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Question Number 95001 by mathocean1 last updated on 22/May/20

Exercise Given a, b ∈ R^∗ and t is a variable real. 1) Solve in R^2 for x,y this system: { ((xsin t−ycos t=−a)),((xcos t+ysin t=b.)) :} 2)/Demonstrate that these solutions can be written like this ( r and θ ∈ R). { ((x=rcos(t+θ))),((y=rsin(t+θ))) :} 3) we suppose now that a=b=1 and θ=(π/(12)) solve this in [0;2π[ { ((rcos(t+θ)≥−1)),((rsin(t+θ)<−1)) :}

$$\mathrm{Exercise} \\ $$$$\mathrm{Given}\:\mathrm{a},\:\mathrm{b}\:\in\:\mathbb{R}^{\ast} \:\mathrm{and}\:{t}\:\mathrm{is}\:\mathrm{a}\:\mathrm{variable}\:\mathrm{real}. \\ $$$$\left.\mathrm{1}\right)\:\mathrm{Solve}\:\mathrm{in}\:\mathbb{R}^{\mathrm{2}} \:\mathrm{for}\:{x},{y}\:\mathrm{this}\:\mathrm{system}:\: \\ $$$$\begin{cases}{{x}\mathrm{sin}\:{t}−{y}\mathrm{cos}\:{t}=−\mathrm{a}}\\{{x}\mathrm{cos}\:{t}+{y}\mathrm{sin}\:{t}={b}.}\end{cases} \\ $$$$ \\ $$$$\left.\mathrm{2}\right)/\mathrm{Demonstrate}\:\mathrm{that}\:\mathrm{these}\:\mathrm{solutions}\:\mathrm{can} \\ $$$$\mathrm{be}\:\mathrm{written}\:\mathrm{like}\:\mathrm{this}\:\left(\:{r}\:\mathrm{and}\:\theta\:\in\:\mathbb{R}\right). \\ $$$$\begin{cases}{{x}={rcos}\left({t}+\theta\right)}\\{{y}={rsin}\left({t}+\theta\right)}\end{cases} \\ $$$$ \\ $$$$\left.\mathrm{3}\right)\:\mathrm{we}\:\mathrm{suppose}\:\mathrm{now}\:\mathrm{that}\:\mathrm{a}=\mathrm{b}=\mathrm{1}\:\mathrm{and}\:\theta=\frac{\pi}{\mathrm{12}} \\ $$$$\mathrm{solve}\:\mathrm{this}\:\mathrm{in}\:\left[\mathrm{0};\mathrm{2}\pi\left[\right.\right. \\ $$$$\begin{cases}{{rcos}\left({t}+\theta\right)\geqslant−\mathrm{1}}\\{{rsin}\left({t}+\theta\right)<−\mathrm{1}}\end{cases} \\ $$

Answered by bobhans last updated on 22/May/20

(1)⇒(i)x^2 sin^2 t−2xysin tcos t+y^2 cos^2 t = a^2 ⇒(ii)x^2 cos^2 t+2xycos tsin t+y^2 sin^2 t = b^2 (i)+(ii) ⇒ x^2 + y^2 −2xy(sin tcos t−cos tsin t) = a^2 +b^2 ⇒ x^2 + y^2 = a^2 + b^2

$$\left(\mathrm{1}\right)\Rightarrow\left(\mathrm{i}\right)\mathrm{x}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \mathrm{t}−\mathrm{2xysin}\:\mathrm{tcos}\:\mathrm{t}+\mathrm{y}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \mathrm{t}\:=\:\mathrm{a}^{\mathrm{2}} \\ $$$$\Rightarrow\left(\mathrm{ii}\right)\mathrm{x}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \mathrm{t}+\mathrm{2xycos}\:\mathrm{tsin}\:\mathrm{t}+\mathrm{y}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \mathrm{t}\:=\:\mathrm{b}^{\mathrm{2}} \\ $$$$\left(\mathrm{i}\right)+\left(\mathrm{ii}\right)\:\Rightarrow\:{x}^{\mathrm{2}} +\:{y}^{\mathrm{2}} −\mathrm{2}{xy}\left(\mathrm{sin}\:{t}\mathrm{cos}\:{t}−\mathrm{cos}\:{t}\mathrm{sin}\:{t}\right)\:=\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:=\:{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \: \\ $$

