🧠:<think>Alright, so I have this problem about Alex, a mechanical engineer restoring a classic car. He wants to optimize the engine's performance without changing the original design. There are two parts to this problem: calculating the engine's displacement volume and maximum theoretical power, and then determining the optimal ignition timing angle for fuel efficiency. Let me tackle each part step by step.Starting with the first part: calculating the engine's displacement volume. The engine is a single-cylinder four-stroke engine with a bore diameter of 88 mm and a stroke length of 89 mm. I remember that the displacement volume for a piston engine is calculated using the formula:[ V_d = pi times left( frac{bore}{2} right)^2 times stroke ]But wait, I need to make sure all units are consistent. The bore and stroke are given in millimeters, so I should convert them to meters since the formula requires cubic meters for volume.Bore diameter is 88 mm, so the radius is half of that, which is 44 mm or 0.044 meters. The stroke is 89 mm, which is 0.089 meters.Plugging these into the formula:[ V_d = pi times (0.044)^2 times 0.089 ]Let me compute that. First, square the radius:0.044 squared is 0.001936.Multiply that by the stroke:0.001936 * 0.089 = 0.000172384.Then multiply by π (approximately 3.1416):0.000172384 * 3.1416 ≈ 0.0005416 cubic meters.So the displacement volume is approximately 0.0005416 m³. But the question asks for the displacement in cubic centimeters (cc). Since 1 cubic meter is 1,000,000 cubic centimeters, I can convert:0.0005416 m³ * 1,000,000 cm³/m³ = 541.6 cm³.So the displacement volume is about 541.6 cubic centimeters. That seems reasonable for a vintage engine.Now, moving on to calculating the maximum theoretical power output. The formula given is:[ P = frac{2pi cdot n cdot V_d cdot P_{mep}}{60} ]Where:- ( n ) is RPM (6000)- ( V_d ) is displacement volume in cubic meters (0.0005416 m³)- ( P_{mep} ) is mean effective pressure in pascals (850 kPa)First, convert the mean effective pressure from kPa to Pa. Since 1 kPa = 1000 Pa, 850 kPa is 850,000 Pa.Now plug the numbers into the formula:[ P = frac{2pi cdot 6000 cdot 0.0005416 cdot 850000}{60} ]Let me compute this step by step.First, calculate the numerator:2 * π ≈ 6.28326.2832 * 6000 = 37,699.237,699.2 * 0.0005416 ≈ Let's compute that:0.0005416 * 37,699.2 ≈ 20.36Then, 20.36 * 850,000 ≈ 17,296,000So the numerator is approximately 17,296,000.Now divide by 60:17,296,000 / 60 ≈ 288,266.67So the power P is approximately 288,266.67 watts. To convert that to kilowatts, divide by 1000:288.26667 kW.So the maximum theoretical power output is approximately 288.27 kW.Wait, that seems quite high for a single-cylinder engine, even with a high RPM and MEP. Let me double-check my calculations.First, displacement volume: 88 mm bore, 89 mm stroke.Radius: 44 mm = 0.044 mStroke: 0.089 mVolume: π * (0.044)^2 * 0.089Calculating (0.044)^2: 0.0019360.001936 * 0.089 = 0.000172384Multiply by π: 0.000172384 * 3.1416 ≈ 0.0005416 m³. That seems correct.Convert to cc: 0.0005416 * 1,000,000 = 541.6 cc. Correct.MEP: 850 kPa = 850,000 Pa. Correct.Power formula:(2 * π * 6000 * 0.0005416 * 850,000) / 60Compute step by step:2 * π ≈ 6.28326.2832 * 6000 = 37,699.237,699.2 * 0.0005416 ≈ Let's compute 37,699.2 * 0.0005 = 18.849637,699.2 * 0.0000416 ≈ 37,699.2 * 0.00004 = 1.507968So total ≈ 18.8496 + 1.507968 ≈ 20.357568Then 20.357568 * 850,000 ≈ 20.357568 * 800,000 = 16,286,054.420.357568 * 50,000 = 1,017,878.4Total ≈ 16,286,054.4 + 1,017,878.4 ≈ 17,303,932.8Divide by 60: 17,303,932.8 / 60 ≈ 288,398.88 W ≈ 288.4 kWYes, that seems consistent. So despite seeming high, the calculation checks out. Maybe it's a high-performance engine.Moving on to the second part: optimizing ignition timing to minimize fuel consumption. The current timing is 12 degrees before TDC, but Alex wants to find the optimal angle θ that minimizes fuel consumption rate of 230 grams per kilowatt-hour. The specific fuel consumption is given by:[ E(theta) = 0.25theta^2 - 12theta + 250 ]Where E(θ) is in grams per kilowatt-hour, and θ is the ignition timing angle in degrees.We need to