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Question-19021

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Question Number 19021 by ajfour last updated on 03/Aug/17
Commented by ajfour last updated on 03/Aug/17
If φ=tan^(−1) [((√2)−1)cot θ]+θ ⇒ tan (φ−θ)=((√2)−1)cot θ ((tan φ−tan θ)/(1+tan φtan θ))=((cot θ)/( (√2)+1)) ⇒ ((√2)+1) (tan φ−tan θ) =cot θ+tan φ (√2)tan φ=(√2)tan θ+(tan θ+cot θ) tan φ=tan θ+(1/( (√2)))(tan θ+cot θ) . (same as the answer i got ) .
$$\mathrm{If}\:\:\:\:\:\:\phi=\mathrm{tan}^{−\mathrm{1}} \left[\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\mathrm{cot}\:\theta\right]+\theta \\ $$$$\Rightarrow\:\mathrm{tan}\:\left(\phi−\theta\right)=\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\mathrm{cot}\:\theta \\ $$$$\:\:\:\frac{\mathrm{tan}\:\phi−\mathrm{tan}\:\theta}{\mathrm{1}+\mathrm{tan}\:\phi\mathrm{tan}\:\theta}=\frac{\mathrm{cot}\:\theta}{\:\sqrt{\mathrm{2}}+\mathrm{1}} \\ $$$$\Rightarrow\:\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)\:\left(\mathrm{tan}\:\phi−\mathrm{tan}\:\theta\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{cot}\:\theta+\mathrm{tan}\:\phi \\ $$$$\:\sqrt{\mathrm{2}}\mathrm{tan}\:\phi=\sqrt{\mathrm{2}}\mathrm{tan}\:\theta+\left(\mathrm{tan}\:\theta+\mathrm{cot}\:\theta\right) \\ $$$$\:\mathrm{tan}\:\phi=\mathrm{tan}\:\theta+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\mathrm{tan}\:\theta+\mathrm{cot}\:\theta\right)\:. \\ $$$$\:\:\left(\mathrm{same}\:\mathrm{as}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{i}\:\mathrm{got}\:\right)\:. \\ $$
Commented by Tinkutara last updated on 03/Aug/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$
Commented by Tinkutara last updated on 03/Aug/17
Yes Sir, indeed!
$$\mathrm{Yes}\:\mathrm{Sir},\:\mathrm{indeed}! \\ $$
Commented by ajfour last updated on 03/Aug/17
Q.18271 i am getting a similar answer but not the same. Let tan φ=p , tan θ=m eqn. of trajectory: C y=px−((gx^2 )/(2u^2 ))(1+p^2 ) equation of left incline: L_1 y=m(x+a) equation of right incline: L_2 y=m(x−a) Point P lies on L_1 and C and is a double root, so px_P −((gx_P ^2 )/(2u^2 ))(1+p^2 )=m(x_P +a) ⇒ ((gx_P ^2 (1+p^2 ))/(2u^2 ))+(m−p)x_P +am=0 we have a double root at P , so (m−p)^2 =4((g/(2u^2 )))(1+p^2 )(am) or (u^2 /(g(1+p^2 )))=((2am)/((m−p)^2 )) .....(i) At point Q, slope (dy/dx)∣_Q =((−1)/m) as projectile hits plane at a right angle, and from eqn. of parabola (dy/dx)∣_Q =p−((gx_Q )/u^2 )(1+p^2 ) = ((−1)/m) ...(ii) from (ii): ⇒ x_Q =(u^2 /(g(1+p^2 )))(p+(1/m)) ....(a) and ((gx_Q ^2 )/(2u^2 ))(1+p^2 )=(x_Q /2)(p+(1/m)) ...(b) also m(x_Q −a)=px_Q −((gx_Q ^2 )/(2u^2 ))(1+p^2 ) using (b), this becomes m(x_Q −a)=px_Q −(x_Q /2)(p+(1/m)) or x_Q (m−p+(p/2)+(1/(2m)))=am x_Q =((2am^2 )/(2m^2 −mp+1)) comparing with (a) we get x_Q =((2am^2 )/(2m^2 −mp+1))=(u^2 /(g(1+p^2 )))(p+(1/m)) using (i) in R.H.S. we get ((2am^2 )/(2m^2 −mp+1))=((2am)/((m−p)^2 ))(((1+mp)/m)) ⇒m^2 (m−p)^2 =(2m^2 −mp+1)(1+mp) m^4 +m^2 p^2 −2pm^3 =2m^2 −mp+1 +2pm^3 −m^2 p^2 +mp 2m^2 p^2 −4pm^3 +m^4 −2m^2 −1=0 ⇒p=((4m^3 +(√(16m^6 −8m^2 (m^4 −2m^2 −1))))/(4m^2 )) p=m+(√((8m^6 +16m^4 +8m^2 )/(16m^4 ))) p=m+(√((m^2 /2)+1+(1/(2m^2 )))) p=m+(1/( (√2)))(m+(1/m)) ⇒ tan φ=tan θ+(1/( (√2)))(tan θ+cot θ) . ....
