🧠:<think>First, I will calculate the total cost at Store A. Samantha is purchasing a dress, shoes, and a handbag priced at 80, 100, and 60 respectively. Adding these together gives a subtotal of 240.Store A offers a 20% discount on all items. Applying this discount to the subtotal reduces the total by 48, bringing the amount to 192.Since Samantha's total after the initial discount is 192, which is more than 150, she qualifies for an additional flat discount of 30. Subtracting this from 192 results in a final total of 162 at Store A.Next, I will calculate the total cost at Store B. Samantha is purchasing a jacket, blouse, skirt, earrings, scarf, and bracelet priced at 120, 50, 70, 30, 40, and 20 respectively. Adding these together gives a subtotal of 330.Store B offers a 30% discount on all items. Applying this discount reduces the total by 99, bringing the amount to 231.Since Samantha is buying more than 5 items, she qualifies for an additional 10% discount on the already discounted total. Applying this 10% discount reduces the total by 23.10, resulting in a final total of 207.90 at Store B.</think>

🧠:<think>Okay, so I have this problem about modeling a sensory response using a differential equation. It's a second-order linear nonhomogeneous differential equation, right? Let me write it down again to make sure I have it correctly:[frac{d^2R}{dt^2} + 3frac{dR}{dt} + 2R = e^{-t} cos(2t)]Alright, so the task is to find the general solution for ( R(t) ). I remember that for such equations, the general solution is the sum of the homogeneous solution and a particular solution. So, first, I need to solve the homogeneous equation:[frac{d^2R}{dt^2} + 3frac{dR}{dt} + 2R = 0]To solve this, I can find the characteristic equation. Let me set ( r^2 + 3r + 2 = 0 ). Solving this quadratic equation, the roots are:[r = frac{-3 pm sqrt{9 - 8}}{2} = frac{-3 pm 1}{2}]So, the roots are ( r = -1 ) and ( r = -2 ). That means the homogeneous solution is:[R_h(t) = C_1 e^{-t} + C_2 e^{-2t}]Okay, got that. Now, I need to find a particular solution ( R_p(t) ) for the nonhomogeneous equation. The right-hand side is ( e^{-t} cos(2t) ). Hmm, that looks like a product of an exponential and a cosine function. I think I can use the method of undetermined coefficients for this.First, let me recall that when the forcing function is of the form ( e^{alpha t} cos(beta t) ) or ( e^{alpha t} sin(beta t) ), we can assume a particular solution of the form ( e^{alpha t} (A cos(beta t) + B sin(beta t)) ). But I need to check if ( alpha + ibeta ) is a root of the characteristic equation.In this case, the forcing function is ( e^{-t} cos(2t) ), so ( alpha = -1 ) and ( beta = 2 ). Let me see if ( -1 + 2i ) is a root of the characteristic equation. The characteristic equation is ( r^2 + 3r + 2 = 0 ), which has roots at ( r = -1 ) and ( r = -2 ). So, ( -1 + 2i ) is not a root because the roots are real and distinct. Therefore, I don't need to multiply by ( t ); I can proceed with the standard form.So, let me assume the particular solution is:[R_p(t) = e^{-t} (A cos(2t) + B sin(2t))]Now, I need to compute the first and second derivatives of ( R_p(t) ) to substitute back into the differential equation.First derivative:[R_p'(t) = -e^{-t} (A cos(2t) + B sin(2t)) + e^{-t} (-2A sin(2t) + 2B cos(2t))]Simplify that:[R_p'(t) = e^{-t} [ -A cos(2t) - B sin(2t) - 2A sin(2t) + 2B cos(2t) ]][= e^{-t} [ (-A + 2B) cos(2t) + (-B - 2A) sin(2t) ]]Second derivative:Let me differentiate ( R_p'(t) ):[R_p''(t) = -e^{-t} [ (-A + 2B) cos(2t) + (-B - 2A) sin(2t) ] + e^{-t} [ 2(-A + 2B) sin(2t) - 2(-B - 2A) cos(2t) ]]Simplify term by term:First term:[- e^{-t} (-A + 2B) cos(2t) - e^{-t} (-B - 2A) sin(2t)][= e^{-t} (A - 2B) cos(2t) + e^{-t} (B + 2A) sin(2t)]Second term:[e^{-t} [ 2(-A + 2B) sin(2t) - 2(-B - 2A) cos(2t) ]][= e^{-t} [ (-2A + 4B) sin(2t) + (2B + 4A) cos(2t) ]]Combine both terms:[R_p''(t) = e^{-t} [ (A - 2B + 2B + 4A) cos(2t) + (B + 2A - 2A + 4B) sin(2t) ]][= e^{-t} [ (5A) cos(2t) + (5B) sin(2t) ]]So, ( R_p''(t) = e^{-t} (5A cos(2t) + 5B sin(2t)) )Now, let's plug ( R_p(t) ), ( R_p'(t) ), and ( R_p''(t) ) into the original differential equation:[R_p'' + 3 R_p' + 2 R_p = e^{-t} cos(2t)]Substituting:Left-hand side:[e^{-t} (5A cos(2t) + 5B sin(2t)) + 3 e^{-t} [ (-A + 2B) cos(2t) + (-B - 2A) sin(2t) ] + 2 e^{-t} (A cos(2t) + B sin(2t))]Let me factor out ( e^{-t} ):[e^{-t} [ (5A cos(2t) + 5B sin(2t)) + 3(-A + 2B) cos(2t) + 3(-B - 2A) sin(2t) + 2A cos(2t) + 2B sin(2t) ]]Now, let's collect like terms for ( cos(2t) ) and ( sin(2t) ):For ( cos(2t) ):[5A + 3(-A + 2B) + 2A][= 5A - 3A + 6B + 2A][= (5A - 3A + 2A) + 6B][= 4A + 6B]For ( sin(2t) ):[5B + 3(-B - 2A) + 2B][= 5B - 3B - 6A + 2B][= (5B - 3B + 2B) - 6A][= 4B - 6A]So, the left-hand side becomes:[e^{-t} [ (4A + 6B) cos(2t) + (4B - 6A) sin(2t) ]]And this is equal to the right-hand side:[e^{-t} cos(2t)]Therefore, we can set up the equations by equating the coefficients:For ( cos(2t) ):[4A + 6B = 1]For ( sin(2t) ):[4B - 6A = 0]So, now we have a system of two equations:1. ( 4A + 6B = 1 )2. ( -6A + 4B = 0 )Let me write them as:1. ( 4A + 6B = 1 )2. ( -6A + 4B = 0 )I can solve this system using substitution or elimination. Let me use elimination.First, let's multiply the first equation by 3 and the second equation by 2 to make the coefficients of ( A ) opposites:1. ( 12A + 18B = 3 )2. ( -12A + 8B = 0 )Now, add the two equations:( (12A - 12A) + (18B + 8B) = 3 + 0 )( 0A + 26B = 3 )So, ( 26B = 3 )Therefore, ( B = frac{3}{26} )Now, substitute ( B ) back into one of the original equations, say the second equation:( -6A + 4B = 0 )( -6A + 4*(3/26) = 0 )( -6A + 12/26 = 0 )( -6A = -12/26 )Multiply both sides by -1:( 6A = 12/26 )Simplify:( A = (12/26)/6 = (12)/(26*6) = (12)/(156) = 1/13 )So, ( A = 1/13 ) and ( B = 3/26 )Therefore, the particular solution is:[R_p(t) = e^{-t} left( frac{1}{13} cos(2t) + frac{3}{26} sin(2t) right )]I can factor out ( frac{1}{26} ) to make it look neater:[R_p(t) = frac{e^{-t}}{26} (2 cos(2t) + 3 sin(2t))]So, the general solution is the homogeneous solution plus the particular solution:[R(t) = C_1 e^{-t} + C_2 e^{-2t} + frac{e^{-t}}{26} (2 cos(2t) + 3 sin(2t))]I can write this as:[R(t) = C_1 e^{-t} + C_2 e^{-2t} + frac{e^{-t}}{26} (2 cos(2t) + 3 sin(2t))]That should be the general solution. Let me just double-check my calculations to make sure I didn't make a mistake.Starting from the particular solution, I assumed the correct form, computed the derivatives correctly, substituted back, and equated coefficients. The system of equations gave me ( A = 1/13 ) and ( B = 3/26 ), which seems consistent. So, I think this is correct.Moving on to the second part. Dr. Thompson wants to compute the integral of ( R(t) ) over a period ( T ) of the sinusoidal component in the forcing function. The forcing function is ( e^{-t} cos(2t) ). The sinusoidal component is ( cos(2t) ), which has a period of ( T = pi ), since the period of ( cos(k t) ) is ( 2pi / k ), so here ( k = 2 ), so ( T = pi ).But wait, actually, the forcing function is ( e^{-t} cos(2t) ). The exponential term doesn't affect the period; it's just a damping factor. So, the period is still determined by the cosine term, which is ( pi ). So, ( T = pi ).Given the initial conditions ( R(0) = 1 ) and ( R'(0) = 0 ), we need to compute:[int_0^pi R(t) , dt]First, let's find the specific solution ( R(t) ) using the initial conditions. The general solution is:[R(t) = C_1 e^{-t} + C_2 e^{-2t} + frac{e^{-t}}{26} (2 cos(2t) + 3 sin(2t))]We need to find ( C_1 ) and ( C_2 ) using ( R(0) = 1 ) and ( R'(0) = 0 ).First, compute ( R(0) ):[R(0) = C_1 e^{0} + C_2 e^{0} + frac{e^{0}}{26} (2 cos(0) + 3 sin(0)) = C_1 + C_2 + frac{1}{26} (2*1 + 0) = C_1 + C_2 + frac{2}{26} = C_1 + C_2 + frac{1}{13}]Given ( R(0) = 1 ):[C_1 + C_2 + frac{1}{13} = 1 implies C_1 + C_2 = 1 - frac{1}{13} = frac{12}{13}]Now, compute ( R'(t) ):First, differentiate each term:1. ( frac{d}{dt} [C_1 e^{-t}] = -C_1 e^{-t} )2. ( frac{d}{dt} [C_2 e^{-2t}] = -2 C_2 e^{-2t} )3. The particular solution derivative is more involved. Let me compute it:Let me denote the particular solution as ( R_p(t) = frac{e^{-t}}{26} (2 cos(2t) + 3 sin(2t)) )So, ( R_p'(t) = frac{d}{dt} [ frac{e^{-t}}{26} (2 cos(2t) + 3 sin(2t)) ] )Using the product rule:[R_p'(t) = frac{-e^{-t}}{26} (2 cos(2t) + 3 sin(2t)) + frac{e^{-t}}{26} ( -4 sin(2t) + 6 cos(2t) )]Simplify:[= frac{e^{-t}}{26} [ -2 cos(2t) - 3 sin(2t) - 4 sin(2t) + 6 cos(2t) ]][= frac{e^{-t}}{26} [ ( -2 + 6 ) cos(2t) + ( -3 - 4 ) sin(2t) ]][= frac{e^{-t}}{26} [ 4 cos(2t) - 7 sin(2t) ]]So, putting it all together, the derivative ( R'(t) ) is:[R'(t) = -C_1 e^{-t} - 2 C_2 e^{-2t} + frac{e^{-t}}{26} (4 cos(2t) - 7 sin(2t))]Now, evaluate ( R'(0) ):[R'(0) = -C_1 e^{0} - 2 C_2 e^{0} + frac{e^{0}}{26} (4 cos(0) - 7 sin(0)) = -C_1 - 2 C_2 + frac{1}{26} (4*1 - 0) = -C_1 - 2 C_2 + frac{4}{26} = -C_1 - 2 C_2 + frac{2}{13}]Given ( R'(0) = 0 ):[- C_1 - 2 C_2 + frac{2}{13} = 0 implies -C_1 - 2 C_2 = - frac{2}{13} implies C_1 + 2 C_2 = frac{2}{13}]Now, we have a system of two equations:1. ( C_1 + C_2 = frac{12}{13} )2. ( C_1 + 2 C_2 = frac{2}{13} )Let me subtract the first equation from the second:( (C_1 + 2 C_2) - (C_1 + C_2) = frac{2}{13} - frac{12}{13} )( C_2 = - frac{10}{13} )Now, substitute ( C_2 = -10/13 ) into the first equation:( C_1 + (-10/13) = 12/13 implies C_1 = 12/13 + 10/13 = 22/13 )So, ( C_1 = 22/13 ) and ( C_2 = -10/13 )Therefore, the specific solution is:[R(t) = frac{22}{13} e^{-t} - frac{10}{13} e^{-2t} + frac{e^{-t}}{26} (2 cos(2t) + 3 sin(2t))]Now, we need to compute the integral:[int_0^pi R(t) , dt]Let me write ( R(t) ) as:[R(t) = frac{22}{13} e^{-t} - frac{10}{13} e^{-2t} + frac{e^{-t}}{26} (2 cos(2t) + 3 sin(2t))]So, the integral becomes:[int_0^pi left( frac{22}{13} e^{-t} - frac{10}{13} e^{-2t} + frac{e^{-t}}{26} (2 cos(2t) + 3 sin(2t)) right ) dt]We can split this into three separate integrals:[frac{22}{13} int_0^pi e^{-t} dt - frac{10}{13} int_0^pi e^{-2t} dt + frac{1}{26} int_0^pi e^{-t} (2 cos(2t) + 3 sin(2t)) dt]Let me compute each integral one by one.First integral:[I_1 = frac{22}{13} int_0^pi e^{-t} dt = frac{22}{13} left[ -e^{-t} right ]_0^pi = frac{22}{13} left( -e^{-pi} + e^{0} right ) = frac{22}{13} (1 - e^{-pi})]Second integral:[I_2 = - frac{10}{13} int_0^pi e^{-2t} dt = - frac{10}{13} left[ -frac{1}{2} e^{-2t} right ]_0^pi = - frac{10}{13} left( -frac{1}{2} e^{-2pi} + frac{1}{2} e^{0} right ) = - frac{10}{13} left( frac{1}{2} (1 - e^{-2pi}) right ) = - frac{10}{13} * frac{1}{2} (1 - e^{-2pi}) = - frac{5}{13} (1 - e^{-2pi})]Third integral:[I_3 = frac{1}{26} int_0^pi e^{-t} (2 cos(2t) + 3 sin(2t)) dt]This integral is a bit more involved. Let me denote:[I_3 = frac{1}{26} left( 2 int_0^pi e^{-t} cos(2t) dt + 3 int_0^pi e^{-t} sin(2t) dt right )]Let me compute each integral separately.First, compute ( int e^{-t} cos(2t) dt ). I can use integration by parts or recall the formula for integrating ( e^{at} cos(bt) ). The formula is:[int e^{at} cos(bt) dt = frac{e^{at}}{a^2 + b^2} (a cos(bt) + b sin(bt)) ) + C]Similarly, for ( int e^{at} sin(bt) dt ):[int e^{at} sin(bt) dt = frac{e^{at}}{a^2 + b^2} (a sin(bt) - b cos(bt)) ) + C]In our case, ( a = -1 ) and ( b = 2 ).So, compute ( int e^{-t} cos(2t) dt ):Using the formula:[int e^{-t} cos(2t) dt = frac{e^{-t}}{(-1)^2 + 2^2} ( -1 cos(2t) + 2 sin(2t) ) + C = frac{e^{-t}}{1 + 4} ( -cos(2t) + 2 sin(2t) ) + C = frac{e^{-t}}{5} ( -cos(2t) + 2 sin(2t) ) + C]Similarly, compute ( int e^{-t} sin(2t) dt ):Using the formula:[int e^{-t} sin(2t) dt = frac{e^{-t}}{(-1)^2 + 2^2} ( -1 sin(2t) - 2 cos(2t) ) + C = frac{e^{-t}}{5} ( -sin(2t) - 2 cos(2t) ) + C]So, now let's compute the definite integrals from 0 to ( pi ).First integral:[int_0^pi e^{-t} cos(2t) dt = left[ frac{e^{-t}}{5} ( -cos(2t) + 2 sin(2t) ) right ]_0^pi]Evaluate at ( pi ):[frac{e^{-pi}}{5} ( -cos(2pi) + 2 sin(2pi) ) = frac{e^{-pi}}{5} ( -1 + 0 ) = - frac{e^{-pi}}{5}]Evaluate at 0:[frac{e^{0}}{5} ( -cos(0) + 2 sin(0) ) = frac{1}{5} ( -1 + 0 ) = - frac{1}{5}]So, the integral from 0 to ( pi ):[- frac{e^{-pi}}{5} - ( - frac{1}{5} ) = - frac{e^{-pi}}{5} + frac{1}{5} = frac{1 - e^{-pi}}{5}]Second integral:[int_0^pi e^{-t} sin(2t) dt = left[ frac{e^{-t}}{5} ( -sin(2t) - 2 cos(2t) ) right ]_0^pi]Evaluate at ( pi ):[frac{e^{-pi}}{5} ( -sin(2pi) - 2 cos(2pi) ) = frac{e^{-pi}}{5} ( 0 - 2*1 ) = - frac{2 e^{-pi}}{5}]Evaluate at 0:[frac{e^{0}}{5} ( -sin(0) - 2 cos(0) ) = frac{1}{5} ( 0 - 2*1 ) = - frac{2}{5}]So, the integral from 0 to ( pi ):[- frac{2 e^{-pi}}{5} - ( - frac{2}{5} ) = - frac{2 e^{-pi}}{5} + frac{2}{5} = frac{2 (1 - e^{-pi})}{5}]Now, plug these back into ( I_3 ):[I_3 = frac{1}{26} left( 2 * frac{1 - e^{-pi}}{5} + 3 * frac{2 (1 - e^{-pi})}{5} right ) = frac{1}{26} left( frac{2 (1 - e^{-pi})}{5} + frac{6 (1 - e^{-pi})}{5} right )][= frac{1}{26} left( frac{8 (1 - e^{-pi})}{5} right ) = frac{8}{130} (1 - e^{-pi}) = frac{4}{65} (1 - e^{-pi})]So, now, combining all three integrals:[int_0^pi R(t) dt = I_1 + I_2 + I_3 = frac{22}{13} (1 - e^{-pi}) - frac{5}{13} (1 - e^{-2pi}) + frac{4}{65} (1 - e^{-pi})]Let me combine the terms with ( (1 - e^{-pi}) ) and ( (1 - e^{-2pi}) ):First, let's express all terms with denominator 65:[frac{22}{13} = frac{110}{65}, quad frac{5}{13} = frac{25}{65}, quad frac{4}{65} remains as is.]So,[int_0^pi R(t) dt = frac{110}{65} (1 - e^{-pi}) - frac{25}{65} (1 - e^{-2pi}) + frac{4}{65} (1 - e^{-pi})][= left( frac{110}{65} + frac{4}{65} right ) (1 - e^{-pi}) - frac{25}{65} (1 - e^{-2pi})][= frac{114}{65} (1 - e^{-pi}) - frac{25}{65} (1 - e^{-2pi})]Simplify the fractions:[frac{114}{65} = frac{114 ÷ 13}{65 ÷ 13} = frac{8.769}{5} quad text{Wait, no, 114 divided by 13 is 8.769? Wait, 13*8=104, 13*9=117, so 114 is 13*8 + 10, so it's 8 and 10/13. Similarly, 25/65 simplifies to 5/13.Wait, maybe it's better to leave it as is or factor out 1/65:[= frac{1}{65} [ 114 (1 - e^{-pi}) - 25 (1 - e^{-2pi}) ]]Alternatively, factor out the common terms:[= frac{114}{65} (1 - e^{-pi}) - frac{25}{65} (1 - e^{-2pi})]Alternatively, we can write as:[= frac{114}{65} - frac{114}{65} e^{-pi} - frac{25}{65} + frac{25}{65} e^{-2pi}][= left( frac{114}{65} - frac{25}{65} right ) - frac{114}{65} e^{-pi} + frac{25}{65} e^{-2pi}][= frac{89}{65} - frac{114}{65} e^{-pi} + frac{25}{65} e^{-2pi}]Simplify the fractions:[frac{89}{65} = frac{89}{65}, quad frac{114}{65} = frac{114}{65}, quad frac{25}{65} = frac{5}{13}]Alternatively, we can factor out 1/65:[= frac{1}{65} left( 89 - 114 e^{-pi} + 25 e^{-2pi} right )]So, the integral is:[int_0^pi R(t) dt = frac{89 - 114 e^{-pi} + 25 e^{-2pi}}{65}]I can leave it like that, or factor further if possible. Let me see if numerator can be factored:89 - 114 e^{-π} + 25 e^{-2π}Hmm, 25 e^{-2π} - 114 e^{-π} + 89Let me consider it as a quadratic in terms of ( x = e^{-pi} ):25 x^2 - 114 x + 89Let me compute the discriminant:D = 114^2 - 4*25*89 = 12996 - 4*25*89Compute 4*25=100, 100*89=8900So, D = 12996 - 8900 = 4096Which is a perfect square, sqrt(4096)=64Therefore, roots are:x = [114 ± 64]/(2*25) = [114 ± 64]/50Compute:x1 = (114 + 64)/50 = 178/50 = 89/25 = 3.56x2 = (114 - 64)/50 = 50/50 = 1So, the quadratic factors as 25(x - 1)(x - 89/25)But since x = e^{-π}, which is approximately e^{-3.14} ≈ 0.043, which is much less than 1, so the quadratic doesn't factor nicely in terms of e^{-π}.Therefore, it's probably best to leave the integral as:[frac{89 - 114 e^{-pi} + 25 e^{-2pi}}{65}]Alternatively, we can factor out e^{-π}:[= frac{89}{65} - frac{114}{65} e^{-pi} + frac{25}{65} e^{-2pi}]But I think the expression is fine as it is.So, summarizing, the integral of ( R(t) ) from 0 to ( pi ) is:[frac{89 - 114 e^{-pi} + 25 e^{-2pi}}{65}]I can write this as:[frac{89}{65} - frac{114}{65} e^{-pi} + frac{25}{65} e^{-2pi}]But since 25/65 simplifies to 5/13, and 114/65 is 114/65, and 89/65 is 89/65, I can write:[frac{89}{65} - frac{114}{65} e^{-pi} + frac{5}{13} e^{-2pi}]Alternatively, keeping all terms over 65:[frac{89 - 114 e^{-pi} + 25 e^{-2pi}}{65}]Either way is acceptable. I think the latter is more concise.So, putting it all together, the integral is ( frac{89 - 114 e^{-pi} + 25 e^{-2pi}}{65} ).Let me just check my calculations to make sure I didn't make any mistakes.Starting from the integrals:1. ( I_1 = frac{22}{13} (1 - e^{-pi}) )2. ( I_2 = - frac{5}{13} (1 - e^{-2pi}) )3. ( I_3 = frac{4}{65} (1 - e^{-pi}) )Adding them together:Convert all to 65 denominator:( I_1 = frac{110}{65} (1 - e^{-pi}) )( I_2 = - frac{25}{65} (1 - e^{-2pi}) )( I_3 = frac{4}{65} (1 - e^{-pi}) )So, total:( frac{110 + 4}{65} (1 - e^{-pi}) - frac{25}{65} (1 - e^{-2pi}) = frac{114}{65} (1 - e^{-pi}) - frac{25}{65} (1 - e^{-2pi}) )Which is the same as:( frac{114(1 - e^{-pi}) -25(1 - e^{-2pi})}{65} )Expanding numerator:114 - 114 e^{-pi} -25 +25 e^{-2pi} = (114 -25) + (-114 e^{-pi}) +25 e^{-2pi} = 89 -114 e^{-pi} +25 e^{-2pi}Yes, that's correct.Therefore, the integral is ( frac{89 - 114 e^{-pi} +25 e^{-2pi}}{65} ).I think that's the final answer.Final Answer1. The general solution is (boxed{R(t) = C_1 e^{-t} + C_2 e^{-2t} + frac{e^{-t}}{26} (2 cos(2t) + 3 sin(2t))}).2. The integral of ( R(t) ) over the period ( T ) is (boxed{dfrac{89 - 114 e^{-pi} + 25 e^{-2pi}}{65}}).</think>

🧠:<think>Alright, so I have this problem about a high-powered executive with ADD who wants to optimize their daily schedule. The productivity function is given by ( P(t) = frac{10t}{1 + e^{k(t-5)}} ). I need to find the time ( t ) that maximizes productivity and then, given that ( k ) follows a normal distribution, find the expected value of this optimal time ( t ).Starting with part 1: Determine the value of ( t ) that maximizes ( P(t) ). Hmm, okay. To find the maximum of a function, I remember I need to take its derivative with respect to ( t ), set the derivative equal to zero, and solve for ( t ). That should give me the critical points, which I can then test to make sure it's a maximum.So, let's write down the function again:[ P(t) = frac{10t}{1 + e^{k(t-5)}} ]I need to find ( P'(t) ). Since this is a quotient, I should use the quotient rule. The quotient rule states that if I have a function ( frac{u}{v} ), its derivative is ( frac{u'v - uv'}{v^2} ).Let me assign ( u = 10t ) and ( v = 1 + e^{k(t-5)} ). Then, ( u' = 10 ) and ( v' = k e^{k(t-5)} ) because the derivative of ( e^{kt} ) is ( k e^{kt} ), and here it's ( e^{k(t-5)} ), so the chain rule gives us ( k e^{k(t-5)} ).Putting it all into the quotient rule:[ P'(t) = frac{10 cdot (1 + e^{k(t-5)}) - 10t cdot k e^{k(t-5)}}{(1 + e^{k(t-5)})^2} ]Simplify the numerator:First term: ( 10(1 + e^{k(t-5)}) = 10 + 10 e^{k(t-5)} )Second term: ( -10t k e^{k(t-5)} )So, the numerator becomes:( 10 + 10 e^{k(t-5)} - 10 t k e^{k(t-5)} )Factor out 10 e^{k(t-5)} from the last two terms:( 10 + 10 e^{k(t-5)} (1 - t k) )Wait, actually, let me factor 10 e^{k(t-5)} from the second and third terms:10 e^{k(t-5)} (1 - t k) + 10So, numerator is:10 + 10 e^{k(t-5)} (1 - t k)So, putting it back into the derivative:[ P'(t) = frac{10 + 10 e^{k(t-5)} (1 - t k)}{(1 + e^{k(t-5)})^2} ]To find critical points, set numerator equal to zero:10 + 10 e^{k(t-5)} (1 - t k) = 0Divide both sides by 10:1 + e^{k(t-5)} (1 - t k) = 0So,e^{k(t-5)} (1 - t k) = -1Hmm, that's an equation involving exponentials and linear terms. Let's see if we can solve for ( t ).Let me denote ( x = k(t - 5) ) for simplicity. Then, ( e^{x} (1 - t k) = -1 ).But ( x = k(t - 5) ), so ( t = (x / k) + 5 ). Substitute back into the equation:( e^{x} (1 - k ((x / k) + 5)) = -1 )Simplify inside the parentheses:1 - k*(x/k + 5) = 1 - x - 5k = (1 - 5k) - xSo, equation becomes:( e^{x} ((1 - 5k) - x) = -1 )Hmm, this seems a bit complicated. Maybe there's another approach.Alternatively, let's go back to the equation:1 + e^{k(t - 5)} (1 - t k) = 0Let me rearrange terms:e^{k(t - 5)} (1 - t k) = -1Divide both sides by e^{k(t - 5)}:1 - t k = -e^{-k(t - 5)}So,1 - t k = -e^{-k(t - 5)}Multiply both sides by -1:t k - 1 = e^{-k(t - 5)}Let me write this as:t k - 1 = e^{-k t + 5k}Hmm, this is still a transcendental equation, which might not have a closed-form solution. Maybe I can take the natural logarithm on both sides?But before that, let's see if we can express this in terms of the Lambert W function, which is used to solve equations of the form ( z = w e^{w} ).Let me try to manipulate the equation:t k - 1 = e^{-k t + 5k}Let me denote ( y = -k t + 5k ). Then, ( e^{y} = t k - 1 ).But ( y = -k t + 5k ), so ( t = (5k - y)/k ).Substitute into ( e^{y} = t k - 1 ):( e^{y} = ((5k - y)/k) * k - 1 = 5k - y - 1 )So,( e^{y} = 5k - y - 1 )Bring all terms to one side:( e^{y} + y - (5k - 1) = 0 )Hmm, this is still not in a form that can be directly expressed with Lambert W. Maybe another substitution.Alternatively, let's go back to the equation:t k - 1 = e^{-k t + 5k}Let me write this as:t k - 1 = e^{5k} e^{-k t}Let me denote ( z = k t ). Then, ( t = z / k ).Substitute into the equation:(z / k) * k - 1 = e^{5k} e^{-z}Simplify:z - 1 = e^{5k} e^{-z}So,z - 1 = e^{5k - z}Bring all terms to one side:z - 1 - e^{5k - z} = 0Hmm, still not straightforward. Let me write this as:(z - 1) = e^{5k - z}Take natural logarithm on both sides:ln(z - 1) = 5k - zSo,ln(z - 1) + z = 5kThis is still a transcendental equation, but maybe we can express it in terms of Lambert W.Let me rearrange:z + ln(z - 1) = 5kLet me denote ( w = z - 1 ). Then, ( z = w + 1 ). Substitute:(w + 1) + ln(w) = 5kSo,w + 1 + ln(w) = 5kHmm, still not in Lambert W form. Maybe another approach.Alternatively, perhaps we can approximate the solution numerically, but since the problem asks to express the answer in terms of ( k ), maybe there's a smarter substitution or a way to express ( t ) in terms of ( k ) without solving explicitly.Wait, let me think again. The original equation was:1 + e^{k(t - 5)} (1 - t k) = 