🧠:<think>Alright, so I have this problem about modeling a foot arch as a parabola. It's divided into two parts. Let me try to tackle them one by one.Starting with part 1: I need to find the coefficients ( a ), ( b ), and ( c ) of the quadratic equation ( y = ax^2 + bx + c ). The given information is that the arch reaches its maximum height of 6 cm at a point 4 cm from the heel, and the height at the heel (x = 0) is 2 cm.Hmm, okay. So, since it's a quadratic equation, and the parabola opens downward because it has a maximum point. That means the coefficient ( a ) should be negative.First, let's recall that the vertex form of a parabola is ( y = a(x - h)^2 + k ), where ( (h, k) ) is the vertex. In this case, the vertex is at (4, 6). So, plugging that in, the equation becomes:( y = a(x - 4)^2 + 6 )But we also know that when ( x = 0 ), ( y = 2 ). So, let's substitute that into the equation to find ( a ).Substituting ( x = 0 ) and ( y = 2 ):( 2 = a(0 - 4)^2 + 6 )Simplify ( (0 - 4)^2 ):( 2 = a(16) + 6 )So, ( 2 = 16a + 6 )Subtract 6 from both sides:( 2 - 6 = 16a )( -4 = 16a )Divide both sides by 16:( a = -4 / 16 )Simplify:( a = -1/4 )Okay, so ( a = -1/4 ). Now, let's write the equation in vertex form:( y = -1/4 (x - 4)^2 + 6 )But the problem asks for the standard form ( y = ax^2 + bx + c ). So, I need to expand this equation.First, expand ( (x - 4)^2 ):( (x - 4)^2 = x^2 - 8x + 16 )Multiply by ( -1/4 ):( -1/4 * x^2 + (-1/4)*(-8x) + (-1/4)*16 )Simplify each term:- ( -1/4 x^2 )- ( + 2x ) (because ( -1/4 * -8 = 2 ))- ( -4 ) (because ( -1/4 * 16 = -4 ))So, putting it all together:( y = -1/4 x^2 + 2x - 4 )But wait, we also have the +6 from the vertex form. So, adding that:( y = -1/4 x^2 + 2x - 4 + 6 )Simplify:( y = -1/4 x^2 + 2x + 2 )So, the standard form is ( y = -1/4 x^2 + 2x + 2 ). Therefore, the coefficients are:( a = -1/4 ), ( b = 2 ), ( c = 2 ).Let me double-check this. When ( x = 4 ), plugging into the equation:( y = -1/4*(16) + 2*4 + 2 = -4 + 8 + 2 = 6 ). That's correct.At ( x = 0 ):( y = 0 + 0 + 2 = 2 ). Correct as well.Okay, so part 1 seems done.Moving on to part 2: Dr. Footman models the pressure distribution ( P(x) ) as a function inversely proportional to the square of the height ( y ) of the arch. Given that the pressure at the maximum height point (x = 4 cm) is 1 N/cm², find the general form of the pressure distribution function ( P(x) ).Alright, so inversely proportional to the square of the height. That means:( P(x) = k / y(x)^2 )Where ( k ) is the constant of proportionality.We know that at ( x = 4 ), ( P(4) = 1 ) N/cm². So, let's find ( k ).First, find ( y(4) ). From part 1, we know that ( y(4) = 6 ) cm.So, substituting into the equation:( 1 = k / (6)^2 )Simplify:( 1 = k / 36 )Multiply both sides by 36:( k = 36 )Therefore, the pressure distribution function is:( P(x) = 36 / y(x)^2 )But we can substitute ( y(x) ) from part 1. Since ( y(x) = -1/4 x^2 + 2x + 2 ), so:( P(x) = 36 / (-1/4 x^2 + 2x + 2)^2 )Alternatively, we can write it as:( P(x) = frac{36}{left(-frac{1}{4}x^2 + 2x + 2right)^2} )But let me see if we can write it in a nicer form. Maybe factor out the negative sign or something.Alternatively, since ( y(x) = -1/4 x^2 + 2x + 2 ), we can factor out the -1/4:( y(x) = -1/4 (x^2 - 8x - 8) )Wait, but that might complicate things more. Alternatively, perhaps we can write the denominator squared as is.Alternatively, since ( y(x) ) is quadratic, we can express ( P(x) ) as:( P(x) = frac{36}{left(-frac{1}{4}x^2 + 2x + 2right)^2} )But perhaps it's better to write it without fractions in the denominator. Let me see.Multiply numerator and denominator by 16 to eliminate the fraction inside the square:( P(x) = frac{36 * 16}{( -x^2 + 8x + 8 )^2} )Wait, let's see:( y(x) = -1/4 x^2 + 2x + 2 )Multiply numerator and denominator by 16:( P(x) = frac{36}{( (-1/4 x^2 + 2x + 2) )^2} = frac{36 * 16}{( -x^2 + 8x + 8 )^2} )Wait, is that correct?Wait, if I have ( (-1/4 x^2 + 2x + 2) ), multiplying numerator and denominator by 16:Wait, actually, if I factor out -1/4 from ( y(x) ):( y(x) = -1/4 (x^2 - 8x - 8) )So, ( y(x)^2 = ( -1/4 )^2 (x^2 - 8x - 8)^2 = 1/16 (x^2 - 8x - 8)^2 )Therefore, ( P(x) = 36 / y(x)^2 = 36 / (1/16 (x^2 - 8x - 8)^2 ) = 36 * 16 / (x^2 - 8x - 8)^2 = 576 / (x^2 - 8x - 8)^2 )So, that might be a cleaner way to write it.But let me verify:Given ( y(x) = -1/4 x^2 + 2x + 2 ), so ( y(x) = (-x^2 + 8x + 8)/4 ). Therefore, ( y(x)^2 = ( (-x^2 + 8x + 8)^2 ) / 16 ). Therefore, ( 1/y(x)^2 = 16 / (-x^2 + 8x + 8)^2 ). Therefore, ( P(x) = 36 * 16 / (-x^2 + 8x + 8)^2 = 576 / (-x^2 + 8x + 8)^2 ).Alternatively, since the denominator is squared, the negative sign can be incorporated:( (-x^2 + 8x + 8)^2 = (x^2 - 8x - 8)^2 ), because squaring removes the negative.Therefore, ( P(x) = 576 / (x^2 - 8x - 8)^2 )Alternatively, factor the denominator if possible.Let me see if ( x^2 - 8x - 8 ) can be factored. The discriminant is ( 64 + 32 = 96 ), which is not a perfect square, so it doesn't factor nicely. So, we can leave it as is.Therefore, the general form is ( P(x) = frac{576}{(x^2 - 8x - 8)^2} ).Alternatively, if we prefer to keep the original quadratic expression, we can write it as:( P(x) = frac{36}{left(-frac{1}{4}x^2 + 2x + 2right)^2} )But the first form with the multiplied-out denominator might be preferable for simplicity.Let me double-check the calculation for ( k ). We had ( P(x) = k / y(x)^2 ), and at ( x = 4 ), ( y = 6 ), so ( P(4) = k / 36 = 1 ), so ( k = 36 ). That seems correct.Therefore, substituting back, ( P(x) = 36 / y(x)^2 ), and since ( y(x) = -1/4 x^2 + 2x + 2 ), that's correct.Alternatively, as I did before, expressing it as ( 576 / (x^2 - 8x - 8)^2 ) is also correct.I think both forms are acceptable, but perhaps the problem expects it in terms of ( y(x) ), so maybe ( P(x) = 36 / y(x)^2 ) is sufficient. But since they ask for the general form, perhaps substituting ( y(x) ) is better.Wait, the problem says "find the general form of the pressure distribution function ( P(x) )". So, they might expect an expression in terms of ( x ), not in terms of ( y(x) ). So, substituting ( y(x) ) into the equation is necessary.So, let's write it as:( P(x) = frac{36}{left(-frac{1}{4}x^2 + 2x + 2right)^2} )Alternatively, as I did earlier, multiply numerator and denominator by 16 to eliminate the fraction inside the square:( P(x) = frac{36 * 16}{(-x^2 + 8x + 8)^2} = frac{576}{(-x^2 + 8x + 8)^2} )But since the denominator is squared, the negative sign can be incorporated:( (-x^2 + 8x + 8)^2 = (x^2 - 8x - 8)^2 )So, ( P(x) = frac{576}{(x^2 - 8x - 8)^2} )I think this is a good form because it's a single fraction without nested fractions, which might be preferable.Let me just verify once more:Given ( y(x) = -1/4 x^2 + 2x + 2 ), so ( y(x) = (-x^2 + 8x + 8)/4 ). Therefore, ( y(x)^2 = ( (-x^2 + 8x + 8)^2 ) / 16 ). Therefore, ( 1/y(x)^2 = 16 / (-x^2 + 8x + 8)^2 ). Hence, ( P(x) = 36 * 16 / (-x^2 + 8x + 8)^2 = 576 / (-x^2 + 8x + 8)^2 ). Since