$$\mathrm{Q}.\mathrm{18271}\: \\ $$$$\mathrm{i}\:\mathrm{am}\:\mathrm{getting}\:\mathrm{a}\:\mathrm{similar}\:\mathrm{answer}\:\mathrm{but} \\ $$$$\mathrm{not}\:\mathrm{the}\:\mathrm{same}.\: \\ $$$$ \\ $$$$\mathrm{Let}\:\:\mathrm{tan}\:\phi=\mathrm{p}\:,\:\mathrm{tan}\:\theta=\mathrm{m} \\ $$$$\mathrm{eqn}.\:\mathrm{of}\:\mathrm{trajectory}:\:\mathrm{C} \\ $$$$\:\:\mathrm{y}=\mathrm{px}−\frac{\mathrm{gx}^{\mathrm{2}} }{\mathrm{2u}^{\mathrm{2}} }\left(\mathrm{1}+\mathrm{p}^{\mathrm{2}} \right) \\ $$$$\mathrm{equation}\:\mathrm{of}\:\mathrm{left}\:\mathrm{incline}:\:\mathrm{L}_{\mathrm{1}} \\ $$$$\:\:\:\:\:\:\mathrm{y}=\mathrm{m}\left(\mathrm{x}+\mathrm{a}\right) \\ $$$$\mathrm{equation}\:\mathrm{of}\:\mathrm{right}\:\mathrm{incline}:\:\mathrm{L}_{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\mathrm{y}=\mathrm{m}\left(\mathrm{x}−\mathrm{a}\right) \\ $$$$\mathrm{Point}\:\mathrm{P}\:\mathrm{lies}\:\mathrm{on}\:\mathrm{L}_{\mathrm{1}} \mathrm{and}\:\mathrm{C}\:\mathrm{and}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{double}\:\mathrm{root},\:\mathrm{so} \\ $$$$\mathrm{px}_{\mathrm{P}} −\frac{\mathrm{gx}_{\mathrm{P}} ^{\mathrm{2}} }{\mathrm{2u}^{\mathrm{2}} }\left(\mathrm{1}+\mathrm{p}^{\mathrm{2}} \right)=\mathrm{m}\left(\mathrm{x}_{\mathrm{P}} +\mathrm{a}\right) \\ $$$$\Rightarrow\:\frac{\mathrm{gx}_{\mathrm{P}} ^{\mathrm{2}} \left(\mathrm{1}+\mathrm{p}^{\mathrm{2}} \right)}{\mathrm{2u}^{\mathrm{2}} }+\left(\mathrm{m}−\mathrm{p}\right)\mathrm{x}_{\mathrm{P}} +\mathrm{am}=\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{a}\:\mathrm{double}\:\mathrm{root}\:\mathrm{at}\:\mathrm{P}\:,\:\mathrm{so} \\ $$$$\:\:\left(\mathrm{m}−\mathrm{p}\right)^{\mathrm{2}} =\mathrm{4}\left(\frac{\mathrm{g}}{\mathrm{2u}^{\mathrm{2}} }\right)\left(\mathrm{1}+\mathrm{p}^{\mathrm{2}} \right)\left(\mathrm{am}\right) \\ $$$$\:\:\mathrm{or}\:\:\:\frac{\mathrm{u}^{\mathrm{2}} }{\mathrm{g}\left(\mathrm{1}+\mathrm{p}^{\mathrm{2}} \right)}=\frac{\mathrm{2am}}{\left(\mathrm{m}−\mathrm{p}\right)^{\mathrm{2}} }\:\:\:\:…..