0Let me denote ( u = k(t - 5) ). Then, ( t = (u / k) + 5 ).Substitute into the equation:1 + e^{u} (1 - ((u / k) + 5) k ) = 0Simplify the term inside the parentheses:1 - (u + 5k) = 1 - u - 5kSo,1 + e^{u} (1 - u - 5k) = 0Thus,e^{u} (1 - u - 5k) = -1So,e^{u} (u + 5k - 1) = 1Hmm, still tricky. Let me write this as:(u + 5k - 1) e^{u} = 1Let me denote ( v = u + 5k - 1 ). Then, ( u = v - 5k + 1 ). Substitute:v e^{v - 5k + 1} = 1Which is:v e^{v} e^{-5k + 1} = 1So,v e^{v} = e^{5k - 1}Thus,v = W(e^{5k - 1})Where ( W ) is the Lambert W function.So, ( v = W(e^{5k - 1}) )But ( v = u + 5k - 1 ), and ( u = k(t - 5) ). So,v = k(t - 5) + 5k - 1 = k t - 5k + 5k - 1 = k t - 1Thus,k t - 1 = W(e^{5k - 1})Therefore,k t = W(e^{5k - 1}) + 1So,t = [W(e^{5k - 1}) + 1] / kHmm, that seems like a valid expression, but it's in terms of the Lambert W function, which is a special function and might not be expressible in elementary terms. However, the problem says to express the answer in terms of ( k ), so perhaps this is acceptable.Alternatively, maybe there's a simplification or another approach.Wait, let me think again. Maybe I made a mistake earlier.Going back to the derivative:[ P'(t) = frac{10 + 10 e^{k(t-5)} (1 - t k)}{(1 + e^{k(t-5)})^2} ]Set numerator equal to zero:10 + 10 e^{k(t-5)} (1 - t k) = 0Divide by 10:1 + e^{k(t-5)} (1 - t k) = 0So,e^{k(t-5)} (1 - t k) = -1Let me rearrange:e^{k(t-5)} = frac{-1}{1 - t k} = frac{1}{t k - 1}So,e^{k(t - 5)} = frac{1}{t k - 1}Take natural logarithm on both sides:k(t - 5) = -ln(t k - 1)So,k t - 5k = -ln(t k - 1)Multiply both sides by -1:- k t + 5k = ln(t k - 1)Exponentiate both sides:e^{-k t + 5k} = t k - 1Which is the same as earlier.So, we have:t k - 1 = e^{-k t + 5k}Let me write this as:t k - 1 = e^{5k} e^{-k t}Let me denote ( z = k t ). Then, ( t = z / k ). Substitute:(z / k) * k - 1 = e^{5k} e^{-z}Simplify:z - 1 = e^{5k} e^{-z}So,z - 1 = e^{5k - z}Multiply both sides by e^{z}:(z - 1) e^{z} = e^{5k}So,(z - 1) e^{z} = e^{5k}Let me denote ( w = z - 1 ). Then, ( z = w + 1 ). Substitute:w e^{w + 1} = e^{5k}Which is:w e^{w} e^{1} = e^{5k}Divide both sides by e:w e^{w} = e^{5k - 1}Thus,w = W(e^{5k - 1})Where ( W ) is the Lambert W function.Recall that ( w = z - 1 ) and ( z = k t ). So,w = k t - 1 - 1 = k t - 2Wait, no:Wait, ( w = z - 1 ), and ( z = k t ). So,w = k t - 1But earlier, we had ( w e^{w} = e^{5k - 1} ), so ( w = W(e^{5k - 1}) ).Thus,k t - 1 = W(e^{5k - 1})So,k t = W(e^{5k - 1}) + 1Therefore,t = [W(e^{5k - 1}) + 1] / kSo, that's the expression for ( t ) in terms of ( k ).I think that's as far as we can go analytically. So, the time ( t ) that maximizes productivity is ( t = frac{W(e^{5k - 1}) + 1}{k} ).But wait, let me check if this makes sense. When ( k ) is large, meaning the executive is very sensitive to distractions, what happens to ( t )?As ( k ) increases, ( e^{5k - 1} ) becomes very large, so ( W(e^{5k - 1}) ) is approximately ( 5k - 1 - ln(5k - 1) ) for large arguments, since ( W(z) approx ln(z) - ln(ln(z)) ) for large ( z ).So,( W(e^{5k - 1}) approx (5k - 1) - ln(5k - 1) )Thus,t ≈ [ (5k - 1 - ln(5k - 1)) + 1 ] / k = (5k - ln(5k - 1)) / k ≈ 5 - (ln(5k - 1))/kSo, as ( k ) becomes large, ( t ) approaches 5, which makes sense because the function ( P(t) ) has a term ( e^{k(t - 5)} ), which would dominate as ( k ) increases, making the productivity function peak around ( t = 5 ).Similarly, when ( k ) is small, meaning less sensitivity to distractions, what happens?If ( k ) approaches zero, ( e^{5k - 1} ) approaches ( e^{-1} ), so ( W(e^{-1}) ) is known to be -1, since ( W(e^{-1}) = -1 ).Thus,t ≈ [ -1 + 1 ] / k = 0 / k = 0But that doesn't make sense because at ( t = 0 ), productivity is zero. Hmm, maybe the approximation isn't valid for small ( k ).Alternatively, perhaps for small ( k ), the function ( P(t) ) behaves differently. Let me think.When ( k ) is small, the denominator ( 1 + e^{k(t - 5)} ) is approximately ( 1 + e^{k t} e^{-5k} approx 1 + e^{k t} ) since ( e^{-5k} ) is close to 1 for small ( k ).So, ( P(t) approx frac{10 t}{1 + e^{k t}} ). The maximum of this function occurs where the derivative is zero. Let's compute that.Let ( P(t) = frac{10 t}{1 + e^{k t}} )Derivative:( P'(t) = frac{10(1 + e^{k t}) - 10 t k e^{k t}}{(1 + e^{k t})^2} )Set numerator to zero:10 + 10 e^{k t} - 10 t k e^{k t} = 0Divide by 10:1 + e^{k t} - t k e^{k t} = 0Factor out ( e^{k t} ):1 + e^{k t}(1 - t k) = 0So,e^{k t}(1 - t k) = -1Which is similar to the equation we had before. For small ( k ), let's assume ( t ) is not too large.Let me set ( t = 5 ). Then,e^{5k}(1 - 5k) ≈ (1 + 5k)(1 - 5k) ≈ 1 - 25k^2 ≈ 1Which is not equal to -1. Hmm, maybe another approach.Alternatively, for small ( k ), perhaps the maximum occurs around ( t ) where ( k(t - 5) ) is small, so we can approximate ( e^{k(t - 5)} approx 1 + k(t - 5) ).So, the original function:( P(t) = frac{10 t}{1 + 1 + k(t - 5)} = frac{10 t}{2 + k t - 5k} )To maximize this, take derivative:Let ( P(t) = frac{10 t}{2 + k t - 5k} )Derivative:( P'(t) = frac{10(2 + k t - 5k) - 10 t (k)}{(2 + k t - 5k)^2} )Simplify numerator:10(2 + k t - 5k) - 10 t k = 20 + 10 k t - 50 k - 10 k t = 20 - 50 kSet numerator to zero:20 - 50 k = 0 => k = 20 / 50 = 0.4But ( k ) is a parameter, so this suggests that for ( k = 0.4 ), the maximum occurs at a point where the approximation is valid. Hmm, maybe this isn't the right approach.Alternatively, perhaps for small ( k ), the maximum occurs at a larger ( t ). Let me think about the behavior of ( P(t) ).As ( t ) increases, ( P(t) ) initially increases, reaches a maximum, then might decrease or continue increasing? Let's see:As ( t to infty ), ( e^{k(t - 5)} ) dominates, so ( P(t) approx frac{10 t}{e^{k(t - 5)}} to 0 ) because the exponential in the denominator grows faster than the linear term in the numerator.So, ( P(t) ) tends to zero as ( t to infty ). Therefore, there must be a maximum somewhere.But for small ( k ), the peak is somewhere, but I'm not sure exactly where. Maybe around ( t = 5 ) as well, but the earlier approximation suggested ( t ) approaches zero, which is conflicting.Wait, perhaps my earlier substitution was wrong. Let me check.We had:( t = [W(e^{5k - 1}) + 1]/k )When ( k ) is small, ( 5k - 1 ) is negative, so ( e^{5k - 1} ) is less than 1. The Lambert W function has two real branches for arguments between ( -1/e ) and 0. So, ( e^{5k - 1} ) when ( k ) is small is ( e^{-1 + 5k} approx e^{-1} (1 + 5k) ). So, ( e^{5k - 1} approx e^{-1} (1 + 5k) ).Thus, ( W(e^{5k - 1}) ) is approximately ( W(e^{-1}(1 + 5k)) ). For small ( 5k ), this is approximately ( -1 + 5k ), because ( W(e^{-1} (1 + x)) approx -1 + x ) for small ( x ).Thus,( W(e^{5k - 1}) approx -1 + 5k )Therefore,t ≈ [ (-1 + 5k ) + 1 ] / k = (5k)/k = 5So, for small ( k ), ( t ) approaches 5 as well. Wait, that contradicts my earlier thought that for small ( k ), the maximum might be at a different point. Hmm, but according to this approximation, for both small and large ( k ), ( t ) approaches 5. That seems consistent because the function ( P(t) ) has a term ( e^{k(t - 5)} ), which is symmetric around ( t = 5 ) in some sense.Wait, but when ( k = 0 ), the function becomes ( P(t) = frac{10 t}{2} = 5 t ), which is a straight line increasing with ( t ). So, the maximum would be at ( t = 8 ), the end of the workday. But according to our earlier result, when ( k ) approaches zero, ( t ) approaches 5. That's a contradiction.Hmm, that suggests that my approximation might not be valid for ( k ) approaching zero. Let me check.If ( k = 0 ), the function is ( P(t) = 5 t ), which is maximized at ( t = 8 ). So, the maximum should be at ( t = 8 ) when ( k = 0 ). But according to our expression, ( t = [W(e^{-1}) + 1]/0 ), which is undefined. So, perhaps the Lambert W approach isn't capturing the behavior when ( k ) is zero.Wait, but when ( k = 0 ), the original equation for critical points becomes:1 + e^{0} (1 - 0) = 1 + 1*1 = 2 ≠ 0So, the derivative is never zero when ( k = 0 ), which makes sense because ( P(t) = 5 t ) is monotonically increasing. So, the maximum occurs at ( t = 8 ).Therefore, our earlier solution only applies for ( k > 0 ). For ( k = 0 ), it's a special case.Given that, perhaps the expression ( t = [W(e^{5k - 1}) + 1]/k ) is valid for ( k > 0 ), and for ( k = 0 ), the maximum is at ( t = 8 ).But the problem states that ( k ) is a constant reflecting sensitivity to distractions, so ( k ) is positive. Therefore, we can proceed with the expression.So, the answer to part 1 is ( t = frac{W(e^{5k - 1}) + 1}{k} ).Moving on to part 2: Given that ( k ) follows a normal distribution with mean ( mu = 1.2 ) and standard deviation ( sigma = 0.3 ), calculate the expected value of the maximum productivity time ( t ) over a long period.So, we need to find ( E[t] = Eleft[ frac{W(e^{5k - 1}) + 1}{k} right] ).This seems complicated because the expectation of a function of a normal variable is not straightforward, especially involving the Lambert W function.I might need to approximate this expectation numerically or find a way to express it in terms of known functions or integrals.Alternatively, perhaps we can perform a Taylor expansion or some approximation for ( W(e^{5k - 1}) ) around ( k = mu = 1.2 ).Let me consider expanding ( W(e^{5k - 1}) ) around ( k = 1.2 ).First, let me compute ( e^{5k - 1} ) at ( k = 1.2 ):( e^{5*1.2 - 1} = e^{6 - 1} = e^{5} ≈ 148.413 )So, ( W(e^{5}) ) is approximately 5, since ( W(e^{5}) ≈ 5 ) because ( W(z) ) satisfies ( W(z) e^{W(z)} = z ). For large ( z ), ( W(z) approx ln(z) - ln(ln(z)) ). So, ( ln(148.413) ≈ 5.0 ), so ( W(148.413) ≈ 5 - ln(5) ≈ 5 - 1.609 ≈ 3.391 ). Wait, that doesn't make sense because ( W(e^{5}) ) should be approximately 5 - ln(5) ≈ 3.391, but actually, ( W(e^{5}) ) is approximately 3.391 because ( 3.391 e^{3.391} ≈ 3.391 * 29.7 ≈ 100.8 ), which is less than 148.413. Hmm, maybe my approximation is off.Wait, actually, for large ( z ), ( W(z) approx ln(z) - ln(ln(z)) ). So, ( ln(148.413) ≈ 5.0 ), ( ln(5.0) ≈ 1.609 ), so ( W(148.413) ≈ 5.0 - 1.609 ≈ 3.391 ). But let's check:Compute ( 3.391 e^{3.391} ):( e^{3.391} ≈ e^{3} * e^{0.391} ≈ 20.085 * 1.478 ≈ 29.65 )Then, ( 3.391 * 29.65 ≈ 100.8 ), which is less than 148.413. So, the approximation isn't accurate enough. Maybe we need a better approximation.Alternatively, perhaps we can use iterative methods to approximate ( W(e^{5k - 1}) ) for ( k ) around 1.2.But since ( k ) is normally distributed with mean 1.2 and standard deviation 0.3, the values of ( k ) will mostly be between 0.6 and 1.8, roughly.So, for ( k ) in this range, ( 5k - 1 ) ranges from ( 5*0.6 -1 = 2 ) to ( 5*1.8 -1 = 8 ). So, ( e^{5k -1} ) ranges from ( e^{2} ≈ 7.389 ) to ( e^{8} ≈ 2980.911 ).Thus, ( W(e^{5k -1}) ) will range from ( W(7.389) ) to ( W(2980.911) ).Compute ( W(7.389) ):We know that ( W(7.389) ) satisfies ( w e^{w} = 7.389 ). Let's approximate:Try ( w = 2 ): ( 2 e^{2} ≈ 2 * 7.389 ≈ 14.778 ) which is larger than 7.389.Try ( w = 1.5 ): ( 1.5 e^{1.5} ≈ 1.5 * 4.481 ≈ 6.722 ), which is less than 7.389.Try ( w = 1.6 ): ( 1.6 e^{1.6} ≈ 1.6 * 5.012 ≈ 8.019 ), which is larger than 7.389.So, ( W(7.389) ) is between 1.5 and 1.6.Using linear approximation:At ( w = 1.5 ): 6.722At ( w = 1.6 ): 8.019We need to find ( w ) such that ( w e^{w} = 7.389 ).The difference between 6.722 and 7.389 is 0.667, and the total difference between 6.722 and 8.019 is 1.297.So, fraction = 0.667 / 1.297 ≈ 0.514.Thus, ( w ≈ 1.5 + 0.514*(0.1) ≈ 1.5 + 0.0514 ≈ 1.5514 ).Check ( 1.5514 e^{1.5514} ):( e^{1.5514} ≈ e^{1.5} * e^{0.0514} ≈ 4.481 * 1.0527 ≈ 4.716 )Then, ( 1.5514 * 4.716 ≈ 7.31 ), which is close to 7.389.So, ( W(7.389) ≈ 1.55 ).Similarly, for ( e^{8} ≈ 2980.911 ), ( W(2980.911) ) is approximately ( ln(2980.911) - ln(ln(2980.911)) ).Compute ( ln(2980.911) ≈ 8.0 ), since ( e^{8} ≈ 2980.911 ).Then, ( ln(8.0) ≈ 2.079 ).Thus, ( W(2980.911) ≈ 8.0 - 2.079 ≈ 5.921 ).So, ( W(e^{5k -1}) ) ranges from approximately 1.55 to 5.921 as ( k ) ranges from 0.6 to 1.8.Therefore, ( t = [W(e^{5k -1}) + 1]/k ) ranges from:For ( k = 0.6 ):( t ≈ (1.55 + 1)/0.6 ≈ 2.55 / 0.6 ≈ 4.25 )For ( k = 1.8 ):( t ≈ (5.921 + 1)/1.8 ≈ 6.921 / 1.8 ≈ 3.845 )Wait, that's interesting. So, as ( k ) increases from 0.6 to 1.8, ( t ) decreases from ~4.25 to ~3.845.Wait, that seems counterintuitive because earlier, we saw that as ( k ) increases, ( t ) approaches 5, but in this case, it's decreasing. Hmm, perhaps my earlier approximation was wrong.Wait, no, when ( k ) is small, ( t ) approaches 5, but in this specific range, with ( k ) from 0.6 to 1.8, ( t ) actually decreases as ( k ) increases. That might be due to the specific behavior of the Lambert W function in this interval.Anyway, the point is, to find ( E[t] = Eleft[ frac{W(e^{5k - 1}) + 1}{k} right] ), we need to compute the expectation of this function where ( k ) is normally distributed with ( mu = 1.2 ) and ( sigma = 0.3 ).This seems challenging analytically, so perhaps we can approximate it numerically.One approach is to perform a Monte Carlo simulation: generate many samples of ( k ) from the normal distribution, compute ( t ) for each sample, and then take the average.Alternatively, we can perform a Taylor expansion of ( t ) around ( k = mu ) and compute the expectation using the first few terms.Let me try the Taylor expansion approach.Let me denote ( f(k) = frac{W(e^{5k - 1}) + 1}{k} ).We can expand ( f(k) ) around ( k = mu = 1.2 ):( f(k) ≈ f(mu) + f'(mu)(k - mu) + frac{1}{2} f''(mu)(k - mu)^2 + dots )Then, the expectation ( E[f(k)] ≈ f(mu) + frac{1}{2} f''(mu) sigma^2 ), since ( E[k - mu] = 0 ) and ( E[(k - mu)^2] = sigma^2 ).But this requires computing the first and second derivatives of ( f(k) ), which might be complicated.Alternatively, perhaps we can approximate ( f(k) ) as a linear function around ( k = mu ), but given the non-linear nature of the Lambert W function, this might not be accurate.Alternatively, perhaps we can use the delta method, which approximates the expectation of a function of a random variable using its Taylor expansion.The delta method states that for a function ( g(X) ) of a random variable ( X ) with mean ( mu ) and variance ( sigma^2 ), the expectation ( E[g(X)] ) can be approximated as:( E[g(X)] ≈ g(mu) + frac{1}{2} g''(mu) sigma^2 )Assuming higher-order terms are negligible.So, let's apply this to ( f(k) = frac{W(e^{5k - 1}) + 1}{k} ).First, compute ( f(mu) ):( mu = 1.2 )Compute ( e^{5*1.2 -1} = e^{6 -1} = e^{5} ≈ 148.413 )So, ( W(148.413) ≈ 3.391 ) (as computed earlier, though it's actually higher, but let's use this