the denominator is squared, the negative sign doesn't matter, so we can write it as ( (x^2 - 8x - 8)^2 ).Yes, that seems correct.So, summarizing:1. The quadratic equation is ( y = -1/4 x^2 + 2x + 2 ), so coefficients are ( a = -1/4 ), ( b = 2 ), ( c = 2 ).2. The pressure distribution function is ( P(x) = 576 / (x^2 - 8x - 8)^2 ).I think that's it. Let me just check if I made any calculation errors.For part 1:- Vertex at (4,6), so equation is ( y = a(x - 4)^2 + 6 ). At x=0, y=2: 2 = a(16) + 6 => 2 = 16a + 6 => 16a = -4 => a = -1/4. Correct.Expanding ( y = -1/4 (x - 4)^2 + 6 ):( (x - 4)^2 = x^2 - 8x + 16 )Multiply by -1/4: ( -1/4 x^2 + 2x - 4 ). Add 6: ( -1/4 x^2 + 2x + 2 ). Correct.For part 2:( P(x) = k / y^2 ). At x=4, y=6, P=1: 1 = k / 36 => k=36. So, ( P(x) = 36 / y^2 ). Substituting y: ( 36 / (-1/4 x^2 + 2x + 2)^2 ). Multiply numerator and denominator by 16: 576 / (-x^2 + 8x + 8)^2 = 576 / (x^2 - 8x - 8)^2. Correct.Yes, I think everything checks out.Final Answer1. The coefficients are ( a = boxed{-dfrac{1}{4}} ), ( b = boxed{2} ), and ( c = boxed{2} ).2. The pressure distribution function is ( P(x) = boxed{dfrac{576}{(x^2 - 8x - 8)^2}} ).</think>

🧠:<think>Alright, so I'm trying to solve this problem about nutrient retention in soil treatments A and B. Let me start with part 1 since it seems more straightforward.Problem 1: Treatment AWe have the differential equation:[frac{dN_A}{dt} = -k_A N_A + c_A]with the initial condition ( N_A(0) = N_0 ). I need to find ( N_A(t) ) in terms of ( N_0 ), ( k_A ), and ( c_A ), and then determine its long-term behavior as ( t to infty ).Hmm, this looks like a linear first-order differential equation. I remember that these can be solved using an integrating factor. The standard form is:[frac{dy}{dt} + P(t)y = Q(t)]In our case, let's rewrite the equation:[frac{dN_A}{dt} + k_A N_A = c_A]So, ( P(t) = k_A ) and ( Q(t) = c_A ). The integrating factor ( mu(t) ) is given by:[mu(t) = e^{int P(t) dt} = e^{int k_A dt} = e^{k_A t}]Multiplying both sides of the differential equation by ( mu(t) ):[e^{k_A t} frac{dN_A}{dt} + k_A e^{k_A t} N_A = c_A e^{k_A t}]The left side is the derivative of ( N_A e^{k_A t} ) with respect to t:[frac{d}{dt} left( N_A e^{k_A t} right) = c_A e^{k_A t}]Now, integrate both sides with respect to t:[int frac{d}{dt} left( N_A e^{k_A t} right) dt = int c_A e^{k_A t} dt]This simplifies to:[N_A e^{k_A t} = frac{c_A}{k_A} e^{k_A t} + C]Where C is the constant of integration. Solving for ( N_A ):[N_A(t) = frac{c_A}{k_A} + C e^{-k_A t}]Now, apply the initial condition ( N_A(0) = N_0 ):[N_0 = frac{c_A}{k_A} + C e^{0} implies N_0 = frac{c_A}{k_A} + C]So, solving for C:[C = N_0 - frac{c_A}{k_A}]Therefore, the solution is:[N_A(t) = frac{c_A}{k_A} + left( N_0 - frac{c_A}{k_A} right) e^{-k_A t}]Simplifying, we can write:[N_A(t) = frac{c_A}{k_A} left( 1 - e^{-k_A t} right) + N_0 e^{-k_A t}]But the first form is probably more useful. Now, to determine the long-term behavior as ( t to infty ). Since ( e^{-k_A t} ) approaches 0 as t becomes large (assuming ( k_A > 0 )), the term with C will vanish. Therefore, the long-term behavior is:[lim_{t to infty} N_A(t) = frac{c_A}{k_A}]So, the nutrient level approaches a steady state of ( c_A / k_A ).Problem 2: Treatment BNow, moving on to treatment B. The function is given by:[N_B(t) = N_0 e^{-k_B t} + int_0^t g(s) e^{-k_B(t - s)} ds]And ( g(t) = c_B sin(omega t) ). We need to analyze the periodic behavior and find the steady-state