\left(\boldsymbol{\mathrm{i}}\right) \\ $$$$\mathrm{At}\:\mathrm{point}\:\mathrm{Q},\:\mathrm{slope}\:\frac{\mathrm{dy}}{\mathrm{dx}}\mid_{\mathrm{Q}} =\frac{−\mathrm{1}}{\mathrm{m}} \\ $$$$\mathrm{as}\:\mathrm{projectile}\:\mathrm{hits}\:\mathrm{plane}\:\mathrm{at}\:\mathrm{a}\:\mathrm{right}\: \\ $$$$\mathrm{angle},\:\mathrm{and}\:\mathrm{from}\:\mathrm{eqn}.\:\mathrm{of}\:\mathrm{parabola} \\ $$$$\:\:\frac{\mathrm{dy}}{\mathrm{dx}}\mid_{\mathrm{Q}} =\mathrm{p}−\frac{\mathrm{gx}_{\mathrm{Q}} }{\mathrm{u}^{\mathrm{2}} }\left(\mathrm{1}+\mathrm{p}^{\mathrm{2}} \right)\:=\:\frac{−\mathrm{1}}{\mathrm{m}}\:\:…\left(\mathrm{ii}\right) \\ $$$$\mathrm{from}\:\left(\mathrm{ii}\right): \\ $$$$\Rightarrow\:\mathrm{x}_{\mathrm{Q}} =\frac{\mathrm{u}^{\mathrm{2}} }{\mathrm{g}\left(\mathrm{1}+\mathrm{p}^{\mathrm{2}} \right)}\left(\mathrm{p}+\frac{\mathrm{1}}{\mathrm{m}}\right)\:\:\:\:\:….\left(\mathrm{a}\right) \\ $$$$\mathrm{and}\:\:\frac{\mathrm{gx}_{\mathrm{Q}} ^{\mathrm{2}} }{\mathrm{2u}^{\mathrm{2}} }\left(\mathrm{1}+\mathrm{p}^{\mathrm{2}} \right)=\frac{\mathrm{x}_{\mathrm{Q}} }{\mathrm{2}}\left(\mathrm{p}+\frac{\mathrm{1}}{\mathrm{m}}\right)\:\:…\left(\mathrm{b}\right) \\ $$$$\mathrm{also}\:\mathrm{m}\left(\mathrm{x}_{\mathrm{Q}} −\mathrm{a}\right)=\mathrm{px}_{\mathrm{Q}} −\frac{\mathrm{gx}_{\mathrm{Q}} ^{\mathrm{2}} }{\mathrm{2u}^{\mathrm{2}} }\left(\mathrm{1}+\mathrm{p}^{\mathrm{2}} \right) \\ $$$$\mathrm{using}\:\left(\mathrm{b}\right),\:\mathrm{this}\:\mathrm{becomes} \\ $$$$\mathrm{m}\left(\mathrm{x}_{\mathrm{Q}} −\mathrm{a}\right)=\mathrm{px}_{\mathrm{Q}} −\frac{\mathrm{x}_{\mathrm{Q}} }{\mathrm{2}}\left(\mathrm{p}+\frac{\mathrm{1}}{\mathrm{m}}\right) \\ $$$$\mathrm{or}\:\:\mathrm{x}_{\mathrm{Q}} \left(\mathrm{m}−\mathrm{p}+\frac{\mathrm{p}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2m}}\right)=\mathrm{am} \\ $$$$\:\:\:\:\:\:\mathrm{x}_{\mathrm{Q}} =\frac{\mathrm{2am}^{\mathrm{2}} }{\mathrm{2m}^{\mathrm{2}} −\mathrm{mp}+\mathrm{1}} \\ $$$$\mathrm{comparing}\:\mathrm{with}\:\left(\mathrm{a}\right)\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{x}_{\mathrm{Q}} =\frac{\mathrm{2am}^{\mathrm{2}} }{\mathrm{2m}^{\mathrm{2}} −\mathrm{mp}+\mathrm{1}}=\frac{\mathrm{u}^{\mathrm{2}} }{\mathrm{g}\left(\mathrm{1}+\mathrm{p}^{\mathrm{2}} \right)}\left(\mathrm{p}+\frac{\mathrm{1}}{\mathrm{m}}\right) \\ $$$$\mathrm{using}\:\left(\boldsymbol{\mathrm{i}}\right)\:\mathrm{in}\:\mathrm{R}.