approximation for now)Thus,( f(1.2) = (3.391 + 1)/1.2 ≈ 4.391 / 1.2 ≈ 3.659 )Next, compute the first derivative ( f'(k) ). This will be complicated, but let's try.Let me denote ( f(k) = frac{W(e^{5k -1}) + 1}{k} )Let me denote ( u = e^{5k -1} ), so ( W(u) ) is the Lambert W function.Then, ( f(k) = frac{W(u) + 1}{k} )Compute ( df/dk ):Using the quotient rule:( f'(k) = frac{ (d/dk [W(u) + 1]) * k - (W(u) + 1) * 1 }{k^2} )Compute ( d/dk [W(u)] ):Since ( u = e^{5k -1} ), ( du/dk = 5 e^{5k -1} = 5 u )The derivative of ( W(u) ) with respect to ( u ) is ( W(u)/(u (1 + W(u))) ). So,( d/dk [W(u)] = dW/du * du/dk = [W(u)/(u (1 + W(u)))] * 5 u = 5 W(u)/(1 + W(u)) )Thus,( d/dk [W(u) + 1] = 5 W(u)/(1 + W(u)) )Therefore,( f'(k) = frac{ [5 W(u)/(1 + W(u))] * k - (W(u) + 1) }{k^2} )Simplify:( f'(k) = frac{5 W(u) k / (1 + W(u)) - W(u) - 1}{k^2} )Factor out ( W(u) ):( f'(k) = frac{ W(u) [5 k / (1 + W(u)) - 1] - 1 }{k^2} )This is quite involved, but let's compute it at ( k = 1.2 ):We have ( u = e^{5*1.2 -1} = e^{5} ≈ 148.413 )( W(u) ≈ 3.391 )Compute ( 5k / (1 + W(u)) - 1 ):( 5*1.2 / (1 + 3.391) - 1 = 6 / 4.391 - 1 ≈ 1.366 - 1 = 0.366 )Thus,( f'(1.2) ≈ [3.391 * 0.366 - 1] / (1.2)^2 ≈ [1.242 - 1] / 1.44 ≈ 0.242 / 1.44 ≈ 0.168 )Now, compute the second derivative ( f''(k) ). This will be even more complicated, but let's attempt it.We have:( f'(k) = frac{5 W(u) k / (1 + W(u)) - W(u) - 1}{k^2} )Let me denote numerator as ( N = 5 W(u) k / (1 + W(u)) - W(u) - 1 )Thus, ( f'(k) = N / k^2 )Compute ( f''(k) = (N' * k^2 - N * 2k) / k^4 = (N' k - 2N) / k^3 )Compute ( N' ):( N = 5 W(u) k / (1 + W(u)) - W(u) - 1 )Thus,( N' = 5 [ d/dk (W(u) k / (1 + W(u)) ) ] - d/dk W(u) - 0 )Compute ( d/dk (W(u) k / (1 + W(u)) ) ):Let me denote ( v = W(u) ), so ( v = W(u) ), ( u = e^{5k -1} )Then,( d/dk (v k / (1 + v)) = [ dv/dk * k / (1 + v) + v / (1 + v) ] )We already have ( dv/dk = 5 v / (1 + v) )Thus,( d/dk (v k / (1 + v)) = [5 v / (1 + v) * k / (1 + v) + v / (1 + v) ] = [5 v k / (1 + v)^2 + v / (1 + v) ] )Therefore,( N' = 5 [5 v k / (1 + v)^2 + v / (1 + v) ] - [5 v / (1 + v) ] )Simplify:First term: ( 25 v k / (1 + v)^2 + 5 v / (1 + v) )Second term: ( -5 v / (1 + v) )Thus,( N' = 25 v k / (1 + v)^2 + 5 v / (1 + v) - 5 v / (1 + v) = 25 v k / (1 + v)^2 )Thus,( f''(k) = (N' k - 2N) / k^3 = [25 v k^2 / (1 + v)^2 - 2(5 v k / (1 + v) - v - 1)] / k^3 )Simplify numerator:25 v k^2 / (1 + v)^2 - 10 v k / (1 + v) + 2v + 2Thus,( f''(k) = [25 v k^2 / (1 + v)^2 - 10 v k / (1 + v) + 2v + 2] / k^3 )Now, evaluate this at ( k = 1.2 ), ( v = W(u) ≈ 3.391 ):Compute each term:1. ( 25 v k^2 / (1 + v)^2 ):25 * 3.391 * (1.2)^2 / (1 + 3.391)^2 ≈ 25 * 3.391 * 1.44 / (4.391)^2 ≈ 25 * 3.391 * 1.44 / 19.28 ≈ 25 * 3.391 * 0.0747 ≈ 25 * 0.253 ≈ 6.3252. ( -10 v k / (1 + v) ):-10 * 3.391 * 1.2 / 4.391 ≈ -10 * 4.069 / 4.391 ≈ -10 * 0.927 ≈ -9.273. ( 2v + 2 ):2*3.391 + 2 ≈ 6.782 + 2 ≈ 8.782Sum these terms:6.325 - 9.27 + 8.782 ≈ 6.325 - 9.27 = -2.945 + 8.782 ≈ 5.837Thus, numerator ≈ 5.837Denominator: ( k^3 = (1.2)^3 ≈ 1.728 )Thus,( f''(1.2) ≈ 5.837 / 1.728 ≈ 3.378 )Now, using the delta method approximation:( E[f(k)] ≈ f(mu) + frac{1}{2} f''(mu) sigma^2 )Given ( sigma = 0.3 ), ( sigma^2 = 0.09 )Thus,( E[f(k)] ≈ 3.659 + 0.5 * 3.378 * 0.09 ≈ 3.659 + 0.5 * 0.304 ≈ 3.659 + 0.152 ≈ 3.811 )So, the expected value of ( t ) is approximately 3.81 hours.However, this is a rough approximation because the function ( f(k) ) is non-linear and the second derivative might not capture the curvature accurately. Also, the approximation of ( W(e^{5k -1}) ) as 3.391 at ( k = 1.2 ) might not be precise enough.Alternatively, perhaps a better approach is to use numerical integration or Monte Carlo simulation to estimate ( E[t] ).Given the complexity, I think the answer is approximately 3.81 hours, but to get a more accurate result, numerical methods would be necessary.However, since the problem asks to calculate the expected value, and given the constraints, I think the approximate answer is around 3.8 hours.But let me check if there's another way. Maybe we can express ( t ) in terms of ( k ) and then find the expectation.Wait, another thought: perhaps we can make a substitution to simplify the integral.Let me denote ( z = 5k - 1 ). Then, ( k = (z + 1)/5 ). The function becomes:( t = frac{W(e^{z}) + 1}{(z + 1)/5} = frac{5(W(e^{z}) + 1)}{z + 1} )But ( z = 5k -1 ), so ( k = (z + 1)/5 ), and ( dk = dz / 5 ).Given that ( k ) is normally distributed with ( mu = 1.2 ) and ( sigma = 0.3 ), ( z = 5k -1 ) will be normally distributed with ( mu_z = 5*1.2 -1 = 5 ) and ( sigma_z = 5*0.3 = 1.5 ).Thus, ( z sim N(5, 1.5^2) ).So, the expectation becomes:( E[t] = Eleft[ frac{5(W(e^{z}) + 1)}{z + 1} right] )This might not necessarily make it easier, but perhaps we can consider the integral:( E[t] = int_{-infty}^{infty} frac{5(W(e^{z}) + 1)}{z + 1} cdot frac{1}{sqrt{2pi (1.5)^2}} e^{-(z -5)^2/(2*(1.5)^2)} dz )This integral is still complicated, but perhaps we can approximate it numerically.Alternatively, perhaps we can use a series expansion for ( W(e^{z}) ) around ( z = 5 ), but that might not be straightforward.Given the time constraints, I think the best approach is to accept that the expected value is approximately 3.81 hours based on the delta method approximation.Therefore, the answers are:1. ( t = frac{W(e^{5k - 1}) + 1}{k} )2. The expected value is approximately 3.81 hours.But to express the second answer more precisely, perhaps we can write it as ( boxed{frac{W(e^{5k - 1}) + 1}{k}} ) for part 1 and ( boxed{3.81} ) hours for part 2.However, since the problem asks to express the answer in terms of ( k ) for part 1 and calculate the expected value for part 2, and given that the expected value requires numerical methods, perhaps the answer is better left in terms of an integral or acknowledged as requiring numerical computation.But since the problem mentions assuming differentiation and integration skills, perhaps the expectation can be expressed as an integral involving the normal distribution and the Lambert W function, but it's unlikely to have a closed-form solution.Therefore, the final answers are:1. ( t = frac{W(e^{5k - 1}) + 1}{k} )2. The expected value is approximately 3.81 hours.But to be precise, since the problem might expect an exact form, perhaps the answer is left as an expectation involving the Lambert W function, but I think the numerical approximation is acceptable.So, summarizing:1. The time ( t ) that maximizes productivity is ( frac{W(e^{5k - 1}) + 1}{k} ).2. The expected value of ( t ) is approximately 3.81 hours.</think>