find the θ that minimizes E(θ). Since this is a quadratic function in terms of θ, it will have a minimum at the vertex of the parabola.For a quadratic function ( ax^2 + bx + c ), the vertex occurs at ( x = -frac{b}{2a} ).Here, a = 0.25, b = -12.So θ = -(-12)/(2*0.25) = 12 / 0.5 = 24 degrees.Wait, so the optimal ignition timing angle is 24 degrees before TDC? But the current timing is 12 degrees before TDC. So Alex needs to advance the timing by another 12 degrees to 24 degrees BTDC.But wait, let me make sure. The function E(θ) is given as 0.25θ² -12θ +250. So the coefficient of θ² is positive, meaning the parabola opens upwards, so the vertex is indeed the minimum point.Calculating θ = -b/(2a) = -(-12)/(2*0.25) = 12 / 0.5 = 24 degrees.So yes, θ = 24 degrees. So Alex should set the ignition timing to 24 degrees before TDC to minimize fuel consumption.But wait, the problem mentions maintaining an average engine temperature of 95°C. Does this affect the calculation? The equation given doesn't include temperature, so perhaps it's just additional context. Maybe in a more detailed model, temperature would influence fuel consumption, but here, since it's not part of the equation, we can ignore it for this optimization.Therefore, the optimal ignition timing angle is 24 degrees before TDC.Let me recap:1. Displacement volume: 541.6 cc, Power ≈ 288.4 kW.2. Optimal ignition timing: 24 degrees BTDC.I think that's it. The calculations seem straightforward, but let me just verify the power calculation one more time because 288 kW seems high. Maybe I made a unit conversion error.Wait, displacement volume was 0.0005416 m³. MEP is 850,000 Pa. RPM is 6000.Power formula:P = (2π * n * Vd * P_mep) / 60Plugging in:(2 * π * 6000 * 0.0005416 * 850000) / 60Calculate step by step:First, 2 * π ≈ 6.28326.2832 * 6000 ≈ 37,699.237,699.2 * 0.0005416 ≈ 20.35720.357 * 850,000 ≈ 17,298,45017,298,450 / 60 ≈ 288,307.5 W ≈ 288.31 kWYes, that's consistent. So unless the formula is incorrect, which I don't think it is, the power is indeed around 288 kW. Maybe it's a very high-performance engine, or perhaps the MEP is unusually high.Alternatively, sometimes MEP is given in bar, but here it's in kPa, so 850 kPa is 8.5 bar, which is quite high for a naturally aspirated engine but possible for a racing engine.So, I think the calculations are correct.Final Answer1. The engine's displacement volume is boxed{541.6} cubic centimeters, and the maximum theoretical power output is boxed{288.3} kilowatts.2. The optimal ignition timing angle is boxed{24} degrees before top dead center.</think>

🧠:<think>Okay, so I have this problem about a Stratovarius concert. The fan went to a concert where they played 12 songs. Each song has a duration, and between each song, there's an interlude of 2 minutes. The total concert time, including all the interludes, was 120 minutes. Part 1 asks me to formulate a system of equations representing the total concert time. Hmm, let's break this down. There are 12 songs, each with their own durations: ( d_1, d_2, ldots, d_{12} ). Between each song, there's a 2-minute interlude. So, how many interludes are there? Well, if there are 12 songs, there are 11 interludes between them, right? Because after the 12th song, there's no interlude needed. So, the total concert time is the sum of all the song durations plus the sum of all the interludes. The interludes are all 2 minutes each, so the total interlude time is ( 11 times 2 = 22 ) minutes. Therefore, the total concert time equation would be the sum of all song durations plus 22 minutes equals 120 minutes. In mathematical terms, that would be:[d_1 + d_2 + d_3 + ldots + d_{12} + 22 = 120]But to write this more concisely, I can use summation notation. So, the sum of all ( d_i ) from ( i = 1 ) to ( 12 ) plus 22 equals 120. That would be:[sum_{i=1}^{12} d_i + 22 = 120]So, that's the system of equations for part 1. It's a single equation because we only have one total time given, but it involves 12 variables, which are the durations of each song.Moving on to part 2. It says that the durations of the songs form an arithmetic progression. The first song ( d_1 ) is 5 minutes, and the common difference ( d ) is unknown. I need to find the value of ( d ) and ( d_{12} ).Alright, arithmetic progression. So, in an arithmetic progression, each term is the previous term plus a common difference. So, ( d_2 = d_1 + d ), ( d_3 = d_2 + d = d_1 + 2d ), and so on, up to ( d_{12} = d_1 + 11d ).Given that ( d_1 = 5 ), so ( d_1 = 5 ). Therefore, the durations are 5, 5 + d, 5 + 2d, ..., up to 5 + 11d.Now, from part 1, we have the equation:[sum_{i=1}^{12} d_i + 22 = 120]Which simplifies to:[sum_{i=1}^{12} d_i = 120 - 22 = 98]So, the sum of the song durations is 98 minutes.Since the durations form an arithmetic progression, the sum of an arithmetic progression can be calculated using the formula:[S_n = frac{n}{2} times (2a + (n - 1)d)]Where:- ( S_n ) is the sum of the first ( n ) terms,- ( a ) is the first term,- ( d ) is the common difference,- ( n ) is the number of terms.In this case, ( n = 12 ), ( a = 5 ), and ( S_n = 98 ). Plugging these into the formula:[98 = frac{12}{2} times (2 times 5 + (12 - 1)d)]Simplify step by step. First, ( frac{12}{2} = 6 ). So,[98 = 6 times (10 + 11d)]Divide both sides by 6 to simplify:[frac{98}{6} = 10 + 11d]Calculating ( frac{98}{6} ). Let me do that division. 6 goes into 98 sixteen times because 16*6=96, with a remainder of 2. So, ( frac{98}{6} = 16 frac{1}{3} ) or approximately 16.3333.So,[16 frac{1}{3} = 10 + 11d]Subtract 10 from both sides:[16 frac{1}{3} - 10 = 11d]Which is:[6 frac{1}{3} = 11d]Convert ( 6 frac{1}{3} ) to an improper fraction. That's ( frac{19}{3} ).So,[frac{19}{3} = 11d]To solve for ( d ), divide both sides by 11:[d = frac{19}{3 times 11} = frac{19}{33}]Simplify ( frac{19}{33} ). Hmm, 19 is a prime number, so it can't be reduced further. So, ( d = frac{19}{33} ) minutes. That's approximately 0.5758 minutes, which is about 34.55 seconds. Seems a bit short, but maybe it's correct.Wait, let me double-check my calculations because 19/33 seems a bit odd. Let me go back.We had:[98 = 6 times (10 + 11d)]So, 98 divided by 6 is indeed 16 and 1/3. Then, subtracting 10 gives 6 and 1/3, which is 19/3. So, 19/3 equals 11d. Therefore, d is 19/(3*11) = 19/33. So, that seems correct.Alternatively, maybe I made a mistake in the formula. Let me recall the arithmetic series sum formula. It is ( S_n = frac{n}{2}(a_1 + a_n) ). Alternatively, ( S_n = frac{n}{2}[2a_1 + (n - 1)d] ). So, both are equivalent.Let me try using the other formula to see if I get the same result.( S_n = frac{n}{2}(a_1 + a_n) )We know ( S_n = 98 ), ( n = 12 ), ( a_1 = 5 ), ( a_n = a_{12} = 5 + 11d ).So,[98 = frac{12}{2} times (5 + (5 + 11d))]Simplify:[98 = 6 times (10 + 11d)]Which is the same equation as before. So, same result. So, d is indeed 19/33.Wait, 19/33 is approximately 0.5758 minutes, which is about 34.55 seconds. That seems a bit short for a song duration difference, but maybe it's correct.Alternatively, maybe I messed up the total sum. Let me check the total concert time.Total concert time is 120 minutes, which includes 12 songs and 11 interludes of 2 minutes each. So, interludes total 22 minutes, so songs total 98 minutes. That's correct.So, the sum of the songs is 98 minutes, which is an arithmetic progression starting at 5, with 12 terms. So, the sum is 98.So, the formula is correct, so d is 19/33.Wait, 19/33 is approximately 0.5758 minutes, which is 34.55 seconds. So, each subsequent song is about 34.55 seconds longer than the previous one.So, the first song is 5 minutes, the second is 5 + 19/33 minutes, which is approximately 5.5758 minutes, the third is 5 + 2*(19/33) ≈ 6.1516 minutes, and so on, up to the 12th song.Let me compute ( d_{12} ). Since ( d_{12} = d_1 + 11d ). So,[d_{12} = 5 + 11 times frac{19}{33}]Simplify 11/33 is 1/3, so:[d_{12} = 5 + frac{19}{3} = 5 + 6 frac{1}{3} = 11 frac{1}{3} text{ minutes}]So, ( d_{12} = frac{34}{3} ) minutes, which is approximately 11.3333 minutes.Wait, 11 and 1/3 minutes is 11 minutes and 20 seconds. That seems reasonable for a song duration.So, let me recap:Given that the total concert time is 120 minutes, with 12 songs and 11 interludes of 2 minutes each, the total song time is 98 minutes. The songs form an arithmetic progression starting at 5 minutes with common difference d. Using the arithmetic series sum formula, we found that d is 19/33 minutes, and the last song ( d_{12} ) is 34/3 minutes or 11 and 1/3 minutes.So, I think that's the solution.Final AnswerThe common difference is ( boxed{dfrac{19}{33}} ) minutes and the duration of the twelfth song is ( boxed{dfrac{34}{3}} ) minutes.</think>

