🧠:<think>Alright, so I have this problem where John, a retired factory worker, is trying to model the number of words in each chapter of the Bible using a sine function. The function given is f(x) = a·sin(bx + c) + d. He has data for the first four chapters: chapter 1 has 120 words, chapter 2 has 140, chapter 3 has 160, and chapter 4 has 110. My task is to figure out the constants a, b, c, and d, and then use that model to calculate the total number of words in the first 10 chapters.Okay, let's start by understanding the function. It's a sine function, which is periodic, so it goes up and down in a wave pattern. The constants a, b, c, and d will determine the amplitude, period, phase shift, and vertical shift of the sine wave, respectively.Given that f(x) = a·sin(bx + c) + d, and we have four data points, I can set up four equations to solve for the four unknowns: a, b, c, d.Let me write down the equations based on the given data:For x = 1: f(1) = 120 = a·sin(b·1 + c) + d For x = 2: f(2) = 140 = a·sin(b·2 + c) + d For x = 3: f(3) = 160 = a·sin(b·3 + c) + d For x = 4: f(4) = 110 = a·sin(b·4 + c) + dSo, we have four equations:1. a·sin(b + c) + d = 120 2. a·sin(2b + c) + d = 140 3. a·sin(3b + c) + d = 160 4. a·sin(4b + c) + d = 110Hmm, solving this system might be a bit tricky because it's nonlinear due to the sine function. I need to find a way to simplify or make some assumptions.First, let's subtract the first equation from the second, third, and fourth to eliminate d.Subtracting equation 1 from equation 2:a·sin(2b + c) - a·sin(b + c) = 140 - 120 = 20 So, a[sin(2b + c) - sin(b + c)] = 20 ...(5)Similarly, subtracting equation 2 from equation 3:a·sin(3b + c) - a·sin(2b + c) = 160 - 140 = 20 So, a[sin(3b + c) - sin(2b + c)] = 20 ...(6)Subtracting equation 3 from equation 4:a·sin(4b + c) - a·sin(3b + c) = 110 - 160 = -50 So, a[sin(4b + c) - sin(3b + c)] = -50 ...(7)Now, equations (5), (6), and (7) are:5. a[sin(2b + c) - sin(b + c)] = 20 6. a[sin(3b + c) - sin(2b + c)] = 20 7. a[sin(4b + c) - sin(3b + c)] = -50Looking at equations 5 and 6, both equal 20. So, the difference between consecutive sine terms is the same for equations 5 and 6. Let's denote:Let’s denote θ = b + c. Then, equation 5 becomes:a[sin(2b + c) - sin(θ)] = 20 But 2b + c = b + (b + c) = b + θ Similarly, equation 6: a[sin(3b + c) - sin(2b + c)] = 20 Which is a[sin(2b + θ) - sin(b + θ)] = 20So, both equations 5 and 6 have the same structure. Maybe we can find a relationship between them.Let me write equation 5 as:a[sin(b + θ) - sin(θ)] = 20 And equation 6 as:a[sin(2b + θ) - sin(b + θ)] = 20So, both differences equal 20. Let me denote Δ1 = sin(b + θ) - sin(θ) and Δ2 = sin(2b + θ) - sin(b + θ). Then, aΔ1 = 20 and aΔ2 = 20. So, Δ1 = Δ2.Therefore, sin(b + θ) - sin(θ) = sin(2b + θ) - sin(b + θ)Let me compute this:sin(b + θ) - sin(θ) = sin(2b + θ) - sin(b + θ)Bring all terms to one side:sin(b + θ) - sin(θ) - sin(2b + θ) + sin(b + θ) = 0 2 sin(b + θ) - sin(θ) - sin(2b + θ) = 0Hmm, let's recall that sin(2b + θ) = sin(b + (b + θ)). Maybe using sine addition formula:sin(A + B) = sin A cos B + cos A sin BSo, sin(2b + θ) = sin(b + (b + θ)) = sin b cos(b + θ) + cos b sin(b + θ)Similarly, sin(b + θ) = sin b cos θ + cos b sin θLet me substitute these into the equation:2[sin b cos θ + cos b sin θ] - sin θ - [sin b cos(b + θ) + cos b sin(b + θ)] = 0Let me expand this:2 sin b cos θ + 2 cos b sin θ - sin θ - sin b cos(b + θ) - cos b sin(b + θ) = 0Hmm, this is getting complicated. Maybe another approach.Alternatively, using the identity sin A - sin B = 2 cos((A+B)/2) sin((A-B)/2)So, let's rewrite the differences:Δ1 = sin(b + θ) - sin θ = 2 cos((b + θ + θ)/2) sin((b + θ - θ)/2) = 2 cos((b + 2θ)/2) sin(b/2)Similarly, Δ2 = sin(2b + θ) - sin(b + θ) = 2 cos((2b + θ + b + θ)/2) sin((2b + θ - (b + θ))/2) = 2 cos((3b + 2θ)/2) sin(b/2)So, from Δ1 = Δ2:2 cos((b + 2θ)/2) sin(b/2) = 2 cos((3b + 2θ)/2) sin(b/2)Assuming sin(b/2) ≠ 0, we can divide both sides by 2 sin(b/2):cos((b + 2θ)/2) = cos((3b + 2θ)/2)So, cos α = cos β implies that α = 2πk ± β, for some integer k.So, (b + 2θ)/2 = 2πk ± (3b + 2θ)/2Let me solve for θ.Case 1: (b + 2θ)/2 = 2πk + (3b + 2θ)/2Multiply both sides by 2:b + 2θ = 4πk + 3b + 2θSubtract 2θ from both sides:b = 4πk + 3bSubtract b:0 = 4πk + 2bSo, 2b = -4πk => b = -2πkBut since b is a constant in the function, and x is chapter number (positive integer), b should be positive. So, k must be negative.Let k = -m, where m is a positive integer.Then, b = -2π(-m) = 2πmBut let's see if this makes sense. If b is 2πm, then the period of the sine function is 2π / b = 1/m. So, the period would be a fraction, which might not make sense for chapters, which are integers. Maybe m=1, so period is 1. But a period of 1 would mean the function repeats every chapter, which might not be the case here. Let's hold onto this thought.Case 2: (b + 2θ)/2 = 2πk - (3b + 2θ)/2Multiply both sides by 2:b + 2θ = 4πk - 3b - 2θBring like terms to one side:b + 3b + 2θ + 2θ = 4πk 4b + 4θ = 4πk Divide both sides by 4:b + θ = πkSo, θ = πk - bBut θ was defined as θ = b + c, so:b + c = πk - b So, c = πk - 2bSo, from case 2, we get c = πk - 2bNow, let's see if we can use this in our equations.From equation 5: a[sin(b + θ) - sin θ] = 20But θ = πk - b, so:sin(b + θ) = sin(b + πk - b) = sin(πk) = 0 Similarly, sin θ = sin(πk - b) = sin(πk)cos b - cos(πk)sin bBut sin(πk) is 0 for integer k, so sin θ = -cos(πk) sin bTherefore, equation 5 becomes:a[0 - (-cos(πk) sin b)] = 20 a[cos(πk) sin b] = 20Similarly, equation 6: a[sin(2b + θ) - sin(b + θ)] = 20But sin(b + θ) = 0 as above, and sin(2b + θ) = sin(2b + πk - b) = sin(b + πk) = sin b cos πk + cos b sin πk = sin b cos πk + 0 = sin b cos πkTherefore, equation 6 becomes:a[sin b cos πk - 0] = 20 a sin b cos πk = 20But from equation 5, we have a sin b cos πk = 20 as well. So, equations 5 and 6 are consistent.Now, let's look at equation 7:a[sin(4b + c) - sin(3b + c)] = -50Again, c = πk - 2b, so:sin(4b + c) = sin(4b + πk - 2b) = sin(2b + πk) = sin 2b cos πk + cos 2b sin πk = sin 2b cos πk Similarly, sin(3b + c) = sin(3b + πk - 2b) = sin(b + πk) = sin b cos πk + cos b sin πk = sin b cos πkTherefore, equation 7 becomes:a[sin 2b cos πk - sin b cos πk] = -50 Factor out sin b cos πk:a cos πk (sin 2b - sin b) = -50But from equation 5, a sin b cos πk = 20, so let's denote:Let’s let’s denote S = sin b, C = cos πkFrom equation 5: a S C = 20 From equation 7: a C (2 sin b cos b - sin b) = -50 Simplify equation 7:a C sin b (2 cos b - 1) = -50But from equation 5, a C sin b = 20, so substitute:20 (2 cos b - 1) = -50 Divide both sides by 20:2 cos b - 1 = -50 / 20 = -2.5 So, 2 cos b = -2.5 + 1 = -1.5 cos b = -1.5 / 2 = -0.75So, cos b = -0.75Therefore, b = arccos(-0.75)Calculating arccos(-0.75). Let's find the angle whose cosine is -0.75.Since cosine is negative, the angle is in the second or third quadrant. Since b is a positive constant, and in the context of a sine function modeling chapters, b is likely between 0 and 2π, so let's take the principal value.arccos(-0.75) ≈ 138.59 degrees or in radians, approximately 2.4189 radians.So, b ≈ 2.4189 radians.Now, let's find sin b:sin b = sqrt(1 - cos² b) = sqrt(1 - 0.75²) = sqrt(1 - 0.5625) = sqrt(0.4375) ≈ 0.6614But since b is in the second quadrant (cos b negative, sin b positive), sin b ≈ 0.6614.Now, from equation 5: a sin b cos πk = 20 We have sin b ≈ 0.6614, cos πk is either 1 or -1 because k is integer.Let’s see, cos πk = (-1)^kSo, cos πk = ±1So, equation 5: a * 0.6614 * (-1)^k = 20So, a = 20 / (0.6614 * (-1)^k)Let’s consider k as integer. Let's test k=0:If k=0, cos πk = 1, so a = 20 / 0.6614 ≈ 30.24If k=1, cos πk = -1, so a = 20 / (-0.6614) ≈ -30.24But a is the amplitude, which is typically positive. So, if k=1, a would be negative, but we can take the absolute value since amplitude is positive. So, let's take k=1, then a ≈ 30.24Wait, but if k=1, cos πk = -1, so a = 20 / (-0.6614) ≈ -30.24, but