Question-88606

Question Number 88606 by TawaTawa1 last updated on 11/Apr/20

Commented by jagoll last updated on 12/Apr/20

the equation of line y = mx ⇒4t−t^2 = m.t m = 4−t ⇒y = (4−t)x line and parabolic intersect ⇒ 4x−x^2 = 4x−tx x^2 −tx=0 , Δ = t^2 area = ((t^2 (√t^2 ))/(6.1)) = (9/2) ⇒ t^3 =27 t = 3

$$\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{line}\: \\ $$$$\mathrm{y}\:=\:\mathrm{mx}\:\Rightarrow\mathrm{4t}−\mathrm{t}^{\mathrm{2}} \:=\:\mathrm{m}.\mathrm{t}\: \\ $$$$\mathrm{m}\:=\:\mathrm{4}−\mathrm{t}\:\Rightarrow\mathrm{y}\:=\:\left(\mathrm{4}−\mathrm{t}\right)\mathrm{x} \\ $$$$\mathrm{line}\:\mathrm{and}\:\mathrm{parabolic}\:\mathrm{intersect} \\ $$$$\Rightarrow\:\mathrm{4x}−\mathrm{x}^{\mathrm{2}} \:=\:\mathrm{4x}−\mathrm{tx} \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{tx}=\mathrm{0}\:,\:\Delta\:=\:\mathrm{t}^{\mathrm{2}} \\ $$$$\mathrm{area}\:=\:\frac{\mathrm{t}^{\mathrm{2}} \sqrt{\mathrm{t}^{\mathrm{2}} }}{\mathrm{6}.\mathrm{1}}\:=\:\frac{\mathrm{9}}{\mathrm{2}}\:\Rightarrow\:\mathrm{t}^{\mathrm{3}} =\mathrm{27} \\ $$$$\mathrm{t}\:=\:\mathrm{3} \\ $$

Commented by TawaTawa1 last updated on 12/Apr/20

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by mahdi last updated on 11/Apr/20

∫_0 ^( t) (4x−x^2 )dx−((t(4t−t^2 ))/2)=(9/2) [2x^2 −(x^3 /3)]_0 ^t −((4t^2 −t^3 )/2)=(t^3 /6)=(9/2)⇒t=3

$$\int_{\mathrm{0}} ^{\:\mathrm{t}} \left(\mathrm{4x}−\mathrm{x}^{\mathrm{2}} \right)\mathrm{dx}−\frac{\mathrm{t}\left(\mathrm{4t}−\mathrm{t}^{\mathrm{2}} \right)}{\mathrm{2}}=\frac{\mathrm{9}}{\mathrm{2}} \\ $$$$\left[\mathrm{2x}^{\mathrm{2}} −\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}\right]_{\mathrm{0}} ^{\mathrm{t}} −\frac{\mathrm{4t}^{\mathrm{2}} −\mathrm{t}^{\mathrm{3}} }{\mathrm{2}}=\frac{\mathrm{t}^{\mathrm{3}} }{\mathrm{6}}=\frac{\mathrm{9}}{\mathrm{2}}\Rightarrow\mathrm{t}=\mathrm{3} \\ $$

Commented by TawaTawa1 last updated on 11/Apr/20

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by mr W last updated on 11/Apr/20

((2t)/3)[((4t)/2)−(t^2 /4)−(1/2)(4t−t^2 )]=(9/2) t^3 =27 ⇒t=3

$$\frac{\mathrm{2}{t}}{\mathrm{3}}\left[\frac{\mathrm{4}{t}}{\mathrm{2}}−\frac{{t}^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{4}{t}−{t}^{\mathrm{2}} \right)\right]=\frac{\mathrm{9}}{\mathrm{2}} \\ $$$${t}^{\mathrm{3}} =\mathrm{27} \\ $$$$\Rightarrow{t}=\mathrm{3} \\ $$

Commented by TawaTawa1 last updated on 11/Apr/20