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Y=θ[SIN(INθ)+COS(INθ)],THEN FIND dy÷dθ.
Solution) Y=θ[SIN(INθ)+COS(INθ)]applying u.v rulethen dy÷dθ={[ SIN(INθ)+COS(INθ) ] dθ÷dθ }+ {θ[ d÷dθ{SIN(INθ)+COS(INθ) ] } => SIN(INθ)+COS(INθ) + θ{ (COS(INθ)÷ θ) - (SIN(INθ)÷θ) } θ is canceled and sin(ln θ ) is also canceled then u will get => 2COS(INθ)
Spurious Correlations - in several rare situations when plotting the data for x and y we may have a group indicating either positive correlation or negative (-ve) correlation
Solve for x. 21x+6
We will firstly notice the undamped case. The differential equation under this case is, mu'' + ku = F(t) It is just a non-homogeneous differential equation and we identify h
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ABCD is a parallelogram in the given figure, AB is divided at P and CD and Q so that AP:PB=3:2 and CQ:QD=4:1. If PQ meets AC at R, prove that AR= 3/7 AC. Ans: ΔAPR ∼ Δ
A reaction following first-order kinetics was studied by determining the reactant concentrations at equal time intervals. Each successive pair of concentrations, [A] o and [A] 1
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y=Θ[sin(lnΘ)+cos(lnΘ)] dy/dΘ=[sin(lnΘ)+cos(lnΘ)] + Θ[cos(lnΘ)-sin(lnΘ)]*1/Θ ---->(Use Multiplication rule) dy/dΘ=2cosΘ.
y=Θ[sin(lnΘ)+cos(lnΘ)]
dy/dΘ=[sin(lnΘ)+cos(lnΘ)] + Θ[cos(lnΘ)-sin(lnΘ)]*1/Θ ---->(Use Multiplication rule)
dy/dΘ=2cosΘ.
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