since amplitude is positive, we can take a=30.24 and adjust the phase shift accordingly.Alternatively, maybe k=0 is acceptable. Let's see.If k=0, then c = π*0 - 2b = -2b ≈ -4.8378 radiansIf k=1, c = π - 2b ≈ 3.1416 - 4.8378 ≈ -1.6962 radiansBut let's see which one makes sense.Also, from equation 7, we had:20 (2 cos b - 1) = -50 Which gave us cos b = -0.75, which is correct.Now, let's find d.From the original equations, let's take equation 1:a·sin(b + c) + d = 120We have a ≈ 30.24, b ≈ 2.4189, c = πk - 2bCase 1: k=0, c = -2b ≈ -4.8378So, sin(b + c) = sin(2.4189 - 4.8378) = sin(-2.4189) ≈ -sin(2.4189) ≈ -0.6614So, equation 1: 30.24*(-0.6614) + d ≈ 120 Calculate 30.24*(-0.6614) ≈ -20So, -20 + d = 120 => d = 140Case 2: k=1, c = π - 2b ≈ 3.1416 - 4.8378 ≈ -1.6962So, sin(b + c) = sin(2.4189 - 1.6962) = sin(0.7227) ≈ 0.6614So, equation 1: 30.24*(0.6614) + d ≈ 120 Calculate 30.24*0.6614 ≈ 20So, 20 + d = 120 => d = 100Wait, so depending on k=0 or k=1, d is 140 or 100.But let's check with equation 2:For k=0, c ≈ -4.8378, a≈30.24, d=140f(2) = 30.24 sin(2*2.4189 -4.8378) +140 = 30.24 sin(4.8378 -4.8378) +140 = 30.24 sin(0) +140 = 0 +140 =140, which matches the given data.For k=1, c≈-1.6962, a≈30.24, d=100f(2) =30.24 sin(2*2.4189 -1.6962) +100 =30.24 sin(4.8378 -1.6962) =30.24 sin(3.1416) ≈30.24*0 +100=100, which does not match the given data of 140.So, k=0 gives us correct value for f(2)=140, while k=1 does not. Therefore, k=0 is the correct choice.So, k=0, c= -2b ≈ -4.8378, a≈30.24, d=140Now, let's verify equation 3:f(3)=30.24 sin(3*2.4189 -4.8378) +140 =30.24 sin(7.2567 -4.8378) =30.24 sin(2.4189) ≈30.24*0.6614≈20 +140=160, which matches.Equation 4:f(4)=30.24 sin(4*2.4189 -4.8378) +140 =30.24 sin(9.6756 -4.8378) =30.24 sin(4.8378) ≈30.24*(-0.6614)≈-20 +140=120, but the given data is 110. Hmm, discrepancy here.Wait, that's a problem. According to our model, f(4)=120, but the given data is 110. So, something is wrong.Wait, let's recalculate f(4):sin(4b + c) = sin(4*2.4189 -4.8378) = sin(9.6756 -4.8378)=sin(4.8378)But sin(4.8378) is sin(π + 1.704) ≈ sin(π + something) = -sin(something). Let's compute sin(4.8378):4.8378 radians is approximately 277 degrees (since π≈3.1416, 4.8378-π≈1.6962≈97 degrees, so 180+97=277 degrees). So, sin(277 degrees)=sin(360-83)= -sin(83)≈-0.9925Wait, but earlier I thought sin(4.8378)≈-0.6614, which was incorrect.Wait, let me compute sin(4.8378) accurately.4.8378 radians is approximately 277 degrees, as above.But let's compute it more precisely.Using calculator:sin(4.8378) ≈ sin(π + 1.6962) = -sin(1.6962)sin(1.6962) ≈ sin(97 degrees) ≈0.9925So, sin(4.8378)≈-0.9925Therefore, f(4)=30.24*(-0.9925)+140≈-30.24*0.9925 +140≈-30 +140=110, which matches the given data.Wait, so earlier I miscalculated sin(4.8378). It's actually approximately -0.9925, not -0.6614. So, f(4)=30.24*(-0.9925)+140≈-30 +140=110, which is correct.So, my mistake earlier was assuming sin(4.8378)≈-0.6614, but it's actually≈-0.9925. Therefore, the model works.So, summarizing:a ≈30.24 b≈2.4189 radians c≈-4.8378 radians d=140But let's express these more precisely.Since cos b = -0.75, b= arccos(-0.75)=2.41885879 radiansSimilarly, sin b= sqrt(1 - 0.75²)=sqrt(0.4375)=sqrt(7/16)=√7 /4≈0.6614So, a=20/(sin b * cos πk)=20/(0.6614*1)=≈30.24But let's compute a more precisely:sin b=√(1 - (3/4)^2)=√(1 - 9/16)=√(7/16)=√7 /4≈0.661438So, a=20 / ( (√7 /4)*1 )=20*(4/√7)=80/√7≈80/2.6458≈30.24So, a=80/√7≈30.24Similarly, b= arccos(-3/4)=2.41885879 radiansc= -2b≈-4.83771758 radiansd=140So, the function is:f(x)= (80/√7) sin(2.41885879 x -4.83771758) +140Alternatively, we can write it as:f(x)= (80/√7) sin(bx + c) +140, where b≈2.4189, c≈-4.8377But perhaps we can express c in terms of b:c= -2b, as we found earlier.So, f(x)= (80/√7) sin(bx -2b) +140= (80/√7) sin(b(x -2)) +140That might be a cleaner way to write it.Now, to answer part 1, we have determined the constants:a=80/√7≈30.24 b= arccos(-3/4)≈2.4189 radians c= -2b≈-4.8377 radians d=140So, that's the solution for part 1.Now, part 2: calculate the total number of words in the first 10 chapters.So, we need to compute f(1)+f(2)+...+f(10)Given f(x)= (80/√7) sin(bx + c) +140, with b≈2.4189, c≈-4.8377But since c= -2b, we can write f(x)= (80/√7) sin(b(x -2)) +140So, f(x)= (80/√7) sin(b(x -2)) +140Let me compute f(x) for x=1 to 10.But since b= arccos(-3/4), let's compute sin(b(x-2)) for each x.Alternatively, since we have the function, we can compute each f(x) numerically.Let me compute each f(x):x=1: f(1)=120 (given) x=2: f(2)=140 (given) x=3: f(3)=160 (given) x=4: f(4)=110 (given) x=5: f(5)= (80/√7) sin(b(5-2)) +140= (80/√7) sin(3b) +140 Similarly, x=6: f(6)= (80/√7) sin(4b) +140 x=7: f(7)= (80/√7) sin(5b) +140 x=8: f(8)= (80/√7) sin(6b) +140 x=9: f(9)= (80/√7) sin(7b) +140 x=10: f(10)= (80/√7) sin(8b) +140But we can compute sin(3b), sin(4b), etc., using the fact that b= arccos(-3/4)Let me recall that cos b= -3/4, so sin b=√7/4≈0.6614We can use multiple-angle formulas to compute sin(3b), sin(4b), etc.Recall that:sin(3b)=3 sin b -4 sin³ b sin(4b)=2 sin(2b) cos(2b)=2*(2 sin b cos b)*(2 cos² b -1)Similarly, sin(5b)=sin(2b +3b)=sin 2b cos 3b + cos 2b sin 3bBut this might get complicated. Alternatively, since we have b≈2.4189, we can compute each sin(n b) numerically.Let me compute b≈2.4189 radiansCompute:sin(b)=sin(2.4189)≈0.6614 sin(2b)=sin(4.8378)≈-0.7490 (since sin(4.8378)=sin(π +1.6962)= -sin(1.6962)≈-0.9925? Wait, earlier we saw sin(4.8378)≈-0.9925, but wait, let me compute accurately.Wait, 2.4189*2=4.8378sin(4.8378)=sin(π +1.6962)= -sin(1.6962)Compute sin(1.6962):1.6962 radians≈97 degrees, sin≈0.9925So, sin(4.8378)= -0.9925Similarly, sin(3b)=sin(7.2567)=sin(2π +1.0145)=sin(1.0145)≈0.8462Wait, 3b≈7.2567, which is more than 2π≈6.2832, so subtract 2π: 7.2567-6.2832≈0.9735 radians≈55.8 degreessin(0.9735)≈0.8290Wait, let me compute sin(7.2567):7.2567 - 2π≈7.2567 -6.2832≈0.9735sin(0.9735)≈sin(55.8 degrees)≈0.8290Similarly, sin(4b)=sin(9.6756)=sin(3π - something)=sin(9.6756 -3π)=sin(9.6756 -9.4248)=sin(0.2508)≈0.2487Wait, 4b≈9.6756, which is more than 3π≈9.4248, so 9.6756-9.4248≈0.2508 radians≈14.37 degreessin(0.2508)≈0.2487Similarly, sin(5b)=sin(12.0945)=sin(4π - something)=sin(12.0945 -12.5664)=sin(-0.4719)= -sin(0.4719)≈-0.4540Wait, 5b≈12.0945, which is less than 4π≈12.5664, so 12.0945-3π≈12.0945-9.4248≈2.6697 radians≈153 degreessin(2.6697)=sin(π -0.4719)=sin(0.4719)≈0.4540, but since it's in the second quadrant, sin is positive. Wait, but 2.6697 radians≈153 degrees, which is in the second quadrant, so sin is positive. So, sin(5b)=sin(2.6697)≈0.4540Wait, but 5b=12.0945, which is 3π + (12.0945-9.4248)=3π +2.6697, but sin(3π +θ)= -sin θ, so sin(5b)=sin(3π +2.6697)= -sin(2.6697)≈-0.4540Wait, that's conflicting. Let me clarify.Wait, 5b≈12.0945 radiansSubtract 2π≈6.2832: 12.0945 -6.2832≈5.8113Subtract another 2π≈6.2832: 5.8113 -6.2832≈-0.4719So, sin(5b)=sin(-0.4719)= -sin(0.4719)≈-0.4540Similarly, sin(6b)=sin(14.5134)=sin(4π + something)=sin(14.5134 -4π)=sin(14.5134 -12.5664)=sin(1.947)≈0.9120Wait, 6b≈14.5134, subtract 4π≈12.5664: 14.5134-12.5664≈1.947 radians≈111.6 degreessin(1.947)≈0.9120Similarly, sin(7b)=sin(17.9323)=sin(5π + something)=sin(17.9323 -5π)=sin(17.9323 -15.70796)=sin(2.2243)≈0.7890Wait, 7b≈17.9323, subtract 5π≈15.70796: 17.9323-15.70796≈2.2243 radians≈127.5 degreessin(2.2243)≈0.7890Similarly, sin(8b)=sin(19.3512)=sin(6π - something)=sin(19.3512 -18.8496)=sin(0.5016)≈0.4800Wait, 8b≈19.3512, subtract 6π≈18.8496: 19.3512-18.8496≈0.5016 radians≈28.7 degreessin(0.5016)≈0.4800So, compiling these:sin(b)=0.6614 sin(2b)= -0.9925 sin(3b)= sin(7.2567)=sin(0.9735)≈0.8290 sin(4b)=sin(9.6756)=sin(0.2508)≈0.2487 sin(5b)=sin(12.0945)=sin(-0.4719)≈-0.4540 sin(6b)=sin(14.5134)=sin(1.947)≈0.9120 sin(7b)=sin(17.9323)=sin(2.2243)≈0.7890 sin(8b)=sin(19.3512)=sin(0.5016)≈0.4800Now, compute f(x) for x=5 to x=10:f(5)= (80/√7)*sin(3b) +140≈(30.24)*0.8290 +140≈25.05 +140≈165.05 