Yahoo Web Search

assuming T _0 is constant..we can solve the differential equation by setting it up like... ln( T - T _0) = -k* t + C raise to the exponent of e T - T _0 = e^(-k* t )*e^C substitute a lim as t approaches inf...

t ( t ^2 - 2t - 3) = 0 => t ( t +1)( t -3) = 0 => t = 0, -1, 3

( t -1)^2=2t^2- t -29 . multiply out the left side t ^2 - 2t + 1 = 2t^2- t -29 . move everything to the right 0 = t ^2 + t - 30 . factor 0 = ( t - 5)( t + 6) . set each factor = to 0 and solve t - 5=0 or t = 5 t + 6 = 0 or t = -6

Subtract 2r from both sides to get: T -2r= 2t Divide both sides by 2 to get: T -2r/2 = t Simplify this because the 2 over the other 2 cancel each other out. Answer: t = T -r

Curvature k( t ) = l r'( t ) x r''( t ) l / l r'( t ) l ^3. Now with r( t ) = < 4*sin( t ), 4*sin( t ), -2*cos( t ) > we have r'( t ) = < 4*cos( t ), 4*cos( t ), 2*sin( t ) > and r''( t ) = <...

f( t ) = 20 cos^3( t ) f ' ( t ) = 60 cos^2( t ) * -sin( t ) f ' ( t ) = -60 sin( t ) cos^2( t ) g( t ) = 20 sin^3( t ) g ' ( t ) = 60 sin^2( t ) * cos( t ) g ' ( t ) = -60 cos( t ) sin^2( t ...

Hello, ∫ e^(m t ) sin(ω t ) dt = let's integrate by ...: ∫ u dv = u v - ∫ v du ∫ e^(m t ) sin(ω t ) dt = e^(m t ) (1/ω)[- cos(ω...e^(m t ) cos(ω t ) + (m/ω²) e^(m t ) sin(ω t ) + C hence: ∫ e^(m t ...

T (n) - T (n-1) = log(n). T (n-1) - T (n-2) = log(n-1) T (n-2) - T (n-3) = log(n-2) T (n-3) - T (n-4) = log(n-3...