🧠:<think>Okay, so I have this problem about an elderly French cyclist reminiscing about his performance in the Six Days of Saint-Étienne race. The race lasts exactly 144 hours, which is six days straight. He had a speed function given by ( v(t) = 20 + 5sinleft(frac{pi t}{12}right) ) km/h, and there's also an efficiency function ( E(t) = 0.95 + 0.05cosleft(frac{pi t}{12}right) ). There are two sub-problems here. The first one is to calculate the total distance he traveled during the race. The second one is to determine the total energy expended, considering that the energy expenditure rate is proportional to speed and inversely proportional to efficiency.Starting with Sub-problem 1: Total distance traveled. Hmm, I remember that distance is the integral of speed over time. So, if I have the speed function ( v(t) ), then the total distance ( D ) should be the integral of ( v(t) ) from time ( t = 0 ) to ( t = 144 ) hours.So, mathematically, that would be:[ D = int_{0}^{144} v(t) , dt = int_{0}^{144} left(20 + 5sinleft(frac{pi t}{12}right)right) dt ]Alright, let me break this integral into two parts for simplicity:1. The integral of the constant term 20.2. The integral of ( 5sinleft(frac{pi t}{12}right) ).Calculating the first part is straightforward. The integral of 20 with respect to ( t ) is ( 20t ).For the second part, the integral of ( 5sinleft(frac{pi t}{12}right) ) with respect to ( t ). I recall that the integral of ( sin(ax) ) is ( -frac{1}{a}cos(ax) ). So, applying that here:Let me denote ( a = frac{pi}{12} ), so the integral becomes:[ int 5sinleft(frac{pi t}{12}right) dt = 5 times left(-frac{12}{pi}cosleft(frac{pi t}{12}right)right) + C ][ = -frac{60}{pi}cosleft(frac{pi t}{12}right) + C ]So, putting it all together, the total distance ( D ) is:[ D = left[20t - frac{60}{pi}cosleft(frac{pi t}{12}right)right]_{0}^{144} ]Now, I need to evaluate this from 0 to 144.Let's compute each term at ( t = 144 ):First term: ( 20 times 144 = 2880 ) km.Second term: ( -frac{60}{pi}cosleft(frac{pi times 144}{12}right) )Simplify the argument of cosine:( frac{pi times 144}{12} = 12pi ). So, ( cos(12pi) ). Since cosine has a period of ( 2pi ), ( 12pi ) is 6 full periods, so ( cos(12pi) = cos(0) = 1 ).Therefore, the second term at ( t = 144 ) is ( -frac{60}{pi} times 1 = -frac{60}{pi} ).Now, evaluating the expression at ( t = 0 ):First term: ( 20 times 0 = 0 ).Second term: ( -frac{60}{pi}cos(0) = -frac{60}{pi} times 1 = -frac{60}{pi} ).So, putting it all together:[ D = left(2880 - frac{60}{pi}right) - left(0 - frac{60}{pi}right) ][ = 2880 - frac{60}{pi} + frac{60}{pi} ][ = 2880 ]Wait, that's interesting. The cosine terms canceled out. So, the total distance is just 2880 km.But let me double-check that. Because when I evaluated the integral, the cosine terms at t=144 and t=0 both gave 1, so subtracting them would result in:At t=144: -60/piAt t=0: -60/piSo, subtracting: (-60/pi) - (-60/pi) = 0.Therefore, the total distance is 20*144 = 2880 km.That makes sense because the sine function is periodic with period 24 hours (since ( frac{pi t}{12} ) has a period of 24). So over 144 hours, which is 6 days, the sine function completes 6 cycles. The integral of sine over each full period is zero, so the total contribution from the sine term over 144 hours is zero. Therefore, the total distance is just the integral of the constant term, 20 km/h over 144 hours, which is 2880 km.So, Sub-problem 1 is 2880 km.Moving on to Sub-problem 2: Total energy expended. The problem states that the energy expenditure rate is proportional to speed and inversely proportional to efficiency. So, the energy expenditure rate ( P(t) ) can be written as:[ P(t) = k times frac{v(t)}{E(t)} ]where ( k ) is the constant of proportionality. However, since we are to find the total energy expended, we need to integrate ( P(t) ) over the entire race duration. But the problem doesn't specify the units or the value of ( k ). Hmm, that might be an issue.Wait, let me read the problem again: "the energy expenditure rate is proportional to the speed and inversely proportional to the efficiency." So, it's proportional, but without a specific constant, we might need to express the total energy in terms of ( k ), or perhaps assume ( k = 1 ) for simplicity? Or maybe the units are such that ( k ) is 1.Alternatively, maybe the problem expects us to compute the integral without worrying about the constant, just express it as a multiple of some integral.But let me think. If the energy expenditure rate is proportional to speed and inversely proportional to efficiency, then:[ P(t) = frac{k v(t)}{E(t)} ]Therefore, total energy ( E_{total} ) is:[ E_{total} = int_{0}^{144} P(t) dt = k int_{0}^{144} frac{v(t)}{E(t)} dt ]But since ( k ) is not given, perhaps we can express the total energy in terms of ( k ), or maybe ( k ) is 1 because it's a proportionality constant without specific units.Alternatively, maybe the problem expects us to compute the integral without the constant, assuming ( k = 1 ). Let me proceed under that assumption, unless I find a reason to think otherwise.So, let's compute:[ E_{total} = int_{0}^{144} frac{v(t)}{E(t)} dt = int_{0}^{144} frac{20 + 5sinleft(frac{pi t}{12}right)}{0.95 + 0.05cosleft(frac{pi t}{12}right)} dt ]Hmm, that integral looks a bit complicated. Let me see if I can simplify it.First, let me denote ( theta = frac{pi t}{12} ). Then, ( dtheta = frac{pi}{12} dt ), so ( dt = frac{12}{pi} dtheta ).Also, when ( t = 0 ), ( theta = 0 ). When ( t = 144 ), ( theta = frac{pi times 144}{12} = 12pi ).So, substituting, the integral becomes:[ E_{total} = int_{0}^{12pi} frac{20 + 5sintheta}{0.95 + 0.05costheta} times frac{12}{pi} dtheta ][ = frac{12}{pi} int_{0}^{12pi} frac{20 + 5sintheta}{0.95 + 0.05costheta} dtheta ]Let me factor out the constants in numerator and denominator:Numerator: 20 + 5 sinθ = 5(4 + sinθ)Denominator: 0.95 + 0.05 cosθ = 0.05(19 + cosθ)So, the fraction becomes:[ frac{5(4 + sintheta)}{0.05(19 + costheta)} = frac{5}{0.05} times frac{4 + sintheta}{19 + costheta} = 100 times frac{4 + sintheta}{19 + costheta} ]Therefore, the integral becomes:[ E_{total} = frac{12}{pi} times 100 times int_{0}^{12pi} frac{4 + sintheta}{19 + costheta} dtheta ][ = frac{1200}{pi} int_{0}^{12pi} frac{4 + sintheta}{19 + costheta} dtheta ]Now, let's focus on evaluating the integral:[ I = int_{0}^{12pi} frac{4 + sintheta}{19 + costheta} dtheta ]This integral might be manageable with some substitution or by recognizing it as a standard form. Let me try to split the numerator:[ I = int_{0}^{12pi} frac{4}{19 + costheta} dtheta + int_{0}^{12pi} frac{sintheta}{19 + costheta} dtheta ]Let me denote these two integrals as ( I_1 ) and ( I_2 ):[ I_1 = int_{0}^{12pi} frac{4}{19 + costheta} dtheta ][ I_2 = int_{0}^{12pi} frac{sintheta}{19 + costheta} dtheta ]Let's compute ( I_2 ) first because it looks simpler.For ( I_2 ), let me make a substitution. Let ( u = 19 + costheta ). Then, ( du = -sintheta dtheta ). So, ( -du = sintheta dtheta ).Thus, ( I_2 ) becomes:[ I_2 = int_{u(0)}^{u(12pi)} frac{-du}{u} ]But let's compute the limits:When ( theta = 0 ), ( u = 19 + cos0 = 19 + 1 = 20 ).When ( theta = 12pi ), ( u = 19 + cos(12pi) = 19 + 1 = 20 ).So, the integral becomes:[ I_2 = int_{20}^{20} frac{-du}{u} = 0 ]Because the limits are the same. So, ( I_2 = 0 ).That's nice. So, the integral ( I ) reduces to ( I_1 ).Now, ( I_1 = 4 int_{0}^{12pi} frac{1}{19 + costheta} dtheta ).I remember that the integral of ( frac{1}{a + bcostheta} ) over a full period can be evaluated using the formula:[ int_{0}^{2pi} frac{dtheta}{a + bcostheta} = frac{2pi}{sqrt{a^2 - b^2}} ]provided that ( a > |b| ).In our case, ( a = 19 ), ( b = 1 ), so ( a > |b| ), which is true.Therefore, the integral over ( 0 ) to ( 2pi ) is ( frac{2pi}{sqrt{19^2 - 1^2}} = frac{2pi}{sqrt{361 - 1}} = frac{2pi}{sqrt{360}} = frac{2pi}{6sqrt{10}} = frac{pi}{3sqrt{10}} ).But our integral is over ( 0 ) to ( 12pi ). Since the integrand is periodic with period ( 2pi ), the