f(6)=30.24*0.2487 +140≈7.51 +140≈147.51 f(7)=30.24*(-0.4540) +140≈-13.72 +140≈126.28 f(8)=30.24*0.9120 +140≈27.63 +140≈167.63 f(9)=30.24*0.7890 +140≈23.83 +140≈163.83 f(10)=30.24*0.4800 +140≈14.52 +140≈154.52So, the f(x) values are approximately:x=1:120 x=2:140 x=3:160 x=4:110 x=5:165.05 x=6:147.51 x=7:126.28 x=8:167.63 x=9:163.83 x=10:154.52Now, let's sum these up:120 +140=260 260+160=420 420+110=530 530+165.05≈695.05 695.05+147.51≈842.56 842.56+126.28≈968.84 968.84+167.63≈1136.47 1136.47+163.83≈1300.3 1300.3+154.52≈1454.82So, the total number of words in the first 10 chapters is approximately 1454.82. Since we're dealing with words, we should round to the nearest whole number, so approximately 1455 words.But let me double-check the calculations for f(5) to f(10):f(5)=30.24*0.8290≈25.05 +140=165.05 f(6)=30.24*0.2487≈7.51 +140=147.51 f(7)=30.24*(-0.4540)≈-13.72 +140=126.28 f(8)=30.24*0.9120≈27.63 +140=167.63 f(9)=30.24*0.7890≈23.83 +140=163.83 f(10)=30.24*0.4800≈14.52 +140=154.52Adding these:165.05 +147.51=312.56 312.56 +126.28=438.84 438.84 +167.63=606.47 606.47 +163.83=770.3 770.3 +154.52=924.82Wait, that's the sum from x=5 to x=10:≈924.82Adding the first four chapters:120+140+160+110=530Total sum:530 +924.82≈1454.82≈1455Yes, that's correct.So, the total number of words in the first 10 chapters according to John's model is approximately 1455 words.</think>

🧠:<think>Alright, so I have this problem about two marketing automation strategies, A and B, and I need to calculate their total engagement over the first 48 hours and also find the time when their engagement levels are equal. Let me try to break this down step by step.First, for part 1, I need to find the total engagement for each strategy over 48 hours. Engagement is given by their respective functions: ( E_A(t) = 100 cdot sinleft(frac{pi}{12} tright) + 300 ) for strategy A and ( E_B(t) = 150 cdot log(t+1) + 250 ) for strategy B. I think total engagement would be the integral of the engagement function over the time period from 0 to 48 hours. So, for strategy A, I need to compute the integral of ( E_A(t) ) from 0 to 48, and similarly for strategy B, the integral of ( E_B(t) ) from 0 to 48.Let me start with strategy A. The function is ( E_A(t) = 100 cdot sinleft(frac{pi}{12} tright) + 300 ). To find the total engagement, I need to integrate this from 0 to 48.So, the integral of ( E_A(t) ) dt from 0 to 48 is:[int_{0}^{48} left(100 cdot sinleft(frac{pi}{12} tright) + 300right) dt]I can split this integral into two parts:[100 int_{0}^{48} sinleft(frac{pi}{12} tright) dt + 300 int_{0}^{48} dt]Let me compute each integral separately.First, the integral of ( sinleft(frac{pi}{12} tright) ). The integral of ( sin(ax) ) is ( -frac{1}{a} cos(ax) ). So here, a is ( frac{pi}{12} ), so the integral becomes:[-frac{12}{pi} cosleft(frac{pi}{12} tright)]Evaluated from 0 to 48, so:[-frac{12}{pi} left[ cosleft(frac{pi}{12} cdot 48right) - cos(0) right]]Simplify ( frac{pi}{12} cdot 48 ). 48 divided by 12 is 4, so it's ( 4pi ). So:[-frac{12}{pi} left[ cos(4pi) - cos(0) right]]I know that ( cos(4pi) ) is 1 because cosine has a period of ( 2pi ), so 4π is two full periods. Similarly, ( cos(0) ) is also 1. So:[-frac{12}{pi} [1 - 1] = 0]So the integral of the sine part is 0. That makes sense because the sine function is oscillating and over a full period, the area cancels out.Now, the second integral is straightforward:[300 int_{0}^{48} dt = 300 cdot (48 - 0) = 300 cdot 48 = 14,400]So, the total engagement for strategy A over 48 hours is 14,400.Wait, that seems too straightforward. Let me double-check. The integral of the sine term over 0 to 48 is indeed zero because it completes an integer number of periods. Since the period of ( sinleft(frac{pi}{12} tright) ) is ( frac{2pi}{pi/12} = 24 ) hours. So over 48 hours, it's two full periods. So yes, the integral over each period is zero, so over two periods, it's still zero. So the total engagement is just the integral of the constant term, which is 300*48=14,400. Okay, that seems correct.Now, moving on to strategy B. The engagement function is ( E_B(t) = 150 cdot log(t + 1) + 250 ). Again, to find the total engagement over 48 hours, I need to integrate this function from 0 to 48.So, the integral is:[int_{0}^{48} left(150 cdot log(t + 1) + 250right) dt]Again, I can split this into two integrals:[150 int_{0}^{48} log(t + 1) dt + 250 int_{0}^{48} dt]Let me compute each part.First, the integral of ( log(t + 1) ). I remember that the integral of ( log(x) ) is ( x log(x) - x ). So, substituting ( u = t + 1 ), then du = dt, and when t=0, u=1, t=48, u=49.So, the integral becomes:[int_{1}^{49} log(u) du = [u log(u) - u]_{1}^{49}]Calculating at 49:[49 log(49) - 49]And at 1:[1 log(1) - 1 = 0 - 1 = -1]So, subtracting:[(49 log(49) - 49) - (-1) = 49 log(49) - 49 + 1 = 49 log(49) - 48]Therefore, the integral of ( log(t + 1) ) from 0 to 48 is ( 49 log(49) - 48 ).So, multiplying by 150:[150 cdot (49 log(49) - 48)]Now, let me compute the second integral:[250 int_{0}^{48} dt = 250 cdot 48 = 12,000]So, the total engagement for strategy B is:[150 cdot (49 log(49) - 48) + 12,000]I need to compute this numerically. Let me calculate each part step by step.First, compute ( log(49) ). Assuming this is the natural logarithm, since in calculus we usually use ln. If it's base 10, the value would be different, but given the context of marketing, it might be base e. Let me confirm: in the problem statement, it's written as log(t+1). In mathematics, log without base specified can sometimes be base 10 or natural log. But in many contexts, especially in calculus, it's natural log. However, in marketing, sometimes log base 10 is used. Hmm, this is a bit ambiguous.Wait, let me check the problem statement again. It says ( E_B(t) = 150 cdot log(t+1) + 250 ). It doesn't specify the base. Hmm. In many applied fields, log can be base 10, but in calculus, it's usually natural log. Since the problem is about integration, which is a calculus concept, I think it's more likely natural logarithm. But to be safe, maybe I should note that. Alternatively, perhaps it's base e.But to be precise, I think I should assume natural logarithm unless specified otherwise. So, let's proceed with natural log.So, ( ln(49) ). 49 is 7 squared, so ( ln(49) = ln(7^2) = 2 ln(7) ). I know that ( ln(7) ) is approximately 1.9459, so ( 2 * 1.9459 ≈ 3.8918 ).So, ( 49 ln(49) = 49 * 3.8918 ≈ 49 * 3.8918 ). Let me compute that:49 * 3 = 14749 * 0.8918 ≈ 49 * 0.8 = 39.2, 49 * 0.0918 ≈ 4.50, so total ≈ 39.2 + 4.50 = 43.7So, total ≈ 147 + 43.7 = 190.7So, ( 49 ln(49) ≈ 190.7 )Then, subtract 48: 190.7 - 48 = 142.7So, ( 49 ln(49) - 48 ≈ 142.7 )Multiply by 150: 150 * 142.7 ≈ 150 * 140 = 21,000 and 150 * 2.7 = 405, so total ≈ 21,000 + 405 = 21,405Then, add the second integral result: 21,405 + 12,000 = 33,405So, the total engagement for strategy B is approximately 33,405.Wait, let me double-check my calculations because 150*(142.7) is 21,405, and adding 12,000 gives 33,405. That seems correct.But let me verify the integral computation again because sometimes constants can be tricky.The integral of ( log(t + 1) ) from 0 to 48 is ( (49 ln(49) - 49) - (1 ln(1) - 1) ). Since ( ln(1) = 0 ), this becomes ( 49 ln(49) - 49 - (-1) = 49 ln(49) - 48 ). So that part is correct.Then, 49 ln(49) is approximately 49 * 3.8918 ≈ 190.7, subtract 48 gives 142.7. Multiply by 150: 142.7 * 150. Let me compute 142.7 * 100 = 14,270; 142.7 * 50 = 7,135. So total is 14,270 + 7,135 = 21,405. Then, add 12,000: 21,405 + 12,000 = 33,405. Okay, that's correct.So, total engagement for strategy A is 14,400 and for strategy B is approximately 33,405.Wait, that's a significant difference. Strategy B seems much more effective in terms of total engagement over 48 hours.Now, moving on to part 2. The consultant wants to find the point in time t (within the first 48 hours) where the engagement levels of both strategies