solution if it exists, considering ( omega ) varies between 0 and ( 2pi ).First, let me substitute ( g(s) ) into the integral:[N_B(t) = N_0 e^{-k_B t} + int_0^t c_B sin(omega s) e^{-k_B(t - s)} ds]Let me factor out the constants:[N_B(t) = N_0 e^{-k_B t} + c_B e^{-k_B t} int_0^t sin(omega s) e^{k_B s} ds]So, the integral becomes:[int_0^t sin(omega s) e^{k_B s} ds]I need to compute this integral. I recall that the integral of ( e^{at} sin(bt) ) dt is a standard integral, which can be solved using integration by parts or using a formula.The formula is:[int e^{at} sin(bt) dt = frac{e^{at}}{a^2 + b^2} (a sin(bt) - b cos(bt)) ) + C]So, applying this formula with ( a = k_B ) and ( b = omega ), the integral from 0 to t is:[left[ frac{e^{k_B s}}{k_B^2 + omega^2} (k_B sin(omega s) - omega cos(omega s)) right]_0^t]Evaluating this from 0 to t:At t:[frac{e^{k_B t}}{k_B^2 + omega^2} (k_B sin(omega t) - omega cos(omega t))]At 0:[frac{e^{0}}{k_B^2 + omega^2} (k_B sin(0) - omega cos(0)) = frac{1}{k_B^2 + omega^2} (0 - omega cdot 1) = -frac{omega}{k_B^2 + omega^2}]So, the integral is:[frac{e^{k_B t}}{k_B^2 + omega^2} (k_B sin(omega t) - omega cos(omega t)) + frac{omega}{k_B^2 + omega^2}]Therefore, plugging this back into ( N_B(t) ):[N_B(t) = N_0 e^{-k_B t} + c_B e^{-k_B t} left[ frac{e^{k_B t}}{k_B^2 + omega^2} (k_B sin(omega t) - omega cos(omega t)) + frac{omega}{k_B^2 + omega^2} right]]Simplify term by term.First, the term with ( e^{-k_B t} times e^{k_B t} ) simplifies to 1:[c_B e^{-k_B t} times frac{e^{k_B t}}{k_B^2 + omega^2} (k_B sin(omega t) - omega cos(omega t)) = frac{c_B}{k_B^2 + omega^2} (k_B sin(omega t) - omega cos(omega t))]The second term:[c_B e^{-k_B t} times frac{omega}{k_B^2 + omega^2} = frac{c_B omega}{k_B^2 + omega^2} e^{-k_B t}]So, putting it all together:[N_B(t) = N_0 e^{-k_B t} + frac{c_B}{k_B^2 + omega^2} (k_B sin(omega t) - omega cos(omega t)) + frac{c_B omega}{k_B^2 + omega^2} e^{-k_B t}]Combine the exponential terms:[N_0 e^{-k_B t} + frac{c_B omega}{k_B^2 + omega^2} e^{-k_B t} = left( N_0 + frac{c_B omega}{k_B^2 + omega^2} right) e^{-k_B t}]So, now ( N_B(t) ) is:[N_B(t) = left( N_0 + frac{c_B omega}{k_B^2 + omega^2} right) e^{-k_B t} + frac{c_B}{k_B^2 + omega^2} (k_B sin(omega t) - omega cos(omega t))]Now, let's analyze the behavior as ( t to infty ). The term ( e^{-k_B t} ) will go to zero if ( k_B > 0 ). So, the first part vanishes, and we are left with:[N_B(t) approx frac{c_B}{k_B^2 + omega^2} (k_B sin(omega t) - omega cos(omega t))]This is a sinusoidal function with amplitude:[sqrt{ left( frac{c_B k_B}{k_B^2 + omega^2} right)^2 + left( frac{c_B (-omega)}{k_B^2 + omega^2} right)^2 } = frac{c_B}{k_B^2 + omega^2} sqrt{ k_B^2 + omega^2 } = frac{c_B}{sqrt{k_B^2 + omega^2}}]So, the steady-state solution is a sinusoidal oscillation with amplitude ( frac{c_B}{sqrt{k_B^2 + omega^2}} ) and phase shift. The frequency remains ( omega ), same as the input.Therefore, the steady-state solution exists and is periodic with the same frequency ( omega ), but with a different amplitude and phase.Wait, but the problem says to consider ( omega ) as a parameter varying between 0 and ( 2pi ). So, depending on ( omega ), the amplitude of the steady-state oscillation changes.If ( omega = 0 ), then the amplitude becomes ( frac{c_B}{k_B} ), which is similar to the steady-state in part 1. As ( omega ) increases, the amplitude decreases because the denominator increases.So, the system responds to the periodic nutrient