🧠:<think>Okay, so I have this problem about the 90s rave scene, and a parent who recorded the number of attendees at each event. The number of attendees is modeled by a trigonometric function, which makes sense because raves probably had periodic attendance due to their schedule or maybe seasonal trends. The first part asks me to calculate the average number of attendees over the first 26 weeks of the year. The function given is:[ A(t) = 500 + 300 sinleft(frac{pi t}{13}right) + 200 cosleft(frac{pi t}{26}right) ]Alright, so to find the average number of attendees over a period, I remember that for periodic functions, the average value over one period is just the average of the constant terms because the sine and cosine functions average out to zero over their periods. But let me think through this step by step to make sure I'm not missing anything. The average value of a function ( A(t) ) over an interval from ( t = a ) to ( t = b ) is given by:[ text{Average} = frac{1}{b - a} int_{a}^{b} A(t) , dt ]In this case, the interval is from ( t = 0 ) to ( t = 26 ) weeks. So, the average number of attendees ( overline{A} ) is:[ overline{A} = frac{1}{26 - 0} int_{0}^{26} A(t) , dt ]Substituting the given function:[ overline{A} = frac{1}{26} int_{0}^{26} left(500 + 300 sinleft(frac{pi t}{13}right) + 200 cosleft(frac{pi t}{26}right)right) dt ]Now, I can split this integral into three separate integrals:[ overline{A} = frac{1}{26} left[ int_{0}^{26} 500 , dt + int_{0}^{26} 300 sinleft(frac{pi t}{13}right) dt + int_{0}^{26} 200 cosleft(frac{pi t}{26}right) dt right] ]Let me compute each integral one by one.First integral: ( int_{0}^{26} 500 , dt )That's straightforward. The integral of a constant is just the constant times the interval length.So, ( 500 times (26 - 0) = 500 times 26 = 13,000 ).Second integral: ( int_{0}^{26} 300 sinleft(frac{pi t}{13}right) dt )To integrate this, I need to find the antiderivative of ( sinleft(frac{pi t}{13}right) ). The integral of ( sin(ax) ) is ( -frac{1}{a} cos(ax) ). So, applying that here:Let ( a = frac{pi}{13} ), so the integral becomes:( 300 times left[ -frac{13}{pi} cosleft(frac{pi t}{13}right) right] ) evaluated from 0 to 26.Calculating this:First, plug in t = 26:( -frac{13}{pi} cosleft(frac{pi times 26}{13}right) = -frac{13}{pi} cos(2pi) )Since ( cos(2pi) = 1 ), this becomes ( -frac{13}{pi} times 1 = -frac{13}{pi} )Now, plug in t = 0:( -frac{13}{pi} cosleft(frac{pi times 0}{13}right) = -frac{13}{pi} cos(0) = -frac{13}{pi} times 1 = -frac{13}{pi} )Subtracting the lower limit from the upper limit:( left(-frac{13}{pi}right) - left(-frac{13}{pi}right) = 0 )So, the second integral is 0.Third integral: ( int_{0}^{26} 200 cosleft(frac{pi t}{26}right) dt )Similarly, the integral of ( cos(ax) ) is ( frac{1}{a} sin(ax) ). So, with ( a = frac{pi}{26} ), the integral becomes:( 200 times left[ frac{26}{pi} sinleft(frac{pi t}{26}right) right] ) evaluated from 0 to 26.Calculating this:First, plug in t = 26:( frac{26}{pi} sinleft(frac{pi times 26}{26}right) = frac{26}{pi} sin(pi) = frac{26}{pi} times 0 = 0 )Now, plug in t = 0:( frac{26}{pi} sinleft(frac{pi times 0}{26}right) = frac{26}{pi} times 0 = 0 )Subtracting the lower limit from the upper limit:( 0 - 0 = 0 )So, the third integral is also 0.Putting it all together:[ overline{A} = frac{1}{26} [13,000 + 0 + 0] = frac{13,000}{26} = 500 ]Wait, that's interesting. So, the average number of attendees is 500. That makes sense because the sine and cosine terms have average values of zero over their periods, so only the constant term contributes to the average.But just to double-check, let me think about the periods of the sine and cosine functions in the given function.The sine term is ( sinleft(frac{pi t}{13}right) ). The period of this function is ( frac{2pi}{pi/13} = 26 ) weeks. Similarly, the cosine term is ( cosleft(frac{pi t}{26}right) ), so its period is ( frac{2pi}{pi/26} = 52 ) weeks.Wait, so the sine term has a period of 26 weeks, which is exactly the interval we're considering. The cosine term has a period of 52 weeks, which is twice the interval. So, over 26 weeks, the cosine term doesn't complete a full period. Hmm, does that affect the average?But when I integrated the cosine term over 0 to 26, I still got zero. Is that correct?Let me re-examine the integral of the cosine term:( int_{0}^{26} 200 cosleft(frac{pi t}{26}right) dt )The antiderivative is ( 200 times frac{26}{pi} sinleft(frac{pi t}{26}right) ), evaluated from 0 to 26.At t=26, ( sinleft(frac{pi times 26}{26}right) = sin(pi) = 0 )At t=0, ( sin(0) = 0 )So, the integral is indeed 0. So, even though the cosine term doesn't complete a full period, the integral over 0 to 26 is zero because it's symmetric around the midpoint.Wait, is that true? Let me think about the graph of ( cosleft(frac{pi t}{26}right) ) from 0 to 26.At t=0, it's 1. At t=13, it's ( cosleft(frac{pi}{2}right) = 0 ). At t=26, it's ( cos(pi) = -1 ). So, the function starts at 1, goes down to 0 at 13, and then to -1 at 26. So, the area under the curve from 0 to 26 is actually not zero, is it?Wait, hold on. If I integrate from 0 to 26, the function starts at 1, goes down to -1, so the area above the x-axis from 0 to 13 is positive, and from 13 to 26 is negative. But are these areas equal in magnitude?Let me compute the integral from 0 to 26:The integral is ( 200 times frac{26}{pi} [sin(pi) - sin(0)] = 0 ). So, mathematically, it's zero. But visually, the area from 0 to 13 is positive and from 13 to 26 is negative, but do they cancel out?Wait, let's compute the integral from 0 to 13 and from 13 to 26 separately.From 0 to 13:( int_{0}^{13} 200 cosleft(frac{pi t}{26}right) dt = 200 times frac{26}{pi} [sin(frac{pi times 13}{26}) - sin(0)] = 200 times frac{26}{pi} [sin(frac{pi}{2}) - 0] = 200 times frac{26}{pi} times 1 = frac{5200}{pi} )From 13 to 26:( int_{13}^{26} 200 cosleft(frac{pi t}{26}right) dt = 200 times frac{26}{pi} [sin(pi) - sin(frac{pi}{2})] = 200 times frac{26}{pi} [0 - 1] = -frac{5200}{pi} )So, adding these two integrals: ( frac{5200}{pi} - frac{5200}{pi} = 0 )Ah, so even though the function isn't completing a full period, the areas above and below the x-axis cancel each other out over the interval 0 to 26. So, the integral is indeed zero. That's why the average of the cosine term is zero over this interval.Therefore, my initial calculation was correct. The average number of attendees is 500.Moving on to the second part. The parent recorded the number of glow sticks distributed at each event, which was proportional to the number of attendees. The number of glow sticks ( G(t) ) is given by:[ G(t) = k cdot A(t) ]where ( k ) is a constant of proportionality. We are told that a total of 15,600 glow sticks were distributed over the first 26 weeks. We need to find ( k ).So, the total number of glow sticks distributed is the integral of ( G(t) ) from 0 to 26 weeks. That is:[ int_{0}^{26} G(t) , dt = 15,600 ]Substituting ( G(t) = k cdot A(t) ):[ int_{0}^{26} k cdot A(t) , dt = 15,600 ]We can factor out the constant ( k ):[ k cdot int_{0}^{26} A(t) , dt = 15,600 ]But from the first part, we already computed ( int_{0}^{26} A(t) , dt ) as 13,000. Wait, let me check:In the first part, the average was 500, which is ( frac{1}{26} times 13,000 ). So, yes, the integral of A(t) over 0 to 26 is 13,000.Therefore:[ k times 13,000 = 15,600 ]Solving for ( k ):[ k = frac{15,600}{13,000} ]Simplify this fraction:Divide numerator and denominator by 100: ( frac{156}{13} )156 divided by 13: 13 x 12 = 156, so 12.Therefore, ( k = 12 ).Wait, let me verify that:15,600 divided by 13,000 is indeed 1.2, but 156 divided by 13 is 12. Wait, that can't be. Wait, 13 x 12 is 156, so 156/13 is 12, which is 12. But 15,600 / 13,000 is 1.2, not 12.Wait, hold on, I think I made a mistake here.Wait, 15,600 divided by 13,000 is equal to 1.2, right? Because 13,000 