are equal. So, set ( E_A(t) = E_B(t) ) and solve for t.So, the equation is:[100 cdot sinleft(frac{pi}{12} tright) + 300 = 150 cdot log(t + 1) + 250]Let me rearrange this equation:[100 cdot sinleft(frac{pi}{12} tright) + 300 - 250 = 150 cdot log(t + 1)]Simplify:[100 cdot sinleft(frac{pi}{12} tright) + 50 = 150 cdot log(t + 1)]Divide both sides by 50 to simplify:[2 cdot sinleft(frac{pi}{12} tright) + 1 = 3 cdot log(t + 1)]So, the equation becomes:[2 sinleft(frac{pi}{12} tright) + 1 = 3 log(t + 1)]This is a transcendental equation, meaning it can't be solved algebraically and likely requires numerical methods. So, I'll need to use some approximation technique, like the Newton-Raphson method, or use a graphing approach to estimate the solution.Alternatively, I can use trial and error by plugging in values of t and seeing where the two sides are equal.Let me first analyze the behavior of both sides.Left side: ( 2 sinleft(frac{pi}{12} tright) + 1 )This is a sine wave with amplitude 2, shifted up by 1. So, it oscillates between -2 + 1 = -1 and 2 + 1 = 3. However, since t is between 0 and 48, let's see how many periods there are.The period of the sine function is ( frac{2pi}{pi/12} = 24 ) hours. So, over 48 hours, it completes two full cycles.Right side: ( 3 log(t + 1) )This is a logarithmic function, which increases slowly over time. At t=0, it's 0. As t increases, it grows, but the growth rate decreases.So, the left side oscillates between -1 and 3, while the right side starts at 0 and increases to ( 3 log(49) ≈ 3 * 3.8918 ≈ 11.675 ). So, the right side is always positive and increasing, while the left side oscillates.Therefore, the equation ( 2 sinleft(frac{pi}{12} tright) + 1 = 3 log(t + 1) ) will have solutions where the oscillating left side intersects the increasing right side.Given that the right side starts at 0 and grows, and the left side starts at 1 (since at t=0, sin(0)=0, so left side is 1). So, at t=0, left side is 1, right side is 0. So, left side is higher.As t increases, the right side increases, while the left side oscillates. So, the first intersection might occur when the left side is decreasing from its peak.Let me try to find approximate values.First, let's compute both sides at various t values.Let me start by checking t=0:Left: 2*sin(0) +1 = 0 +1 =1Right: 3*log(1)=0So, left > right.t=12:Left: 2*sin(π/12 *12)=2*sin(π)=0 +1=1Right: 3*log(13)≈3*2.5649≈7.6947So, left=1, right≈7.6947. So, left < right.So, between t=0 and t=12, the left side goes from 1 to 1, but in between, it peaks at t=6.Wait, let's check t=6:Left: 2*sin(π/12 *6)=2*sin(π/2)=2*1=2 +1=3Right: 3*log(7)≈3*1.9459≈5.8377So, left=3, right≈5.8377. So, left < right.Wait, but at t=0, left=1, right=0. At t=6, left=3, right≈5.8377. So, left increases to 3, right increases to ~5.84. So, left is still less than right at t=6.Wait, but at t=0, left=1 > right=0. At t=6, left=3 < right≈5.84. So, the left side crosses the right side somewhere between t=0 and t=6?Wait, no. Because at t=0, left=1, right=0. So, left > right. At t=6, left=3, right≈5.84. So, left < right. So, the functions cross somewhere between t=0 and t=6.Wait, but the left side is increasing from t=0 to t=6, while the right side is also increasing. So, perhaps they cross once between t=0 and t=6.Wait, let me check t=3:Left: 2*sin(π/12 *3)=2*sin(π/4)=2*(√2/2)=√2≈1.4142 +1≈2.4142Right: 3*log(4)=3*1.3863≈4.1589So, left≈2.4142 < right≈4.1589.So, left is still less than right at t=3.Wait, but at t=0, left=1 > right=0. So, the left side starts above, then at t=3, left≈2.4142 < right≈4.1589. So, they must cross somewhere between t=0 and t=3.Wait, let me check t=1:Left: 2*sin(π/12 *1)=2*sin(π/12)≈2*0.2588≈0.5176 +1≈1.5176Right: 3*log(2)=3*0.6931≈2.0794So, left≈1.5176 < right≈2.0794.So, at t=1, left < right.At t=0, left=1 > right=0.So, the crossing is between t=0 and t=1.Wait, let me check t=0.5:Left: 2*sin(π/12 *0.5)=2*sin(π/24)≈2*0.1305≈0.261 +1≈1.261Right: 3*log(1.5)=3*0.4055≈1.2165So, left≈1.261 > right≈1.2165.So, at t=0.5, left > right.At t=1, left≈1.5176 < right≈2.0794.So, the crossing is between t=0.5 and t=1.Let me try t=0.75:Left: 2*sin(π/12 *0.75)=2*sin(π/16)≈2*0.1951≈0.3902 +1≈1.3902Right: 3*log(1.75)=3*0.5596≈1.6788So, left≈1.3902 < right≈1.6788.So, at t=0.75, left < right.At t=0.5, left≈1.261 > right≈1.2165.So, crossing between t=0.5 and t=0.75.Let me try t=0.6:Left: 2*sin(π/12 *0.6)=2*sin(π/20)≈2*0.1564≈0.3128 +1≈1.3128Right: 3*log(1.6)=3*0.4700≈1.4100So, left≈1.3128 < right≈1.4100.So, at t=0.6, left < right.At t=0.5, left≈1.261 > right≈1.2165.So, crossing between t=0.5 and t=0.6.Let me try t=0.55:Left: 2*sin(π/12 *0.55)=2*sin(π/12 *0.55). Let's compute π/12≈0.2618, so 0.2618*0.55≈0.144 radians.sin(0.144)≈0.1433. So, 2*0.1433≈0.2866 +1≈1.2866Right: 3*log(1.55)=3*0.4383≈1.3149So, left≈1.2866 < right≈1.3149.So, left < right at t=0.55.At t=0.5, left≈1.261 > right≈1.2165.So, the crossing is between t=0.5 and t=0.55.Let me try t=0.525:Left: 2*sin(π/12 *0.525)=2*sin(0.2618*0.525)=2*sin(0.1373)≈2*0.1367≈0.2734 +1≈1.2734Right: 3*log(1.525)=3*0.4224≈1.2672So, left≈1.2734 > right≈1.2672.So, at t=0.525, left > right.At t=0.55, left≈1.2866 < right≈1.3149.Wait, no, at t=0.525, left≈1.2734 > right≈1.2672.At t=0.55, left≈1.2866 < right≈1.3149.Wait, that can't be. Because at t=0.525, left is slightly higher, but at t=0.55, left is lower. So, the crossing is between t=0.525 and t=0.55.Wait, let me compute at t=0.53:Left: 2*sin(π/12 *0.53)=2*sin(0.2618*0.53)=2*sin(0.1388)≈2*0.1383≈0.2766 +1≈1.2766Right: 3*log(1.53)=3*0.4274≈1.2822So, left≈1.2766 < right≈1.2822.So, at t=0.53, left < right.At t=0.525, left≈1.2734 > right≈1.2672.So, crossing between t=0.525 and t=0.53.Let me try t=0.5275:Left: 2*sin(π/12 *0.5275)=2*sin(0.2618*0.5275)=2*sin(0.1382)≈2*0.1379≈0.2758 +1≈1.2758Right: 3*log(1.5275)=3*0.4253≈1.2759So, left≈1.2758 ≈ right≈1.2759.Wow, that's very close. So, t≈0.5275 hours.So, approximately 0.5275 hours is about 31.65 minutes.So, the engagement levels are equal at approximately t≈0.5275 hours, or about 31.65 minutes after the campaign starts.But let me check if there are more intersections beyond this point.Given that the left side is a sine wave with period 24 hours, it will oscillate up and down, while the right side is a slowly increasing logarithmic function.So, after the first intersection near t≈0.5275, the left side will go up to 3 at t=6, then back down to 1 at t=12, then up to 3 at t=18, down to 1 at t=24, and so on.Meanwhile, the right side keeps increasing. So, after t=0.5275, the left side peaks at 3, while the right side at t=6 is about 5.8377, which is higher than 3. So, the left side never catches up again in the first peak.Wait, at t=6, left=3, right≈5.8377. So, left < right.At t=12, left=1, right≈7.6947. So, left < right.At t=18, left=3, right≈3*log(19)=3*2.9444≈8.8332. So, left < right.At t=24, left=1, right≈3*log(25)=3*3.2189≈9.6567.At t=30, left=3, right≈3*log(31)=3*3.4339≈10.3017.At t=36, left=1, right≈3*log(37)=3*3.6109≈10.8327.At t=42, left=3, right≈3*log(43)=3*3.7612≈11.2836.At t=48, left=1, right≈3*log(49)=3*3.8918≈11.6754.So, the left side oscillates between 1 and 3, while the right side increases from 0 to ~11.6754. So, the only intersection is near t≈0.5275 hours.Therefore, the point in time where the engagement levels are equal is approximately t≈0.5275 hours, which is about 31.65 minutes.But to be precise, let me use a more accurate method, like the Newton-Raphson method, to find the root of the equation ( f(t) = 2 sinleft(frac{pi}{12} tright) + 1 - 3 log(t + 1) = 0 ).We can use the Newton-Raphson formula:[t_{n+1} = t_n - frac{f(t_n)}{f'(t_n)}]First, let's define f(t):[f(t) = 2 sinleft(frac{pi}{12} tright) + 1 - 3 log(t + 1)]Compute f'(t):[f'(t) = 2 cdot frac{pi}{12} cosleft(frac{pi}{12} tright) - 3 cdot frac{1}{t + 1}]Simplify:[f'(t) = frac{pi}{6} cosleft(frac{pi}{12} tright) - frac{3}{t + 1}]We already have an initial guess t₀=0.5275, where f(t₀)≈0.Let me compute f(t₀):t₀=0.5275Compute ( frac{pi}{12} t₀ ≈ 0.2618 * 0.5275 ≈ 0.1382 ) radians.sin(0.1382)≈0.1379, so 2*sin≈0.2758So, f(t₀)=0.2758 +1 -3*log(1.5275)Compute log(1.5275)=ln(1.5275)≈0.4253So, 3*0.4253≈1.2759Thus, f(t₀)=1.2758 +1 -1.2759≈1.2758 +1 -1.2759≈1.0000 -0.0001≈0.9999? Wait, that can't be.Wait, no. Wait, f(t)=2 sin(...) +1 -3 log(...). So, 2 sin(...)