\mathrm{H}.\mathrm{S}.\:\mathrm{we}\:\mathrm{get} \\ $$$$\frac{\mathrm{2am}^{\mathrm{2}} }{\mathrm{2m}^{\mathrm{2}} −\mathrm{mp}+\mathrm{1}}=\frac{\mathrm{2am}}{\left(\mathrm{m}−\mathrm{p}\right)^{\mathrm{2}} }\left(\frac{\mathrm{1}+\mathrm{mp}}{\mathrm{m}}\right) \\ $$$$\Rightarrow\mathrm{m}^{\mathrm{2}} \left(\mathrm{m}−\mathrm{p}\right)^{\mathrm{2}} =\left(\mathrm{2m}^{\mathrm{2}} −\mathrm{mp}+\mathrm{1}\right)\left(\mathrm{1}+\mathrm{mp}\right) \\ $$$$\mathrm{m}^{\mathrm{4}} +\mathrm{m}^{\mathrm{2}} \mathrm{p}^{\mathrm{2}} −\mathrm{2pm}^{\mathrm{3}} =\mathrm{2m}^{\mathrm{2}} −\mathrm{mp}+\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{2pm}^{\mathrm{3}} −\mathrm{m}^{\mathrm{2}} \mathrm{p}^{\mathrm{2}} +\mathrm{mp} \\ $$$$\:\mathrm{2m}^{\mathrm{2}} \mathrm{p}^{\mathrm{2}} −\mathrm{4pm}^{\mathrm{3}} +\mathrm{m}^{\mathrm{4}} −\mathrm{2m}^{\mathrm{2}} −\mathrm{1}=\mathrm{0}\:\: \\ $$$$\Rightarrow\boldsymbol{\mathrm{p}}=\frac{\mathrm{4}\boldsymbol{\mathrm{m}}^{\mathrm{3}} +\sqrt{\mathrm{16}\boldsymbol{\mathrm{m}}^{\mathrm{6}} −\mathrm{8}\boldsymbol{\mathrm{m}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{m}}^{\mathrm{4}} −\mathrm{2}\boldsymbol{\mathrm{m}}^{\mathrm{2}} −\mathrm{1}\right)}}{\mathrm{4}\boldsymbol{\mathrm{m}}^{\mathrm{2}} } \\ $$$$\boldsymbol{\mathrm{p}}=\boldsymbol{\mathrm{m}}+\sqrt{\frac{\mathrm{8}\boldsymbol{\mathrm{m}}^{\mathrm{6}} +\mathrm{16}\boldsymbol{\mathrm{m}}^{\mathrm{4}} +\mathrm{8}\boldsymbol{\mathrm{m}}^{\mathrm{2}} }{\mathrm{16}\boldsymbol{\mathrm{m}}^{\mathrm{4}} }}\: \\ $$$$\boldsymbol{\mathrm{p}}=\boldsymbol{\mathrm{m}}+\sqrt{\frac{\boldsymbol{\mathrm{m}}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}\boldsymbol{\mathrm{m}}^{\mathrm{2}} }}\: \\ $$$$\boldsymbol{\mathrm{p}}=\boldsymbol{\mathrm{m}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\boldsymbol{\mathrm{m}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{m}}}\right) \\ $$$$\Rightarrow\:\mathrm{tan}\:\phi=\mathrm{tan}\:\theta+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\mathrm{tan}\:\theta+\mathrm{cot}\:\theta\right)\:. \\ $$$$…. \\ $$

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