x 1.2 = 15,600.But 156 divided by 13 is 12, but 15,600 divided by 13,000 is 1.2.So, perhaps I messed up the decimal places.Wait, 15,600 / 13,000 = (15,600 ÷ 100) / (13,000 ÷ 100) = 156 / 130 = 1.2.Yes, so 15,600 / 13,000 = 1.2, so k = 1.2.Wait, but 156 / 13 is 12, but 15,600 / 13,000 is 1.2. So, the correct value is 1.2.Wait, so where did I go wrong earlier? I thought 156 / 13 is 12, but that's 156 divided by 13. But in the equation, it's 15,600 divided by 13,000, which is 1.2.So, k = 1.2.Wait, but to be precise, 15,600 divided by 13,000:15,600 ÷ 13,000 = (15,600 ÷ 100) ÷ (13,000 ÷ 100) = 156 ÷ 130 = 1.2.Yes, that's correct.So, k is 1.2.But just to make sure, let's write it as a fraction.1.2 is equal to 6/5 or 12/10, which simplifies to 6/5.So, k = 6/5 or 1.2.Therefore, the constant of proportionality is 1.2.So, summarizing:1. The average number of attendees over the first 26 weeks is 500.2. The constant ( k ) is 1.2.Final Answer1. The average number of attendees is boxed{500}.2. The value of ( k ) is boxed{1.2}.</think>

🧠:<think>Alright, so I have this problem about protein concentration in the bloodstream following an exponential decay model. Let me try to figure out how to solve both parts step by step.Starting with part 1: I need to find the time ( t ) it takes for the concentration to reduce from 200 mg/L to 50 mg/L, given the decay constant ( k = 0.3 ) hours(^{-1}). The formula provided is ( C(t) = C_0 e^{-kt} ).First, let me write down what I know:- Initial concentration, ( C_0 = 200 ) mg/L- Decay constant, ( k = 0.3 ) per hour- Final concentration, ( C(t) = 50 ) mg/LI need to solve for ( t ). So, plugging the known values into the formula:( 50 = 200 e^{-0.3 t} )Hmm, okay. Let me divide both sides by 200 to isolate the exponential part:( frac{50}{200} = e^{-0.3 t} )Simplifying the left side:( 0.25 = e^{-0.3 t} )Now, to solve for ( t ), I should take the natural logarithm of both sides. Remember, ( ln(e^{x}) = x ). So:( ln(0.25) = ln(e^{-0.3 t}) )Simplifying the right side:( ln(0.25) = -0.3 t )Now, I can solve for ( t ) by dividing both sides by -0.3:( t = frac{ln(0.25)}{-0.3} )Let me compute ( ln(0.25) ). I remember that ( ln(1) = 0 ) and ( ln(e) = 1 ), but 0.25 is less than 1, so the natural log should be negative. Specifically, ( ln(0.25) ) is approximately ( -1.3863 ).So plugging that in:( t = frac{-1.3863}{-0.3} )The negatives cancel out, so:( t = frac{1.3863}{0.3} )Calculating that division:( 1.3863 ÷ 0.3 ) is approximately 4.621 hours.Wait, let me double-check that division. 0.3 goes into 1.3863 how many times?0.3 * 4 = 1.2, which leaves 0.1863 remaining.0.3 * 0.6 = 0.18, so that's 4.6 so far, with 0.0063 left.0.3 * 0.021 = 0.0063, so total is 4.621.Yes, that seems right. So approximately 4.621 hours.But let me see if I can express this more precisely. Since ( ln(0.25) = ln(1/4) = -ln(4) ). So,( t = frac{-ln(4)}{-0.3} = frac{ln(4)}{0.3} )Calculating ( ln(4) ) is approximately 1.386294, so:( t = frac{1.386294}{0.3} ≈ 4.62098 ) hours.So, rounding to a reasonable decimal place, maybe 4.62 hours or approximately 4.62 hours.But let me check if the question expects an exact form or a decimal. Since it's a time, decimal makes sense. So, 4.62 hours is about 4 hours and 37 minutes (since 0.62 * 60 ≈ 37.2 minutes). But unless they specify, decimal is fine.So, that's part 1. I think I did that correctly.Moving on to part 2: The gym enthusiast consumes a new type of protein shake every day, where the initial concentration ( C_0(n) = 200 + 10n ), with ( n ) being the day number starting from ( n=1 ). The decay constant ( k ) remains the same, which is 0.3 per hour. I need to find the total concentration of protein in the bloodstream 2 hours after consuming the shake on the third day.Wait, total concentration? Hmm. So, does this mean that each day's shake contributes to the concentration, and we have to sum up the concentrations from each day? Or is it just the concentration from the third day's shake?Wait, the wording says: "the total concentration of protein in the bloodstream 2 hours after consuming the shake on the third day."So, perhaps each shake is consumed on a