≈0.2758, plus 1 is 1.2758, minus 3 log(...)≈1.2759. So, 1.2758 -1.2759≈-0.0001.So, f(t₀)=≈-0.0001.So, f(t₀)=≈-0.0001.Compute f'(t₀):( frac{pi}{6} cos(0.1382) - frac{3}{1.5275} )cos(0.1382)≈0.9903So, ( frac{pi}{6} *0.9903≈0.5236 *0.9903≈0.5185 )( frac{3}{1.5275}≈1.964 )So, f'(t₀)=0.5185 -1.964≈-1.4455So, Newton-Raphson update:t₁ = t₀ - f(t₀)/f'(t₀) ≈0.5275 - (-0.0001)/(-1.4455)≈0.5275 - (0.0001/1.4455)≈0.5275 -0.000069≈0.52743So, t₁≈0.52743Compute f(t₁):t=0.52743Compute ( frac{pi}{12} t≈0.2618*0.52743≈0.1382 )sin(0.1382)≈0.1379, 2*sin≈0.2758log(1.52743)=ln(1.52743)≈0.42533*log≈1.2759So, f(t)=0.2758 +1 -1.2759≈1.2758 -1.2759≈-0.0001Wait, same as before. Hmm, perhaps due to the precision of my calculations.Alternatively, maybe the function is very flat near the root, so Newton-Raphson isn't converging quickly.Alternatively, perhaps using linear approximation between t=0.525 and t=0.53.At t=0.525, f(t)=1.2734 -1.2672≈0.0062Wait, no, earlier I had:At t=0.525, left≈1.2734, right≈1.2672, so f(t)=left - right≈0.0062At t=0.53, left≈1.2766, right≈1.2822, so f(t)=1.2766 -1.2822≈-0.0056So, f(t) crosses zero between t=0.525 and t=0.53.Using linear approximation:Between t1=0.525, f(t1)=0.0062t2=0.53, f(t2)=-0.0056The change in t is 0.005, and the change in f is -0.0118.We need to find t where f(t)=0.So, the fraction is 0.0062 / 0.0118≈0.5254So, t≈t1 + (0 - f(t1))*(t2 - t1)/(f(t2)-f(t1))=0.525 + (0 -0.0062)*(0.005)/(-0.0118)=0.525 + (0.0062*0.005)/0.0118≈0.525 + (0.000031)/0.0118≈0.525 +0.00263≈0.52763So, t≈0.5276 hours.Which is consistent with our earlier estimate.So, t≈0.5276 hours, which is approximately 0.5276*60≈31.656 minutes.So, about 31.66 minutes.Therefore, the point in time where the engagement levels are equal is approximately t≈0.5276 hours, or 31.66 minutes.But to express this in hours, it's approximately 0.5276 hours.Alternatively, if we need more precision, we can use more iterations of Newton-Raphson, but given the small value, it's probably sufficient.So, summarizing:1. Total engagement for strategy A: 14,400Total engagement for strategy B: approximately 33,4052. The engagement levels are equal at approximately t≈0.5276 hours, or about 31.66 minutes.But let me check if the total engagement for strategy B is indeed 33,405.Wait, earlier I had:Integral of strategy B: 150*(49 ln49 -48) +12,000≈150*(142.7)+12,000≈21,405 +12,000=33,405.But let me compute 49 ln49 more accurately.ln(49)=ln(7^2)=2 ln7≈2*1.9459101≈3.8918202So, 49*3.8918202≈49*3.8918202.Compute 49*3=14749*0.8918202≈49*0.8=39.2; 49*0.0918202≈4.499So, total≈39.2 +4.499≈43.699So, 147 +43.699≈190.699So, 49 ln49≈190.699Subtract 48:190.699 -48=142.699Multiply by 150:142.699*150=21,404.85Add 12,000:21,404.85 +12,000=33,404.85≈33,405.So, correct.Therefore, the answers are:1. Strategy A: 14,400; Strategy B: 33,4052. t≈0.5276 hours, or about 31.66 minutes.But to present the answer, perhaps we can write t≈0.53 hours or 31.66 minutes.Alternatively, if we need more decimal places, but probably 0.53 hours is sufficient.So, final answers:1. Total engagement: A=14,400; B≈33,4052. t≈0.53 hours</think>

🧠:<think>Alright, so I've got these two problems to solve related to ergonomic office chair design. Let me tackle them one by one.Starting with the first problem: It involves sitting height (H) which follows a normal distribution with a mean of 90 cm and a standard deviation of 7 cm. I need to find the probability that a randomly selected individual has a sitting height within the range that's 1.5 standard deviations below the mean to 2 standard deviations above the mean.Hmm, okay. So, since it's a normal distribution, I can use the Z-score to standardize the values and then use the standard normal distribution table to find the probabilities.First, let me figure out the actual sitting height values corresponding to 1.5 standard deviations below and 2 standard deviations above the mean.The mean (μ) is 90 cm, and the standard deviation (σ) is 7 cm.1.5 standard deviations below the mean would be:μ - 1.5σ = 90 - 1.5*7 = 90 - 10.5 = 79.5 cm.2 standard deviations above the mean would be:μ + 2σ = 90 + 2*7 = 90 + 14 = 104 cm.So, I need the probability that H is between 79.5 cm and 104 cm.To find this probability, I can convert these values to Z-scores.Z-score formula is Z = (X - μ)/σ.For 79.5 cm:Z1 = (79.5 - 90)/7 = (-10.5)/7 = -1.5.For 104 cm:Z2 = (104 - 90)/7 = 14/7 = 2.So, now I need to find the probability that Z is between -1.5 and 2.In the standard normal distribution, the probability between two Z-scores can be found by subtracting the cumulative probability of the lower Z from the cumulative probability of the upper Z.I remember that the cumulative probability for Z = -1.5 is the area to the left of -1.5, and for Z = 2, it's the area to the left of 2.Looking up Z = -1.5 in the standard normal table, the cumulative probability is approximately 0.0668.For Z = 2, the cumulative probability is approximately 0.9772.So, the probability between -1.5 and 2 is 0.9772 - 0.0668 = 0.9104.Therefore, the probability is approximately 91.04%.Wait, let me double-check that. Sometimes, I get confused with the areas. So, if I have Z = -1.5, that's in the left tail, and Z = 2 is in the right tail. The area between them should indeed be the difference between the two cumulative probabilities. Yeah, that seems right.Moving on to the second problem: It's about thigh circumference (C) and buttock-popliteal length (L). They are linearly related by the equation C = kL + b, where k = 0.95 and b = 12 cm. The average L is 50 cm with a standard deviation of 5 cm. I need to determine the expected thigh circumference and its variability for an individual with a buttock-popliteal length that is 1 standard deviation above the mean.Alright, so first, let's find the value of L that is 1 standard deviation above the mean.Mean of L (μ_L) is 50 cm, standard deviation (σ_L) is 5 cm.So, 1 standard deviation above the mean is μ_L + σ_L = 50 + 5 = 55 cm.Now, plug this into the linear equation to find the expected C.C = 0.95*L + 12.So, substituting L = 55:C = 0.95*55 + 12.Let me calculate that:0.95*55: 55*1 = 55, 55*0.95 = 55 - 55*0.05 = 55 - 2.75 = 52.25.Then, adding 12: 52.25 + 12 = 64.25 cm.So, the expected thigh circumference is 64.25 cm.Now, for the variability, I think this refers to the standard deviation of C, given the variability in L. Since C is a linear function of L, the variability (standard deviation) of C can be found using the formula:σ_C = |k| * σ_L.Because in a linear transformation, the standard deviation scales by the absolute value of the slope.Given that k = 0.95, which is positive, so σ_C = 0.95 * σ_L.σ_L is 5 cm, so σ_C = 0.95 * 5 = 4.75 cm.Therefore, the variability (standard deviation) of the thigh circumference is 4.75 cm.Wait, but hold on. Is this the correct approach? Because in regression analysis, the standard deviation of the residuals is also a factor, but the problem mentions that the sum of squared residuals is minimized, which is the least squares regression. However, it doesn't provide the standard error of the estimate or the residual standard deviation. So, maybe the variability here is just referring to the standard deviation of C given L, which is the standard deviation explained by the regression line, which is indeed |k|*σ_L.Alternatively, if we consider that the total variability of C would be the variability explained by L plus the residual variability. But since the problem doesn't provide the residual standard error, I think it's safe to assume that they just want the variability due to L, which is 4.75 cm.So, summarizing:1. The probability is approximately 91.04%.2. The expected thigh circumference is 64.25 cm with a variability (standard deviation) of 4.75 cm.I think that's it. Let me just make sure I didn't make any calculation errors.For the first problem:Z1 = (79.5 - 90)/7 = -1.5, correct.Z2 = (104 - 90)/7 = 2, correct.Cumulative probabilities: Z=-1.5 is 0.0668, Z=2 is 0.9772, difference is 0.9104, so 91.04%, correct.For the second problem:L = 55 cm, C = 0.95*55 + 12 = 52.25 + 12 = 64.25 cm, correct.σ_C = 0.95*5 = 4.75 cm, correct.Yes, that seems solid.Final Answer1. The probability is boxed{0.9104}.2. The expected thigh circumference is boxed{64.25} cm with a variability of boxed{4.75} cm.</think>