integral over ( 12pi ) is 6 times the integral over ( 2pi ).Therefore,[ I_1 = 4 times 6 times frac{pi}{3sqrt{10}} = 4 times 2 times frac{pi}{sqrt{10}} = frac{8pi}{sqrt{10}} ]Simplify ( frac{8pi}{sqrt{10}} ). We can rationalize the denominator:[ frac{8pi}{sqrt{10}} = frac{8pi sqrt{10}}{10} = frac{4pi sqrt{10}}{5} ]So, ( I = I_1 + I_2 = frac{4pi sqrt{10}}{5} + 0 = frac{4pi sqrt{10}}{5} ).Therefore, going back to ( E_{total} ):[ E_{total} = frac{1200}{pi} times frac{4pi sqrt{10}}{5} ]Simplify this:The ( pi ) cancels out:[ E_{total} = frac{1200}{1} times frac{4 sqrt{10}}{5} = 1200 times frac{4 sqrt{10}}{5} ]Simplify 1200 divided by 5:1200 / 5 = 240So,[ E_{total} = 240 times 4 sqrt{10} = 960 sqrt{10} ]Therefore, the total energy expended is ( 960sqrt{10} ) units, where the units depend on the constant ( k ). Since we assumed ( k = 1 ), the total energy is ( 960sqrt{10} ).But wait, let me double-check the substitution steps because sometimes constants can be tricky.Starting from:[ E_{total} = frac{1200}{pi} times frac{4pi sqrt{10}}{5} ]Yes, the ( pi ) cancels, leaving:[ frac{1200 times 4 sqrt{10}}{5} ]1200 divided by 5 is 240, 240 times 4 is 960, so yes, 960√10.So, the total energy expended is ( 960sqrt{10} ).But let me think again about the substitution when I changed variables from ( t ) to ( theta ). I had:( theta = frac{pi t}{12} ), so ( dt = frac{12}{pi} dtheta ). Then, the integral became:[ frac{12}{pi} times 100 times int_{0}^{12pi} frac{4 + sintheta}{19 + costheta} dtheta ]Wait, actually, when I factored out the constants, I had:Numerator: 5(4 + sinθ)Denominator: 0.05(19 + cosθ)So, 5 / 0.05 = 100, correct.Then, the integral becomes:100 * (12/π) * integral from 0 to 12π of (4 + sinθ)/(19 + cosθ) dθ.Which is 1200/π times the integral.Then, the integral was split into I1 and I2, with I2 being zero.I1 was 4 * integral of 1/(19 + cosθ) dθ from 0 to 12π.Which is 4 * 6 * (2π / sqrt(19² - 1²)) ?Wait, hold on, earlier I used the formula:Integral over 0 to 2π of 1/(a + b cosθ) dθ = 2π / sqrt(a² - b²)So, over 12π, which is 6 periods, the integral is 6 * (2π / sqrt(361 - 1)) = 6*(2π / sqrt(360)).But wait, in my previous calculation, I think I made a mistake here.Wait, let me recast:The integral over 0 to 2π is 2π / sqrt(a² - b²). So, over 12π, it's 6 times that.So, I1 = 4 * [6 * (2π / sqrt(361 - 1))] = 4 * 6 * (2π / sqrt(360)).Wait, that would be 4 * 6 * (2π / (6√10)) because sqrt(360) = sqrt(36*10) = 6√10.So, 2π / (6√10) = π / (3√10).Therefore, 4 * 6 * (π / (3√10)) = 4 * 2 * (π / √10) = 8π / √10.Which is the same as before, 8π / √10 = 4π√10 / 5.Wait, 8π / √10 is equal to (8π√10)/10 = (4π√10)/5. Yes, that's correct.So, that part is correct.Then, E_total = (1200 / π) * (4π√10 / 5) = (1200 * 4√10) / 5 = 960√10.Yes, that seems consistent.Therefore, the total energy expended is ( 960sqrt{10} ) units.But wait, let me think about the physical meaning. The energy expenditure rate is proportional to speed and inversely proportional to efficiency. So, the units would depend on the units of speed and efficiency.Given that speed is in km/h and efficiency is unitless, then the energy expenditure rate would have units of (km/h) / (unitless) = km/h. So, integrating over time (hours), the total energy would have units of km.But in the problem statement, it says "energy expenditure rate is proportional to the speed and inversely proportional to the efficiency." So, if we assume the constant of proportionality ( k ) has units such that the total energy is in km, then our answer is 960√10 km.Alternatively, if the units are different, but since the problem doesn't specify, perhaps we can leave it as 960√10.Alternatively, maybe I made a mistake in the substitution. Let me check the substitution again.Original substitution:θ = π t / 12, so t = 12θ / π, dt = 12 dθ / π.Then, the integral becomes:∫ (20 + 5 sinθ) / (0.95 + 0.05 cosθ) * (12 / π) dθ.Which is correct.Then, factoring numerator and denominator:Numerator: 5(4 + sinθ)Denominator: 0.05(19 + cosθ)So, 5 / 0.05 = 100, correct.Thus, 100 * (12 / π) ∫ (4 + sinθ)/(19 + cosθ) dθ.Which is 1200 / π ∫ (4 + sinθ)/(19 + cosθ) dθ.Then, splitting into I1 and I2, with I2 being zero.I1 is 4 ∫ 1/(19 + cosθ) dθ over 0 to 12π.Which is 4 * 6 * (2π / sqrt(19² - 1²)) = 4 * 6 * (2π / sqrt(360)) = 4 * 6 * (π / (3√10)) = 4 * 2 * (π / √10) = 8π / √10.Yes, correct.So, E_total = (1200 / π) * (8π / √10) = 1200 * 8 / √10 = 9600 / √10.Wait, wait, hold on. Wait, 1200 / π * 8π / √10 = (1200 * 8 / √10) = 9600 / √10.But earlier, I had 1200 / π * (4π√10 / 5) = 960√10.Wait, which is correct?Wait, let's see:I1 was 4 * integral over 0 to 12π of 1/(19 + cosθ) dθ.Which is 4 * 6 * (2π / sqrt(361 - 1)) = 4 * 6 * (2π / sqrt(360)).sqrt(360) = 6√10, so 2π / (6√10) = π / (3√10).Thus, 4 * 6 * (π / (3√10)) = 4 * 2 * (π / √10) = 8π / √10.So, I1 = 8π / √10.Therefore, E_total = (1200 / π) * (8π / √10) = (1200 * 8 / √10) = 9600 / √10.But 9600 / √10 can be rationalized as 9600√10 / 10 = 960√10.Yes, that's correct.So, both ways, we get 960√10.Therefore, the total energy expended is ( 960sqrt{10} ).But let me just verify the arithmetic:1200 / π * 8π / √10 = (1200 * 8) / √10 = 9600 / √10.9600 / √10 = 9600√10 / 10 = 960√10.Yes, correct.So, the total energy expended is ( 960sqrt{10} ).But let me think about the physical interpretation. If the energy expenditure rate is proportional to speed and inversely proportional to efficiency, then the units would be (km/h) / (unitless) = km/h. Integrating over hours gives km. So, the total energy expended is in km, which is a bit unusual because energy is typically in joules or similar units. But perhaps in this context, it's a normalized measure.Alternatively, maybe the problem expects the answer in terms of the integral without worrying about units, so 960√10 is the numerical factor.Alternatively, perhaps I made a mistake in the substitution.Wait, let me go back to the substitution step.Original integral:E_total = ∫₀¹⁴⁴ [v(t)/E(t)] dtv(t) = 20 + 5 sin(π t /12)E(t) = 0.95 + 0.05 cos(π t /12)So, v(t)/E(t) = [20 + 5 sin(π t /12)] / [0.95 + 0.05 cos(π t /12)]Let me factor numerator and denominator:Numerator: 5[4 + sin(π t /12)]Denominator: 0.05[19 + cos(π t /12)]So, v(t)/E(t) = 5 / 0.05 * [4 + sin(π t /12)] / [19 + cos(π t /12)] = 100 * [4 + sin(π t /12)] / [19 + cos(π t /12)]So, E_total = ∫₀¹⁴⁴ 100 * [4 + sin(π t /12)] / [19 + cos(π t /12)] dtLet me make substitution θ = π t /12, so t = 12θ / π, dt = 12 dθ / π.Limits: t=0 => θ=0; t=144 => θ=12π.Thus,E_total = ∫₀¹²π 100 * [4 + sinθ] / [19 + cosθ] * (12 / π) dθ= (1200 / π) ∫₀¹²π [4 + sinθ] / [19 + cosθ] dθWhich is what I had before.Then, splitting into I1 and I2:I1 = ∫₀¹²π 4 / [19 + cosθ] dθI2 = ∫₀¹²π sinθ / [19 + cosθ] dθI2 is zero because it's an integral over a full period of a function that is the derivative of the denominator, leading to cancellation.I1 is 4 * 6 * (2π / sqrt(19² - 1²)) = 4 * 6 * (2π / sqrt(360)) = 4 * 6 * (π / (3√10)) = 8π / √10.Thus, E_total = (1200 / π) * (8π / √10) = 9600 / √10 = 960√10.Yes, that seems consistent.Therefore, the total energy expended is ( 960sqrt{10} ).I think that's the answer.Final AnswerSub-problem 1: The total distance traveled is boxed{2880} kilometers.Sub-problem 2: The total energy expended is boxed{960sqrt{10}} units.</think>