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Find-the-slope-of-the-line-tangent-to-the-graph-r-3cos-2-2-at-pi-6-

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Question Number 120463 by bramlexs22 last updated on 31/Oct/20
Find the slope of the line tangent to the graph r=3cos^2 (2θ) at θ=(π/6).
$${Find}\:{the}\:{slope}\:{of}\:{the}\:{line}\:{tangent} \\ $$$${to}\:{the}\:{graph}\:{r}=\mathrm{3cos}\:^{\mathrm{2}} \left(\mathrm{2}\theta\right)\:{at} \\ $$$$\theta=\frac{\pi}{\mathrm{6}}.\: \\ $$
Answered by mr W last updated on 31/Oct/20
(dr/dθ)=−12 cos (2θ) sin (2θ)=−6 sin (4θ) x=r cos θ (dx/dθ)=−r sin θ+(dr/dθ) cos θ =−3 cos^2 (2θ) sin θ−6 sin (4θ) cos θ y=r sin θ (dy/dθ)=r cos θ+(dr/dθ) sin θ =3 cos^2 (2θ) cos θ−6 sin (4θ) sin θ (dy/dx)=((dy/dθ)/(dx/dθ))=((3 cos^2 (2θ) cos θ−6 sin (4θ) sin θ)/(−3 cos^2 (2θ) sin θ−6 sin (4θ) cos θ)) =((cos^2 (2θ) cos θ−2 sin (4θ) sin θ)/(−cos^2 (2θ) sin θ−2 sin (4θ) cos θ)) =−((cos^2 (2θ)−2 sin (4θ) tan θ)/(cos^2 (2θ) tan θ+2 sin (4θ))) =−((cos^2 ((π/3))−2 sin (((2π)/3)) tan (π/6))/(cos^2 ((π/3)) tan (π/6)+2 sin (((2π)/3)))) =−(((1/4)−2×((√3)/2)×(1/( (√3))))/((1/(4(√3)))+2×((√3)/2))) =((3(√3))/(13))
$$\frac{{dr}}{{d}\theta}=−\mathrm{12}\:\mathrm{cos}\:\left(\mathrm{2}\theta\right)\:\mathrm{sin}\:\left(\mathrm{2}\theta\right)=−\mathrm{6}\:\mathrm{sin}\:\left(\mathrm{4}\theta\right) \\ $$$${x}={r}\:\mathrm{cos}\:\theta \\ $$$$\frac{{dx}}{{d}\theta}=−{r}\:\mathrm{sin}\:\theta+\frac{{dr}}{{d}\theta}\:\mathrm{cos}\:\theta \\ $$$$=−\mathrm{3}\:\mathrm{cos}^{\mathrm{2}} \:\left(\mathrm{2}\theta\right)\:\mathrm{sin}\:\theta−\mathrm{6}\:\mathrm{sin}\:\left(\mathrm{4}\theta\right)\:\mathrm{cos}\:\theta \\ $$$${y}={r}\:\mathrm{sin}\:\theta \\ $$$$\frac{{dy}}{{d}\theta}={r}\:\mathrm{cos}\:\theta+\frac{{dr}}{{d}\theta}\:\mathrm{sin}\:\theta \\ $$$$=\mathrm{3}\:\mathrm{cos}^{\mathrm{2}} \:\left(\mathrm{2}\theta\right)\:\mathrm{cos}\:\theta−\mathrm{6}\:\mathrm{sin}\:\left(\mathrm{4}\theta\right)\:\mathrm{sin}\:\theta \\ $$$$\frac{{dy}}{{dx}}=\frac{\frac{{dy}}{{d}\theta}}{\frac{{dx}}{{d}\theta}}=\frac{\mathrm{3}\:\mathrm{cos}^{\mathrm{2}} \:\left(\mathrm{2}\theta\right)\:\mathrm{cos}\:\theta−\mathrm{6}\:\mathrm{sin}\:\left(\mathrm{4}\theta\right)\:\mathrm{sin}\:\theta}{−\mathrm{3}\:\mathrm{cos}^{\mathrm{2}} \:\left(\mathrm{2}\theta\right)\:\mathrm{sin}\:\theta−\mathrm{6}\:\mathrm{sin}\:\left(\mathrm{4}\theta\right)\:\mathrm{cos}\:\theta} \\ $$$$=\frac{\mathrm{cos}^{\mathrm{2}} \:\left(\mathrm{2}\theta\right)\:\mathrm{cos}\:\theta−\mathrm{2}\:\mathrm{sin}\:\left(\mathrm{4}\theta\right)\:\mathrm{sin}\:\theta}{−\mathrm{cos}^{\mathrm{2}} \:\left(\mathrm{2}\theta\right)\:\mathrm{sin}\:\theta−\mathrm{2}\:\mathrm{sin}\:\left(\mathrm{4}\theta\right)\:\mathrm{cos}\:\theta} \\ $$$$=−\frac{\mathrm{cos}^{\mathrm{2}} \:\left(\mathrm{2}\theta\right)−\mathrm{2}\:\mathrm{sin}\:\left(\mathrm{4}\theta\right)\:\mathrm{tan}\:\theta}{\mathrm{cos}^{\mathrm{2}} \:\left(\mathrm{2}\theta\right)\:\mathrm{tan}\:\theta+\mathrm{2}\:\mathrm{sin}\:\left(\mathrm{4}\theta\right)} \\ $$$$=−\frac{\mathrm{cos}^{\mathrm{2}} \:\left(\frac{\pi}{\mathrm{3}}\right)−\mathrm{2}\:\mathrm{sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\:\mathrm{tan}\:\frac{\pi}{\mathrm{6}}}{\mathrm{cos}^{\mathrm{2}} \:\left(\frac{\pi}{\mathrm{3}}\right)\:\mathrm{tan}\:\frac{\pi}{\mathrm{6}}+\mathrm{2}\:\mathrm{sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)} \\ $$$$=−\frac{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{2}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}{\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{3}}}+\mathrm{2}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \\ $$$$=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{13}} \\ $$
Commented by mr W last updated on 31/Oct/20
Answered by john santu last updated on 01/Nov/20
The slope of the tangent to the graph of r = f(θ) at (r,θ) given by m = (dy/dx)= ((tan θ.(dr/dθ) +r)/((dr/dθ)−r tan θ)). Here f(θ)=3cos^2 θ →f ′(θ)=−6sin 4θ. now f(π/6)=3cos^2 (π/6)=(3/4) and f ′(π/6)=−3(√3). Thus the formula gives m = (dy/dx) = ((−3(√3) tan (π/6)+(3/4))/(−3(√3)−((3/4))tan (π/6))) = ((3(√3))/(13))
$${The}\:{slope}\:{of}\:{the}\:{tangent}\:{to}\:{the}\:{graph} \\ $$$${of}\:{r}\:=\:{f}\left(\theta\right)\:{at}\:\left({r},\theta\right)\:{given}\:{by}\: \\ $$$${m}\:=\:\frac{{dy}}{{dx}}=\:\frac{\mathrm{tan}\:\theta.\frac{{dr}}{{d}\theta}\:+{r}}{\frac{{dr}}{{d}\theta}−{r}\:\mathrm{tan}\:\theta}. \\ $$$${Here}\:{f}\left(\theta\right)=\mathrm{3cos}\:^{\mathrm{2}} \theta\:\rightarrow{f}\:'\left(\theta\right)=−\mathrm{6sin}\:\mathrm{4}\theta. \\ $$$${now}\:{f}\left(\pi/\mathrm{6}\right)=\mathrm{3cos}\:^{\mathrm{2}} \left(\pi/\mathrm{6}\right)=\frac{\mathrm{3}}{\mathrm{4}}\:{and}\: \\ $$$${f}\:'\left(\pi/\mathrm{6}\right)=−\mathrm{3}\sqrt{\mathrm{3}}.\:{Thus}\:{the}\:{formula} \\ $$$${gives}\:{m}\:=\:\frac{{dy}}{{dx}}\:=\:\frac{−\mathrm{3}\sqrt{\mathrm{3}}\:\mathrm{tan}\:\left(\pi/\mathrm{6}\right)+\frac{\mathrm{3}}{\mathrm{4}}}{−\mathrm{3}\sqrt{\mathrm{3}}−\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\mathrm{tan}\:\left(\pi/\mathrm{6}\right)} \\ $$$$\:=\:\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{13}} \\ $$

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