different day, and each shake's concentration decays over time, but when considering the third day, we have to account for the shakes consumed on day 1, day 2, and day 3, each decaying for a different amount of time.Wait, but the wording is a bit ambiguous. Let me read it again:"find the total concentration of protein in the bloodstream 2 hours after consuming the shake on the third day."So, does that mean only the concentration from the third day's shake, 2 hours after consumption? Or does it mean the sum of concentrations from all shakes consumed up to the third day, each decaying for 2 hours?Hmm. The wording says "the total concentration... 2 hours after consuming the shake on the third day." So, if we're considering the third day, and 2 hours after consuming the shake on that day, the total concentration would include the contributions from shakes consumed on previous days as well, each decaying for more than 2 hours.Wait, but if the shakes are consumed every day, starting from day 1, then on day 3, the person has consumed shakes on day 1, day 2, and day 3. Each shake will have been in the bloodstream for a different amount of time when we measure at 2 hours after day 3's shake.So, let's clarify:- Shake on day 1: consumed at time t = 0 (relative to day 1). When we measure 2 hours after day 3's shake, that would be 2 + 2*24 hours after day 1's shake? Wait, no.Wait, actually, if each day is a separate day, then the shakes are consumed 24 hours apart. So, shake on day 1 is consumed at t = 0, shake on day 2 is consumed at t = 24 hours, shake on day 3 is consumed at t = 48 hours. Then, 2 hours after day 3's shake is at t = 50 hours.Therefore, the concentration from day 1's shake at t = 50 hours is ( C_1(50) = C_0(1) e^{-k*50} ).Similarly, concentration from day 2's shake at t = 50 hours is ( C_2(50 - 24) = C_0(2) e^{-k*(50 - 24)} = C_0(2) e^{-k*26} ).Concentration from day 3's shake at t = 50 hours is ( C_3(50 - 48) = C_0(3) e^{-k*2} ).Therefore, the total concentration is the sum of these three.So, let me write that down:Total concentration ( C_{total}(50) = C_1(50) + C_2(26) + C_3(2) )Where:- ( C_1(50) = C_0(1) e^{-0.3*50} )- ( C_2(26) = C_0(2) e^{-0.3*26} )- ( C_3(2) = C_0(3) e^{-0.3*2} )Given that ( C_0(n) = 200 + 10n ), so:- ( C_0(1) = 200 + 10*1 = 210 ) mg/L- ( C_0(2) = 200 + 10*2 = 220 ) mg/L- ( C_0(3) = 200 + 10*3 = 230 ) mg/LSo, plugging these into the equations:1. ( C_1(50) = 210 e^{-0.3*50} )2. ( C_2(26) = 220 e^{-0.3*26} )3. ( C_3(2) = 230 e^{-0.3*2} )Now, let's compute each term.Starting with ( C_1(50) ):Compute exponent: ( -0.3 * 50 = -15 )So, ( C_1(50) = 210 e^{-15} )I know that ( e^{-15} ) is a very small number. Let me compute it:( e^{-15} ≈ 3.059023205 × 10^{-7} )So, ( C_1(50) ≈ 210 * 3.059023205 × 10^{-7} ≈ 6.42394873 × 10^{-5} ) mg/LThat's approximately 0.0000642 mg/L. That's negligible, practically zero.Moving on to ( C_2(26) ):Compute exponent: ( -0.3 * 26 = -7.8 )So, ( C_2(26) = 220 e^{-7.8} )Calculating ( e^{-7.8} ):I know that ( e^{-7} ≈ 0.000911882 ), and ( e^{-8} ≈ 0.00033546 ). So, ( e^{-7.8} ) is between these two.Using a calculator, ( e^{-7.8} ≈ 0.0004493 )Therefore, ( C_2(26) ≈ 220 * 0.0004493 ≈ 0.098846 ) mg/LApproximately 0.0988 mg/L.Now, ( C_3(2) ):Compute exponent: ( -0.3 * 2 = -0.6 )So, ( C_3(2) = 230 e^{-0.6} )Calculating ( e^{-0.6} ):I remember that ( e^{-0.5} ≈ 0.6065 ) and ( e^{-0.6} ≈ 0.5488 ). Let me verify:Using calculator: ( e^{-0.6} ≈ 0.5488116 )So, ( C_3(2) ≈ 230 * 0.5488116 ≈ 126.226668 ) mg/LApproximately 126.23 mg/L.Now, adding up all three concentrations:( C_{total} ≈ 0.0000642 + 0.098846 + 126.226668 ≈ 126.325578 ) mg/LSo, approximately 126.33 mg/L.But let me check if I did that correctly. The concentration from day 1 is negligible, day 2 is about 0.0988, and day 3 is about 126.23. So, adding them together gives roughly 126.33 mg/L.But wait, is this the correct interpretation? Because the problem says "the total concentration of protein in the bloodstream 2 hours after consuming the shake on the third day."Alternatively, maybe it's considering that each shake is consumed on the same day, but that doesn't make much sense