addition with a sinusoidal steady-state, whose amplitude depends inversely on the square root of ( k_B^2 + omega^2 ).Therefore, the steady-state solution is:[N_B(t) = frac{c_B}{k_B^2 + omega^2} (k_B sin(omega t) - omega cos(omega t))]Alternatively, this can be written in terms of a single sine function with a phase shift, but the above form is sufficient for analysis.Summary of Thoughts:For Treatment A, the solution is an exponential decay towards a steady state of ( c_A / k_A ). For Treatment B, the solution has a transient exponential decay term and a steady-state oscillation that depends on the frequency ( omega ). The amplitude of this oscillation decreases as ( omega ) increases, which means higher frequency nutrient additions result in smaller steady-state oscillations in nutrient levels.I think I covered all the steps, but let me double-check if I made any mistakes. In the integral, I used the standard formula correctly, and the substitution seems right. The combining of terms also looks correct. So, I think my solutions are accurate.Final Answer1. The nutrient level for treatment A is ( boxed{N_A(t) = frac{c_A}{k_A} + left( N_0 - frac{c_A}{k_A} right) e^{-k_A t}} ) and it approaches ( boxed{frac{c_A}{k_A}} ) as ( t to infty ).2. The steady-state solution for treatment B is ( boxed{frac{c_B}{k_B^2 + omega^2} (k_B sin(omega t) - omega cos(omega t))} ).</think>

🧠:<think>Alright, so I have this problem about a coastal artist creating resin art with circular jewelry pieces. The artist has a rectangular mold that's 30 cm by 20 cm, and she wants to embed seven circles with radii ranging from 1 cm up to 4 cm. The first part is to figure out the maximum area these circles can occupy without overlapping or going beyond the mold. The second part is about calculating the weight of the resin mold after embedding, considering the resin's density and thickness.Starting with the first problem: maximizing the area covered by the circles. I know that the area of a circle is πr², so if I can fit all seven circles without overlapping, the total area would be the sum of each circle's area. But wait, is it possible to fit all seven circles in a 30x20 cm rectangle? I should check the sizes.The radii are 1, 1.5, 2, 2.5, 3, 3.5, and 4 cm. So the diameters would be double that: 2, 3, 4, 5, 6, 7, and 8 cm. The largest circle has a diameter of 8 cm, which is manageable in a 30x20 cm rectangle. But arranging all seven might be tricky because of their varying sizes.I remember that circle packing is a complex problem, especially in rectangles. There are algorithms and formulas, but since this is a small number of circles, maybe I can figure it out manually or by some logical arrangement.First, let's calculate the total area of all seven circles:1. Radius 1 cm: Area = π(1)² = π cm²2. Radius 1.5 cm: Area = π(1.5)² = 2.25π cm²3. Radius 2 cm: Area = π(2)² = 4π cm²4. Radius 2.5 cm: Area = π(2.5)² = 6.25π cm²5. Radius 3 cm: Area = π(3)² = 9π cm²6. Radius 3.5 cm: Area = π(3.5)² = 12.25π cm²7. Radius 4 cm: Area = π(4)² = 16π cm²Adding these up: π + 2.25π + 4π + 6.25π + 9π + 12.25π + 16π = Let's compute the coefficients:1 + 2.25 = 3.253.25 + 4 = 7.257.25 + 6.25 = 13.513.5 + 9 = 22.522.5 + 12.25 = 34.7534.75 + 16 = 50.75So total area is 50.75π cm². Approximately, π is about 3.1416, so 50.75 * 3.1416 ≈ 159.5 cm².But the area of the mold is 30*20=600 cm². So the circles only take up about 26.6% of the mold's area. That seems low, but maybe