathrm{2}\right)\:{w}.{r}.{t}.\:\alpha \\ $$$$\frac{{dy}}{{d}\alpha}=\left({a}+{b}\right)\mathrm{cos}\:\:\alpha−{b}\left(\frac{{a}+{b}}{{b}}\right)\mathrm{cos}\:\:\left(\frac{{a}+{b}}{{b}}\right)\alpha \\ $$$$=\left({a}+{b}\right)\left\{\mathrm{cos}\:\:\alpha−\mathrm{cos}\:\:\left(\frac{{a}+{b}}{{b}}\right)\alpha\right\} \\ $$$$=\left({a}+{b}\right).\mathrm{2sin}\:\left(\frac{{a}+\mathrm{2}{b}}{{b}}\right)\alpha\mathrm{sin}\:\frac{{a}}{{b}}\alpha…\left(\mathrm{4}\right) \\ $$$$\therefore\:\frac{{dy}}{{dx}}=\frac{\frac{{dy}}{{d}\alpha}}{\frac{{dx}}{{d}\alpha}} \\ $$$$=\mathrm{tan}\:\frac{{a}+\mathrm{2}{b}}{{b}}\alpha \\ $$

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