because it says "every day." So, the shakes are consumed on different days, each day starting fresh.Wait, but if the shakes are consumed every day, then on day 3, 2 hours after consuming the shake, the previous shakes (from day 1 and day 2) have been in the system for 48 + 2 = 50 hours and 24 + 2 = 26 hours respectively.So, yes, my initial interpretation seems correct.Therefore, the total concentration is approximately 126.33 mg/L.But let me verify the calculations step by step to make sure I didn't make a mistake.First, ( C_1(50) = 210 e^{-15} ). As I calculated, ( e^{-15} ) is about 3.059e-7, so 210 * 3.059e-7 ≈ 6.423e-5, which is 0.0000642 mg/L. That seems correct.Next, ( C_2(26) = 220 e^{-7.8} ). ( e^{-7.8} ≈ 0.0004493 ), so 220 * 0.0004493 ≈ 0.098846 mg/L. Correct.Then, ( C_3(2) = 230 e^{-0.6} ≈ 230 * 0.5488116 ≈ 126.226668 mg/L. Correct.Adding them up: 0.0000642 + 0.098846 ≈ 0.09891, then 0.09891 + 126.226668 ≈ 126.325578 mg/L. So, approximately 126.33 mg/L.But let me consider if the question expects the concentration only from the third day's shake, which would be 126.23 mg/L. But the wording says "total concentration," which implies all sources. So, including the previous days' shakes.But wait, another thought: If the shakes are consumed every day, but the person is only considering the third day, maybe they are only looking at the shake consumed on the third day, 2 hours later. But the problem says "the total concentration," which might include all shakes consumed up to that point.But given that the shakes are consumed every day, and each shake's concentration decays over time, the total concentration at any time is the sum of the concentrations from all shakes consumed on previous days, each decayed appropriately.Therefore, yes, my initial calculation is correct, adding up all three contributions.But just to make sure, let me think about it again. If on day 3, 2 hours after consuming the shake, the person has shakes from day 1, day 2, and day 3 in their system. Each of these shakes has been decaying since their respective consumption times.So, day 1's shake has been decaying for 48 + 2 = 50 hours.Day 2's shake has been decaying for 24 + 2 = 26 hours.Day 3's shake has been decaying for 2 hours.Therefore, yes, we need to calculate each of these and sum them up.So, the total concentration is approximately 126.33 mg/L.But let me compute the exact values without rounding too early to see if the approximation holds.First, compute ( C_1(50) = 210 e^{-15} ). Let's compute ( e^{-15} ) more accurately.Using a calculator, ( e^{-15} ≈ 3.059023205 × 10^{-7} ). So, 210 * 3.059023205e-7 = 210 * 0.0000003059023205 ≈ 0.0000642394873 mg/L.Next, ( C_2(26) = 220 e^{-7.8} ). Calculating ( e^{-7.8} ):Using a calculator, ( e^{-7.8} ≈ 0.00044932899 ). So, 220 * 0.00044932899 ≈ 0.098852378 mg/L.Then, ( C_3(2) = 230 e^{-0.6} ). Calculating ( e^{-0.6} ≈ 0.548811636 ). So, 230 * 0.548811636 ≈ 126.226676 mg/L.Adding them up:0.0000642394873 + 0.098852378 + 126.226676 ≈First, add the first two: 0.0000642394873 + 0.098852378 ≈ 0.0989166175Then add to the third: 0.0989166175 + 126.226676 ≈ 126.3255926 mg/L.So, approximately 126.3256 mg/L, which rounds to 126.33 mg/L.Therefore, the total concentration is approximately 126.33 mg/L.But let me consider if the problem expects an exact expression or if it's okay to leave it in terms of exponentials. However, since the question asks for the concentration, it's more practical to provide a numerical value.Alternatively, maybe I can express it in terms of exact exponentials, but given the context, a numerical answer is more appropriate.So, summarizing:1. The time for the concentration to reduce to 50 mg/L is approximately 4.62 hours.2. The total concentration 2 hours after the third day's shake is approximately 126.33 mg/L.I think that's it. I don't see any mistakes in my calculations, so I'm confident with these answers.Final Answer1. The time it takes for the concentration to reduce to 50 mg/L is boxed{4.62} hours.2. The total concentration of protein in the bloodstream 2 hours after consuming the shake on the third day is boxed{126.33} mg/L.</think>