it's because the circles can't be perfectly packed. However, the question is about the maximum possible area, so if all circles can fit without overlapping, that's the maximum.But can all seven circles fit? Let's think about the arrangement.The largest circle has a diameter of 8 cm, so along the 30 cm side, we can fit 30 / 8 ≈ 3.75, so 3 circles. Along the 20 cm side, 20 / 8 = 2.5, so 2 circles. So in a 3x2 grid, we can fit 6 circles, but we have 7. So maybe we need a different arrangement.Alternatively, maybe arranging them in a way that smaller circles fill the gaps between larger ones.Let me try to visualize:- Place the largest circle (4 cm radius, 8 cm diameter) in one corner, say the bottom left.- Next to it, along the 30 cm side, place the next largest, 3.5 cm radius (7 cm diameter). So from 0 to 8 cm on the x-axis, then 8 to 15 cm for the next circle.- Then, the next one could be 3 cm radius (6 cm diameter), from 15 to 21 cm.Wait, but 8 + 7 + 6 = 21 cm, which is less than 30 cm. So along the length, we have 21 cm used, leaving 9 cm.Similarly, vertically, the height is 20 cm. The largest circle is 8 cm in diameter, so vertically, we can stack another circle on top. Maybe a smaller one.Alternatively, maybe arrange the largest circles along the length and use the remaining space for smaller ones.Alternatively, arrange the circles in a hexagonal packing, but in a rectangle, it's more efficient, but with varying sizes, it's complicated.Alternatively, perhaps place the largest circle in the corner, then next to it, place the next largest, and so on, using the space efficiently.Alternatively, think about the total width and height required.Wait, another approach is to calculate the minimum bounding rectangle for all seven circles. But since the mold is fixed at 30x20, we need to see if all seven can fit within that.Alternatively, think about the sum of the diameters in each direction.But since circles can be placed in both x and y directions, it's not just a linear sum.Alternatively, think about the maximum width and height required.The largest circle has diameter 8 cm, so it needs at least 8 cm in both x and y.But with multiple circles, the total width and height required would be the sum of the maximum diameters in each row and column.Wait, maybe it's better to try to arrange them step by step.Let me try to sketch an arrangement:1. Place the 4 cm radius circle (8 cm diameter) at the bottom left corner.2. To its right, place the 3.5 cm radius circle (7 cm diameter). So from x=0 to x=8, then x=8 to x=15.3. Above the 4 cm circle, place the 3 cm radius circle (6 cm diameter). So y=0 to y=8, then y=8 to y=14.Wait, but the mold is only 20 cm tall, so y=14 is still within 20 cm.But wait, the 3 cm circle is placed above the 4 cm circle, so it would be centered at (4, 8 + 3) = (4, 11). But the 3.5 cm circle is placed to the right of the 4 cm circle, centered at (8 + 3.5, 4) = (11.5, 4). So the distance between centers is sqrt((11.5 - 4)^2 + (4 - 11)^2) = sqrt(7.5² + (-7)^2) = sqrt(56.25 + 49) = sqrt(105.25) ≈ 10.26 cm. The sum of their radii is 4 + 3.5 = 7.5 cm. Since 10.26 > 7.5, they don't overlap. Good.4. Next, to the right of the 3.5 cm circle, we have x from 15 to 30, which is 15 cm. Maybe place the 2.5 cm radius circle (5 cm diameter) there. So centered at (15 + 2.5, 4) = (17.5, 4). Check distance from 3.5 cm circle: sqrt((17.5 - 11.5)^2 + (4 - 4)^2) = sqrt(6² + 0) = 6 cm. Sum of radii is 3.5 + 2.5 = 6 cm. So