🧠:<think>Okay, so I have this problem about a drone's path defined by parametric equations, and I need to find two things: the total length of the path over a certain interval and the total energy consumed based on a power function. Hmm, let's take it step by step.First, the parametric equations are given as:[ x(t) = a sin(bt + c) ][ y(t) = a cos(bt + c) ][ z(t) = d sin^2(bt + c) ]And the interval is from ( t = 0 ) to ( t = frac{2pi}{b} ). Starting with the first part: finding the total length of the path. I remember that the formula for the length of a parametric curve from ( t = a ) to ( t = b ) is the integral of the square root of the sum of the squares of the derivatives of each component with respect to ( t ). So, the formula is:[ L = int_{a}^{b} sqrt{left( frac{dx}{dt} right)^2 + left( frac{dy}{dt} right)^2 + left( frac{dz}{dt} right)^2} , dt ]Alright, so I need to compute the derivatives of ( x(t) ), ( y(t) ), and ( z(t) ) with respect to ( t ), square each, add them up, take the square root, and then integrate over the interval.Let me compute each derivative one by one.First, ( frac{dx}{dt} ):[ x(t) = a sin(bt + c) ]So, derivative is:[ frac{dx}{dt} = a cdot b cos(bt + c) ]Similarly, ( frac{dy}{dt} ):[ y(t) = a cos(bt + c) ]Derivative is:[ frac{dy}{dt} = -a cdot b sin(bt + c) ]Now, ( frac{dz}{dt} ):[ z(t) = d sin^2(bt + c) ]Hmm, derivative of ( sin^2(u) ) is ( 2 sin(u) cos(u) cdot u' ), so:[ frac{dz}{dt} = d cdot 2 sin(bt + c) cos(bt + c) cdot b ]Simplify that:[ frac{dz}{dt} = 2bd sin(bt + c) cos(bt + c) ]Alternatively, I can use the double-angle identity: ( sin(2u) = 2 sin u cos u ), so:[ frac{dz}{dt} = bd sin(2(bt + c)) ]But maybe I'll keep it as is for now.So, now I have all three derivatives:1. ( frac{dx}{dt} = ab cos(bt + c) )2. ( frac{dy}{dt} = -ab sin(bt + c) )3. ( frac{dz}{dt} = 2bd sin(bt + c) cos(bt + c) )Now, let's square each of these:1. ( left( frac{dx}{dt} right)^2 = a^2 b^2 cos^2(bt + c) )2. ( left( frac{dy}{dt} right)^2 = a^2 b^2 sin^2(bt + c) )3. ( left( frac{dz}{dt} right)^2 = 4b^2 d^2 sin^2(bt + c) cos^2(bt + c) )Adding these up:[ left( frac{dx}{dt} right)^2 + left( frac{dy}{dt} right)^2 + left( frac{dz}{dt} right)^2 = a^2 b^2 (cos^2(bt + c) + sin^2(bt + c)) + 4b^2 d^2 sin^2(bt + c) cos^2(bt + c) ]I remember that ( cos^2 theta + sin^2 theta = 1 ), so that simplifies the first part:[ a^2 b^2 (1) + 4b^2 d^2 sin^2(bt + c) cos^2(bt + c) ]So, the expression becomes:[ a^2 b^2 + 4b^2 d^2 sin^2(bt + c) cos^2(bt + c) ]Hmm, that looks a bit complicated, but maybe I can simplify the second term. Let's see:I recall that ( sin(2theta) = 2 sin theta cos theta ), so ( sin^2(2theta) = 4 sin^2 theta cos^2 theta ). Therefore, ( sin^2 theta cos^2 theta = frac{1}{4} sin^2(2theta) ).So, substituting that in:[ 4b^2 d^2 sin^2(bt + c) cos^2(bt + c) = 4b^2 d^2 cdot frac{1}{4} sin^2(2(bt + c)) ]Simplify:[ b^2 d^2 sin^2(2(bt + c)) ]So, now the expression under the square root becomes:[ a^2 b^2 + b^2 d^2 sin^2(2(bt + c)) ]Factor out ( b^2 ):[ b^2 left( a^2 + d^2 sin^2(2(bt + c)) right) ]So, the integrand simplifies to:[ sqrt{b^2 left( a^2 + d^2 sin^2(2(bt + c)) right)} = b sqrt{a^2 + d^2 sin^2(2(bt + c))} ]Therefore, the integral for the length ( L ) becomes:[ L = int_{0}^{frac{2pi}{b}} b sqrt{a^2 + d^2 sin^2(2(bt + c))} , dt ]Hmm, that still looks a bit tricky. Let me see if I can make a substitution to simplify it.Let me set ( u = bt + c ). Then, ( du = b dt ), which implies ( dt = frac{du}{b} ).When ( t = 0 ), ( u = c ), and when ( t = frac{2pi}{b} ), ( u = 2pi + c ).So, substituting into the integral:[ L = int_{c}^{2pi + c} b sqrt{a^2 + d^2 sin^2(2u)} cdot frac{du}{b} ]Simplify:[ L = int_{c}^{2pi + c} sqrt{a^2 + d^2 sin^2(2u)} , du ]Now, since the integrand is periodic with period ( pi ) (because of the ( sin^2(2u) ) term), and the interval of integration is ( 2pi ), which is two periods. Therefore, the integral over ( [c, 2pi + c] ) is the same as twice the integral over one period, say from ( 0 ) to ( pi ).But actually, since the function is periodic, the integral over any interval of length ( 2pi ) will be the same. So, we can shift the interval to start at 0 without changing the value:[ L = int_{0}^{2pi} sqrt{a^2 + d^2 sin^2(2u)} , du ]Hmm, but ( sin^2(2u) ) has a period of ( pi ), so integrating over ( 0 ) to ( pi ) and doubling it would give the same result as integrating over ( 0 ) to ( 2pi ). Let me confirm:Since ( sin^2(2u) ) has period ( pi ), the function ( sqrt{a^2 + d^2 sin^2(2u)} ) is also periodic with period ( pi ). Therefore, integrating over ( 0 ) to ( 2pi ) is the same as twice the integral from ( 0 ) to ( pi ).So, let's write:[ L = 2 int_{0}^{pi} sqrt{a^2 + d^2 sin^2(2u)} , du ]But even better, let's make another substitution to simplify the integral.Let me set ( v = 2u ). Then, ( dv = 2 du ), so ( du = frac{dv}{2} ).When ( u = 0 ), ( v = 0 ); when ( u = pi ), ( v = 2pi ).So, substituting:[ L = 2 int_{0}^{2pi} sqrt{a^2 + d^2 sin^2(v)} cdot frac{dv}{2} ]Simplify:[ L = int_{0}^{2pi} sqrt{a^2 + d^2 sin^2(v)} , dv ]Hmm, this is a standard elliptic integral, isn't it? The integral of ( sqrt{a^2 + d^2 sin^2(v)} ) over a full period. I think there's a formula for this.Wait, actually, the integral ( int_{0}^{2pi} sqrt{a^2 + d^2 sin^2(v)} , dv ) is related to the complete elliptic integral of the second kind. Let me recall.The complete elliptic integral of the second kind is defined as:[ E(k) = int_{0}^{frac{pi}{2}} sqrt{1 - k^2 sin^2 theta} , dtheta ]But in our case, the integrand is ( sqrt{a^2 + d^2 sin^2(v)} ). Let me factor out ( a^2 ):[ sqrt{a^2 + d^2 sin^2(v)} = a sqrt{1 + left( frac{d}{a} right)^2 sin^2(v)} ]So, let me set ( k = frac{d}{a} ). Then, the integrand becomes ( a sqrt{1 + k^2 sin^2(v)} ). Hmm, but the standard elliptic integral has ( 1 - k^2 sin^2 theta ), not ( 1 + k^2 sin^2 theta ). So, it's a bit different.Wait, but maybe we can adjust it. Let me think.Alternatively, perhaps we can express this in terms of the elliptic integral by a substitution.Let me consider the substitution ( phi = v ), but I don't see an immediate way. Alternatively, maybe we can use a trigonometric identity or another substitution.Wait, another thought: the integral over ( 0 ) to ( 2pi ) can be expressed as four times the integral from ( 0 ) to ( frac{pi}{2} ), since the function is symmetric and periodic.But I'm not sure if that helps. Alternatively, perhaps we can use the identity ( sin^2(v) = frac{1 - cos(2v)}{2} ). Let's try that.So, rewrite the integrand:[ sqrt{a^2 + d^2 sin^2(v)} = sqrt{a^2 + frac{d^2}{2} (1 - cos(2v))} ]Simplify:[ sqrt{a^2 + frac{d^2}{2} - frac{d^2}{2} cos(2v)} ]Let me denote ( A = a^2 + frac{d^2}{2} ) and ( B = frac{d^2}{2} ), so the expression becomes:[ sqrt{A - B cos(2v)} ]Hmm, that's an integral of the form ( sqrt{A - B cos(2v)} ). I think this is another standard form, but I don't remember exactly.Wait, perhaps using a substitution for ( 2v ). Let me set ( w = 2v ), so ( dw = 2 dv ), ( dv = frac{dw}{2} ). Then, when ( v = 0 ), ( w = 0 ); when ( v = 2pi ), ( w = 4pi ). But integrating from ( 0 ) to ( 4pi ) is the same as twice the integral from ( 0 ) to ( 2pi ), since the function is periodic with period ( 2pi ).Wait, but maybe not. Let's see:[ int_{0}^{2pi} sqrt{A - B cos(2v)} , dv = frac{1}{2} int_{0}^{4pi} sqrt{A - B cos(w)} , dw ]But ( cos(w) ) has period ( 2pi ), so integrating over ( 0 ) to ( 4pi ) is twice the integral over ( 0 ) to ( 2pi ). So,[ frac{1}{2} times 2 int_{0}^{2pi} sqrt{A - B cos(w)} , dw = int_{0}^{2pi} sqrt{A - B cos(w)} , dw ]So, the integral becomes:[ int_{0}^{2pi} sqrt{A - B cos(w)} , dw ]Which is a standard form. I think this is related to the complete elliptic integral of the second kind as well, but let me recall.The complete elliptic integral of the second kind is:[ E(k) = int_{0}^{frac{pi}{2}} sqrt{1 - k^2 sin^2 theta} , dtheta ]But our integral is over ( 0 ) to ( 2pi ) and has ( cos(w) ) instead of ( sin^2 theta ). Hmm.Wait, maybe we can express ( sqrt{A - B cos(w)} ) in terms of ( sin ) or ( cos ) squared.Alternatively, perhaps use the identity ( cos(w) = 1 - 2 sin^2(w/2) ). Let's try that.So,[ A - B cos(w) = A - B (1 - 2 sin^2(w/2)) = (A - B) + 2B sin^2(w/2) ]So, the integrand becomes:[ sqrt{(A - B) + 2B sin^2(w/2)} ]Let me factor out ( (A - B) ):[ sqrt{(A - B) left( 1 + frac{2B}{A - B} sin^2(w/2) right)} = sqrt{A - B} sqrt{1 + frac{2B}{A - B} sin^2(w/2)} ]Let me denote ( k^2 = frac{2B}{A - B} ), so:[ sqrt{A - B} sqrt{1 + k^2 sin^2(w/2)} ]Hmm, but this is similar to the form of the elliptic integral, except it's ( 1 + k^2 sin^2 ) instead of ( 1 - k^2 sin^2 ). I think that might complicate things because the standard elliptic integrals have ( 1 - k^2 sin^2 ).Alternatively, maybe we can use another substitution. Let me set ( theta = w/2 ), so ( dtheta = dw/2 ), ( dw = 2 dtheta ).When ( w = 0 ), ( theta = 0 ); when ( w = 2pi ), ( theta = pi ).So, the integral becomes:[ int_{0}^{pi} sqrt{(A - B) + 2B sin^2(theta)} cdot 2 dtheta ]Wait, no, let's go back.Wait, I think I messed up the substitution steps. Let me retrace.We had:[ sqrt{A - B cos(w)} = sqrt{(A - B) + 2B sin^2(w/2)} ]So, the integral becomes:[ int_{0}^{2pi} sqrt{(A - B) + 2B sin^2(w/2)} , dw ]Let me set ( theta = w/2 ), so ( w = 2theta ), ( dw = 2 dtheta ). When ( w = 0 ), ( theta = 0 ); when ( w = 2pi ), ( theta = pi ).So, substituting:[ int_{0}^{pi} sqrt{(A - B) + 2B sin^2(theta)} cdot 2 dtheta ]Simplify:[ 2 int_{0}^{pi} sqrt{(A - B) + 2B sin^2(theta)} , dtheta ]But ( sin^2(theta) ) is symmetric around ( pi/2 ), so the integral from ( 0 ) to ( pi ) is twice the integral from ( 0 ) to ( pi/2 ):[ 2 times 2 int_{0}^{pi/2} sqrt{(A - B) + 2B sin^2(theta)} , dtheta ][ = 4 int_{0}^{pi/2} sqrt{(A - B) + 2B sin^2(theta)} , dtheta ]Now, let's factor out ( (A - B) ):[ 4 sqrt{A - B} int_{0}^{pi/2} sqrt{1 + frac{2B}{A - B} sin^2(theta)} , dtheta ]Let me denote ( k^2 = frac{2B}{A - B} ), so:[ 4 sqrt{A - B} int_{0}^{pi/2} sqrt{1 + k^2 sin^2(theta)} , dtheta ]Hmm, this is similar to the elliptic integral but with a plus sign inside the square root. The standard elliptic integral of the second kind is ( int sqrt{1 - k^2 sin^2 theta} , dtheta ). I think the integral with a plus sign doesn't have a standard name, but maybe it can be expressed in terms of elliptic integrals.Alternatively, perhaps we can use a substitution to convert it into a standard form.Let me consider the substitution ( phi = theta ), but I don't see an immediate way. Alternatively, maybe use hyperbolic substitution? Hmm, not sure.Wait, another idea: perhaps express ( sqrt{1 + k^2 sin^2 theta} ) as ( sqrt{1 - (-k^2) sin^2 theta} ). Then, it's similar to the standard form with ( k^2 ) replaced by ( -k^2 ). But elliptic integrals are typically defined for ( |k| < 1 ), so negative ( k^2 ) might complicate things.Alternatively, perhaps express it in terms of the imaginary modulus. I think that might be a stretch, though.Wait, maybe it's better to just express the integral in terms of the elliptic integral of the second kind with a parameter. Let me check.Wait, actually, I found a resource that says:The integral ( int_{0}^{pi/2} sqrt{1 + k^2 sin^2 theta} , dtheta ) is equal to ( frac{pi}{2} sqrt{1 + k^2} ) only when ( k ) is small, but in general, it's another elliptic integral. Wait, no, that's not correct.Wait, actually, no, that integral doesn't have a simple closed-form expression. It's another type of elliptic integral, sometimes called the "imaginary" case.Wait, perhaps I can express it in terms of the standard elliptic integral by a substitution.Let me set ( sin phi = frac{sin theta}{sqrt{1 + k^2}} ). Then, ( sin theta = sqrt{1 + k^2} sin phi ), and ( dtheta = sqrt{1 + k^2} cos phi dphi / sqrt{1 - (1 + k^2) sin^2 phi} ). Hmm, this seems complicated.Alternatively, maybe use a substitution ( phi = arctan(sqrt{k^2} tan theta) ). Not sure.Wait, perhaps it's better to accept that this integral doesn't have an elementary form and express it in terms of the elliptic integral of the second kind with a parameter.But maybe I can relate it to the standard form.Wait, let me think differently. Let's go back to the original integral:[ L = int_{0}^{2pi} sqrt{a^2 + d^2 sin^2(v)} , dv ]Let me factor out ( a ):[ L = a int_{0}^{2pi} sqrt{1 + left( frac{d}{a} right)^2 sin^2(v)} , dv ]Let ( k = frac{d}{a} ), so:[ L = a int_{0}^{2pi} sqrt{1 + k^2 sin^2(v)} , dv ]This is similar to the standard elliptic integral but with a plus sign. I think this is sometimes called the "incomplete" elliptic integral of the second kind with a parameter, but I need to check.Wait, actually, I found that the integral ( int sqrt{1 + k^2 sin^2 theta} , dtheta ) is related to the elliptic integral of the second kind with an imaginary modulus. Specifically, it can be expressed as ( i E(i theta, ik) ), but that might be more complex than needed.Alternatively, perhaps we can express it in terms of the complete elliptic integral of the second kind by a substitution.Wait, another approach: use the binomial expansion for the square root, but that would result in an infinite series, which might not be helpful here.Alternatively, perhaps use a trigonometric identity to express ( sqrt{1 + k^2 sin^2 v} ) in terms of multiple angles.Wait, another thought: maybe express ( sqrt{1 + k^2 sin^2 v} ) as ( sqrt{1 + k^2} sqrt{1 - frac{k^2}{1 + k^2} cos^2 v} ). Let me check:[ 1 + k^2 sin^2 v = 1 + k^2 (1 - cos^2 v) = 1 + k^2 - k^2 cos^2 v = (1 + k^2) left( 1 - frac{k^2}{1 + k^2} cos^2 v right) ]So,[ sqrt{1 + k^2 sin^2 v} = sqrt{1 + k^2} sqrt{1 - frac{k^2}{1 + k^2} cos^2 v} ]Therefore, the integral becomes:[ L = a sqrt{1 + k^2} int_{0}^{2pi} sqrt{1 - frac{k^2}{1 + k^2} cos^2 v} , dv ]Hmm, now this is similar to the standard elliptic integral, but with ( cos^2 v ) instead of ( sin^2 v ). However, since ( cos^2 v = sin^2(v + pi/2) ), we can adjust the integral accordingly.But regardless, the integral ( int_{0}^{2pi} sqrt{1 - m^2 cos^2 v} , dv ) is known and can be expressed in terms of the complete elliptic integral of the second kind.Specifically, the complete elliptic integral of the second kind is defined as:[ E(m) = int_{0}^{frac{pi}{2}} sqrt{1 - m^2 sin^2 theta} , dtheta ]But our integral is over ( 0 ) to ( 2pi ) with ( cos^2 v ). Let me use the identity ( cos^2 v = sin^2(v + pi/2) ), so shifting the variable:Let ( phi = v + pi/2 ), then when ( v = 0 ), ( phi = pi/2 ); when ( v = 2pi ), ( phi = 5pi/2 ). But integrating over a full period, shifting the variable doesn't change the value, so:[ int_{0}^{2pi} sqrt{1 - m^2 cos^2 v} , dv = int_{pi/2}^{5pi/2} sqrt{1 - m^2 sin^2 phi} , dphi ]But since the integrand is periodic with period ( 2pi ), the integral from ( pi/2 ) to ( 5pi/2 ) is the same as the integral from ( 0 ) to ( 2pi ). So,[ int_{0}^{2pi} sqrt{1 - m^2 cos^2 v} , dv = int_{0}^{2pi} sqrt{1 - m^2 sin^2 phi} , dphi ]Which is four times the integral from ( 0 ) to ( pi/2 ):[ 4 int_{0}^{pi/2} sqrt{1 - m^2 sin^2 phi} , dphi = 4 E(m) ]Therefore, putting it all together:[ L = a sqrt{1 + k^2} times 4 Eleft( sqrt{frac{k^2}{1 + k^2}} right) ]Wait, let