Answered by mathmax by abdo last updated on 22/May/20

1) { ((sint x−cost y =−a)),((cost x +sint y =b)) :} Δ_s = determinant (((sint −cost)),((cost sint)))=sin^2 t +cos^2 t?=1 ≠0 ⇒ x =(Δ_x /Δ) and y =((Δy)/Δ) we have Δ_x = determinant (((−a −cost)),((b sint)))=−asint +bcost Δ_y = determinant (((sint −a)),((cost b)))=bsint +acost ⇒ x=bcost −asint and y =acost +bsint 2) we have x =(√(a^2 +b^2 ))((b/(√(a^2 +b^2 ))) cost−(a/(√(a^2 +b^2 )))sint) let cosθ =(b/(√(a^2 +b^2 ))) and sinθ =(a/(√(a^2 +b^2 ))) and r=(√(a^2 +b^2 )) ⇒x =r(cost cosθ−sint sinθ) =r cos(t+θ) y =(√(a^2 +b^2 ))((a/(√(a^2 +b^2 ))) cost +(b/(√(a^2 +b^2 )))sint) =r(sinθ cost +cosθ sint) =r sin(t+θ)

$$\left.\mathrm{1}\right)\:\:\begin{cases}{\mathrm{sint}\:\mathrm{x}−\mathrm{cost}\:\mathrm{y}\:=−\mathrm{a}}\\{\mathrm{cost}\:\mathrm{x}\:+\mathrm{sint}\:\mathrm{y}\:=\mathrm{b}}\end{cases} \\ $$$$\Delta_{\mathrm{s}} =\begin{vmatrix}{\mathrm{sint}\:\:\:\:\:\:\:−\mathrm{cost}}\\{\mathrm{cost}\:\:\:\:\:\:\:\:\:\mathrm{sint}}\end{vmatrix}=\mathrm{sin}^{\mathrm{2}} \mathrm{t}\:+\mathrm{cos}^{\mathrm{2}} \mathrm{t}?=\mathrm{1}\:\neq\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{x}\:=\frac{\Delta_{\mathrm{x}} }{\Delta}\:\:\mathrm{and}\:\mathrm{y}\:=\frac{\Delta\mathrm{y}}{\Delta}\:\:\mathrm{we}\:\mathrm{have}\:\Delta_{\mathrm{x}} =\begin{vmatrix}{−\mathrm{a}\:\:\:\:\:\:\:−\mathrm{cost}}\\{\mathrm{b}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{sint}}\end{vmatrix}=−\mathrm{asint}\:\:+\mathrm{bcost} \\ $$$$\Delta_{\mathrm{y}} =\begin{vmatrix}{\mathrm{sint}\:\:\:\:\:\:\:−\mathrm{a}}\\{\mathrm{cost}\:\:\:\:\:\:\:\:\:\:\:\mathrm{b}}\end{vmatrix}=\mathrm{bsint}\:+\mathrm{acost}\:\Rightarrow \\ $$$$\mathrm{x}=\mathrm{bcost}\:−\mathrm{asint}\:\:\mathrm{and}\:\mathrm{y}\:=\mathrm{acost}\:+\mathrm{bsint} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{we}\:\mathrm{have}\:\mathrm{x}\:=\sqrt{\mathrm{a}^{\mathrm{2}} \:+\mathrm{b}^{\mathrm{2}} }\left(\frac{\mathrm{b}}{\sqrt{\mathrm{a}^{\mathrm{2}} \:+\mathrm{b}^{\mathrm{2}} }}\:\mathrm{cost}−\frac{\mathrm{a}}{\sqrt{\mathrm{a}^{\mathrm{2}} \:+\mathrm{b}^{\mathrm{2}} }}\mathrm{sint}\right) \\ $$$$\mathrm{let}\:\mathrm{cos}\theta\:=\frac{\mathrm{b}}{\sqrt{\mathrm{a}^{\mathrm{2}} \:+\mathrm{b}^{\mathrm{2}} }}\:\:\mathrm{and}\:\mathrm{sin}\theta\:=\frac{\mathrm{a}}{\sqrt{\mathrm{a}^{\mathrm{2}} \:+\mathrm{b}^{\mathrm{2}} }}\:\mathrm{and}\:\mathrm{r}=\sqrt{\mathrm{a}^{\mathrm{2}} \:+\mathrm{b}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{x}\:=\mathrm{r}\left(\mathrm{cost}\:\mathrm{cos}\theta−\mathrm{sint}\:\mathrm{sin}\theta\right)\:=\mathrm{r}\:\mathrm{cos}\left(\mathrm{t}+\theta\right) \\ $$$$\mathrm{y}\:=\sqrt{\mathrm{a}^{\mathrm{2}} \:+\mathrm{b}^{\mathrm{2}} }\left(\frac{\mathrm{a}}{\sqrt{\mathrm{a}^{\mathrm{2}} \:+\mathrm{b}^{\mathrm{2}} }}\:\mathrm{cost}\:+\frac{\mathrm{b}}{\sqrt{\mathrm{a}^{\mathrm{2}} \:+\mathrm{b}^{\mathrm{2}} }}\mathrm{sint}\right)\:=\mathrm{r}\left(\mathrm{sin}\theta\:\mathrm{cost}\:+\mathrm{cos}\theta\:\mathrm{sint}\right) \\ $$$$=\mathrm{r}\:\mathrm{sin}\left(\mathrm{t}+\theta\right) \\ $$$$ \\ $$

Commented by mathocean1 last updated on 08/Jun/20

thanks sir

$${thanks}\:{sir} \\ $$

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