they just touch, no overlap. Good.5. Above the 3 cm circle, we have y from 14 to 20, which is 6 cm. Maybe place the 2 cm radius circle (4 cm diameter) there. So centered at (4, 14 + 2) = (4, 16). Check distance from 3 cm circle: sqrt((4 - 4)^2 + (16 - 11)^2) = sqrt(0 + 25) = 5 cm. Sum of radii is 3 + 2 = 5 cm. Again, just touching, no overlap.6. Next, to the right of the 2 cm circle, we have x from 4 + 4 = 8 cm to 30 cm. Wait, no, the 2 cm circle is at x=4, so to its right is x=8 to 30. Maybe place the 1.5 cm radius circle (3 cm diameter) there. Centered at (8 + 1.5, 16) = (9.5, 16). Check distance from 2 cm circle: sqrt((9.5 - 4)^2 + (16 - 16)^2) = sqrt(5.5² + 0) = 5.5 cm. Sum of radii is 2 + 1.5 = 3.5 cm. 5.5 > 3.5, so no overlap.7. Finally, the 1 cm radius circle (2 cm diameter). Where can we place it? Maybe above the 2.5 cm circle. The 2.5 cm circle is at (17.5, 4), so above it, y from 4 + 2.5 = 6.5 to 20. Maybe place the 1 cm circle centered at (17.5, 6.5 + 1) = (17.5, 7.5). Check distance from 2.5 cm circle: sqrt((17.5 - 17.5)^2 + (7.5 - 4)^2) = sqrt(0 + 3.5²) = 3.5 cm. Sum of radii is 2.5 + 1 = 3.5 cm. So they just touch, no overlap.Wait, but is there space for the 1 cm circle? Let me check the vertical space. The 2.5 cm circle is at y=4, so the 1 cm circle is at y=7.5, which is within the 20 cm height.Alternatively, maybe place it somewhere else, but this seems to work.So in this arrangement, all seven circles are placed without overlapping and within the 30x20 cm mold. Therefore, the maximum area is the sum of their areas, which is 50.75π cm².But wait, let me double-check the arrangement:- 4 cm at (4,4)- 3.5 cm at (11.5,4)- 2.5 cm at (17.5,4)- 3 cm at (4,11)- 2 cm at (4,16)- 1.5 cm at (9.5,16)- 1 cm at (17.5,7.5)Wait, the 1 cm circle is at (17.5,7.5). Is that within the mold? Yes, x=17.5 is within 0-30, y=7.5 is within 0-20.Also, check distances between all circles:- 4 cm and 3.5 cm: distance ≈10.26 > 7.5, good.- 4 cm and 2.5 cm: distance between (4,4) and (17.5,4) is 13.5 cm. Sum of radii 4+2.5=6.5. 13.5 >6.5, good.- 4 cm and 3 cm: distance 5 cm, sum 7 cm. Wait, 5 <7, which would mean overlapping. Wait, no, the 3 cm circle is at (4,11), so distance from (4,4) is 7 cm. Sum of radii is 4+3=7 cm. So they just touch, no overlap. Good.- 4 cm and 2 cm: distance from (4,4) to (4,16) is 12 cm. Sum of radii 4+2=6 cm. 12 >6, good.- 4 cm and 1.5 cm: distance from (4,4) to (9.5,16) is sqrt(5.5² +12²)=sqrt(30.25+144)=sqrt(174.25)≈13.2 cm. Sum of radii 4+1.5=5.5 cm. 13.2>5.5, good.- 4 cm and 1 cm: distance from (4,4) to (17.5,7.5) is sqrt(13.5² +3.5²)=sqrt(182.25+12.25)=sqrt(194.5)≈13.95 cm. Sum of radii 4+1=5 cm. 13.95>5, good.Now, 3.5 cm circle at (11.5,4):- With 2.5 cm: distance 6 cm, sum 3.5+2.5=6 cm. Just touching, good.- With 3 cm: distance from (11.5,4) to (4,11) is sqrt(7.5² +7²)=sqrt(56.25+49)=sqrt(105.25)≈10.26 cm. Sum of radii 3.5+3=6.5 cm. 10.26>6.5, good.- With 2 cm: distance from (11.5,4) to (4,16) is sqrt(7.5² +12²)=sqrt(56.25+144)=sqrt(200.25)=14.15 cm. Sum of radii 3.5+2=5.5 cm. 14.15>5.5, good.- With 1.5 cm: distance from (11.5,4) to (9.5,16) is sqrt(2² +12²)=sqrt(4+144)=sqrt(148)≈12.17 cm. Sum of radii 3.5+1.5=5 cm. 12.17>5, good.- With 1 cm: distance from (11.5,4) to (17.5,7.5) is sqrt(6² +3.5²)=sqrt(36+12.25)=sqrt(48.25)≈6.95 cm. Sum of radii 3.5+1=4.5 cm. 6.95>4.5, good.Next, 2.5 cm circle at (17.5,4):- With 3 cm: distance from (17.5,4) to (4,11) is sqrt(13.5² +7²)=sqrt(182.25+49)=sqrt(231.25)≈15.2 cm. Sum of radii 2.5+3=5.5 cm. 15.2>5.5, good.