me clarify:We had:[ L = a sqrt{1 + k^2} int_{0}^{2pi} sqrt{1 - frac{k^2}{1 + k^2} cos^2 v} , dv ]And we found that:[ int_{0}^{2pi} sqrt{1 - m^2 cos^2 v} , dv = 4 E(m) ]Where ( m = sqrt{frac{k^2}{1 + k^2}} ). Therefore,[ L = a sqrt{1 + k^2} times 4 Eleft( sqrt{frac{k^2}{1 + k^2}} right) ]But ( k = frac{d}{a} ), so substituting back:[ L = a sqrt{1 + left( frac{d}{a} right)^2} times 4 Eleft( sqrt{frac{left( frac{d}{a} right)^2}{1 + left( frac{d}{a} right)^2}} right) ]Simplify ( sqrt{1 + left( frac{d}{a} right)^2} ):[ sqrt{1 + frac{d^2}{a^2}} = frac{sqrt{a^2 + d^2}}{a} ]So,[ L = a times frac{sqrt{a^2 + d^2}}{a} times 4 Eleft( frac{d}{sqrt{a^2 + d^2}} right) ]Simplify:[ L = sqrt{a^2 + d^2} times 4 Eleft( frac{d}{sqrt{a^2 + d^2}} right) ]Wait, but hold on. The substitution earlier led us to:[ L = a sqrt{1 + k^2} times 4 E(m) ]Where ( m = sqrt{frac{k^2}{1 + k^2}} ). So, ( m = frac{k}{sqrt{1 + k^2}} = frac{d/a}{sqrt{1 + (d/a)^2}} = frac{d}{sqrt{a^2 + d^2}} ).So, yes, that's correct.Therefore, the total length is:[ L = 4 sqrt{a^2 + d^2} , Eleft( frac{d}{sqrt{a^2 + d^2}} right) ]But wait, let me confirm the constants. Earlier, we had:[ L = a sqrt{1 + k^2} times 4 E(m) ]But ( a sqrt{1 + k^2} = a sqrt{1 + (d/a)^2} = sqrt{a^2 + d^2} ). So, yes, that's correct.Therefore, the total length is:[ L = 4 sqrt{a^2 + d^2} , Eleft( frac{d}{sqrt{a^2 + d^2}} right) ]Hmm, but I need to make sure about the factor of 4. Let me retrace:We had:1. Original integral: ( L = int_{0}^{2pi} sqrt{a^2 + d^2 sin^2(v)} , dv )2. Factored out ( a ): ( L = a int_{0}^{2pi} sqrt{1 + k^2 sin^2(v)} , dv ) where ( k = d/a )3. Then, expressed it as ( a sqrt{1 + k^2} times 4 E(m) ) where ( m = frac{k}{sqrt{1 + k^2}} )4. So, ( L = a sqrt{1 + k^2} times 4 E(m) )5. Substituted ( a sqrt{1 + k^2} = sqrt{a^2 + d^2} )6. Therefore, ( L = sqrt{a^2 + d^2} times 4 Eleft( frac{d}{sqrt{a^2 + d^2}} right) )Yes, that seems correct.Alternatively, I found another resource that states:The integral ( int_{0}^{2pi} sqrt{a^2 + b^2 sin^2 t} , dt = 4 sqrt{a^2 + b^2} , Eleft( frac{b}{sqrt{a^2 + b^2}} right) )Which matches our result here. So, that's reassuring.Therefore, the total length of the path is:[ L = 4 sqrt{a^2 + d^2} , Eleft( frac{d}{sqrt{a^2 + d^2}} right) ]Where ( E(k) ) is the complete elliptic integral of the second kind.Okay, so that's the first part done.Now, moving on to the second part: calculating the total energy consumed by the drone over the same interval. The power function is given by:[ P(t) = k (x(t)^2 + y(t)^2 + z(t)^2) ]So, the total energy ( E ) is the integral of power over time:[ E = int_{0}^{frac{2pi}{b}} P(t) , dt = k int_{0}^{frac{2pi}{b}} (x(t)^2 + y(t)^2 + z(t)^2) , dt ]So, I need to compute ( x(t)^2 + y(t)^2 + z(t)^2 ) and integrate it over the interval.Let me compute each term:1. ( x(t)^2 = a^2 sin^2(bt + c) )2. ( y(t)^2 = a^2 cos^2(bt + c) )3. ( z(t)^2 = d^2 sin^4(bt + c) )So, adding them up:[ x(t)^2 + y(t)^2 + z(t)^2 = a^2 (sin^2(bt + c) + cos^2(bt + c)) + d^2 sin^4(bt + c) ]Again, ( sin^2 theta + cos^2 theta = 1 ), so this simplifies to:[ a^2 + d^2 sin^4(bt + c) ]Therefore, the integral becomes:[ E = k int_{0}^{frac{2pi}{b}} left( a^2 + d^2 sin^4(bt + c) right) dt ]Let me split this into two integrals:[ E = k left( a^2 int_{0}^{frac{2pi}{b}} dt + d^2 int_{0}^{frac{2pi}{b}} sin^4(bt + c) , dt right) ]Compute each integral separately.First integral:[ int_{0}^{frac{2pi}{b}} dt = frac{2pi}{b} ]Second integral:[ int_{0}^{frac{2pi}{b}} sin^4(bt + c) , dt ]Let me make a substitution here as well. Let ( u = bt + c ), so ( du = b dt ), ( dt = frac{du}{b} ).When ( t = 0 ), ( u = c ); when ( t = frac{2pi}{b} ), ( u = 2pi + c ).So, the integral becomes:[ int_{c}^{2pi + c} sin^4(u) cdot frac{du}{b} = frac{1}{b} int_{c}^{2pi + c} sin^4(u) , du ]Again, since ( sin^4(u) ) is periodic with period ( pi ), the integral over an interval of length ( 2pi ) is twice the integral over ( 0 ) to ( pi ). Alternatively, since the function is periodic, the integral over ( c ) to ( 2pi + c ) is the same as the integral over ( 0 ) to ( 2pi ).So,[ frac{1}{b} int_{0}^{2pi} sin^4(u) , du ]Now, I need to compute ( int_{0}^{2pi} sin^4(u) , du ). I remember that the integral of ( sin^n u ) over ( 0 ) to ( 2pi ) can be found using reduction formulas or known identities.The formula for ( int_{0}^{2pi} sin^4(u) , du ) is:[ frac{3pi}{4} ]Wait, let me verify that.Using the power-reduction formula:[ sin^4 u = left( sin^2 u right)^2 = left( frac{1 - cos(2u)}{2} right)^2 = frac{1}{4} left( 1 - 2 cos(2u) + cos^2(2u) right) ]Then, ( cos^2(2u) = frac{1 + cos(4u)}{2} ), so:[ sin^4 u = frac{1}{4} left( 1 - 2 cos(2u) + frac{1 + cos(4u)}{2} right) = frac{1}{4} left( frac{3}{2} - 2 cos(2u) + frac{1}{2} cos(4u) right) ]Simplify:[ sin^4 u = frac{3}{8} - frac{1}{2} cos(2u) + frac{1}{8} cos(4u) ]Therefore, integrating over ( 0 ) to ( 2pi ):[ int_{0}^{2pi} sin^4 u , du = int_{0}^{2pi} left( frac{3}{8} - frac{1}{2} cos(2u) + frac{1}{8} cos(4u) right) du ]Integrate term by term:1. ( int_{0}^{2pi} frac{3}{8} du = frac{3}{8} times 2pi = frac{3pi}{4} )2. ( int_{0}^{2pi} -frac{1}{2} cos(2u) du = -frac{1}{2} times left[ frac{sin(2u)}{2} right]_0^{2pi} = -frac{1}{4} [ sin(4pi) - sin(0) ] = 0 )3. ( int_{0}^{2pi} frac{1}{8} cos(4u) du = frac{1}{8} times left[ frac{sin(4u)}{4} right]_0^{2pi} = frac{1}{32} [ sin(8pi) - sin(0) ] = 0 )So, the integral is ( frac{3pi}{4} ).Therefore, the second integral is:[ frac{1}{b} times frac{3pi}{4} = frac{3pi}{4b} ]Putting it all together, the total energy is:[ E = k left( a^2 times frac{2pi}{b} + d^2 times frac{3pi}{4b} right) ]Simplify:Factor out ( frac{pi}{b} ):[ E = k times frac{pi}{b} left( 2a^2 + frac{3d^2}{4} right) ]Alternatively, write it as:[ E = frac{k pi}{b} left( 2a^2 + frac{3d^2}{4} right) ]Or, combining the terms:[ E = frac{k pi}{b} left( frac{8a^2 + 3d^2}{4} right) = frac{k pi (8a^2 + 3d^2)}{4b} ]But perhaps it's better to leave it as:[ E = frac{k pi}{b} left( 2a^2 + frac{3d^2}{4} right) ]Either way is fine, but maybe the first expression is clearer.So, summarizing:1. The total length ( L ) is ( 4 sqrt{a^2 + d^2} , Eleft( frac{d}{sqrt{a^2 + d^2}} right) ), where ( E ) is the complete elliptic integral of the second kind.2. The total energy ( E ) is ( frac{k pi}{b} left( 2a^2 + frac{3d^2}{4} right) ).Wait, hold on, I just realized that I used ( E ) for both the total energy and the elliptic integral. That's confusing. Let me correct that.Let me denote the complete elliptic integral of the second kind as ( E(m) ), and the total energy as ( E_{text{total}} ).So, revising:1. The total length ( L ) is:[ L = 4 sqrt{a^2 + d^2} , Eleft( frac{d}{sqrt{a^2 + d^2}} right) ]2. The total energy ( E_{text{total}} ) is:[ E_{text{total}} = frac{k pi}{b} left( 2a^2 + frac{3d^2}{4} right) ]Yes, that's better.So, to recap:For part 1, after computing the derivatives, simplifying, and recognizing the integral as an elliptic integral, we arrived at the expression involving ( E ).For part 2, by expanding the power function, integrating term by term, and using known integrals, we found a closed-form expression for the total energy.I think that's it. I don't see any mistakes in the steps, but let me just quickly verify the energy calculation.We had:[ E_{text{total}} = k left( a^2 times frac{2pi}{b} + d^2 times frac{3pi}{4b} right) ]Yes, that's correct because:- ( int x^2 + y^2 ) over the interval is ( a^2 times frac{2pi}{b} )- ( int z^2 ) over the interval is ( d^2 times frac{3pi}{4b} )So, adding them and multiplying by ( k ) gives the total energy.Therefore, the final answers are:1. The total length is ( 4 sqrt{a^2 + d^2} , Eleft( frac{d}{sqrt{a^2 + d^2}} right) ).2. The total energy is ( frac{k pi}{b} left( 2a^2 + frac{3d^2}{4} right) ).Final Answer1. The total length of the path is (boxed{4 sqrt{a^2 + d^2} , Eleft( frac{d}{sqrt{a^2 + d^2}} right)}).2. The total energy consumed by the drone is (boxed{frac{k pi (8a^2 + 3d^2)}{4b}}).</think>