- With 2 cm: distance from (17.5,4) to (4,16) is sqrt(13.5² +12²)=sqrt(182.25+144)=sqrt(326.25)≈18.06 cm. Sum of radii 2.5+2=4.5 cm. 18.06>4.5, good.- With 1.5 cm: distance from (17.5,4) to (9.5,16) is sqrt(8² +12²)=sqrt(64+144)=sqrt(208)≈14.42 cm. Sum of radii 2.5+1.5=4 cm. 14.42>4, good.- With 1 cm: distance from (17.5,4) to (17.5,7.5) is 3.5 cm. Sum of radii 2.5+1=3.5 cm. So they just touch, good.3 cm circle at (4,11):- With 2 cm: distance from (4,11) to (4,16) is 5 cm. Sum of radii 3+2=5 cm. Just touching, good.- With 1.5 cm: distance from (4,11) to (9.5,16) is sqrt(5.5² +5²)=sqrt(30.25+25)=sqrt(55.25)≈7.43 cm. Sum of radii 3+1.5=4.5 cm. 7.43>4.5, good.- With 1 cm: distance from (4,11) to (17.5,7.5) is sqrt(13.5² +3.5²)=sqrt(182.25+12.25)=sqrt(194.5)≈13.95 cm. Sum of radii 3+1=4 cm. 13.95>4, good.2 cm circle at (4,16):- With 1.5 cm: distance from (4,16) to (9.5,16) is 5.5 cm. Sum of radii 2+1.5=3.5 cm. 5.5>3.5, good.- With 1 cm: distance from (4,16) to (17.5,7.5) is sqrt(13.5² +8.5²)=sqrt(182.25+72.25)=sqrt(254.5)≈15.95 cm. Sum of radii 2+1=3 cm. 15.95>3, good.1.5 cm circle at (9.5,16):- With 1 cm: distance from (9.5,16) to (17.5,7.5) is sqrt(8² +8.5²)=sqrt(64+72.25)=sqrt(136.25)≈11.67 cm. Sum of radii 1.5+1=2.5 cm. 11.67>2.5, good.So, all circles are placed without overlapping and within the mold. Therefore, the maximum area is indeed the sum of their areas, which is 50.75π cm².Now, for the second part: calculating the total weight of the resin mold after embedding the jewelry pieces. The mold has a uniform thickness of 1 cm, and the density is 1.2 g/cm³. The volume of the circles is negligible.First, calculate the volume of the resin mold. Since the mold is a rectangular prism (box) with length 30 cm, width 20 cm, and thickness 1 cm, the volume is 30*20*1 = 600 cm³.But wait, the jewelry pieces are embedded into the mold. Since their volume is negligible, we don't subtract anything. So the total volume of the resin is still 600 cm³.Then, the weight is volume multiplied by density. So 600 cm³ * 1.2 g/cm³ = 720 grams.Wait, but the problem says "after the jewelry pieces have been embedded, considering the volume of the circles (jewelry pieces) is negligible compared to the volume of the resin mold." So yes, we don't subtract the volume of the circles. Therefore, the total weight is 720 grams.But wait, let me think again. The mold is 30x20x1 cm, so 600 cm³. The jewelry pieces are embedded, but their volume is negligible, so the resin volume remains 600 cm³. Therefore, weight is 600 * 1.2 = 720 grams.Alternatively, if the mold's volume is considered without the jewelry, but the problem says the volume of the circles is negligible, so it doesn't affect the resin volume. So yes, 720 grams.So, to summarize:1. The maximum area is 50.75π cm², which is approximately 159.5 cm², but since the question asks for the exact value, it's 50.75π cm².2. The total weight is 720 grams.</think>

🧠:<think>Alright, so I've got this problem about a retired jingle writer analyzing his famous jingle using Fourier series and stuff. It's been a while since I did Fourier series, but let me try to work through this step by step.First, the problem is divided into two parts. Part 1 is about expressing the jingle's waveform as a Fourier series, and Part 2 is about calculating the average energy per beat. Let's tackle them one by one.Problem 1: Expressing the waveform using Fourier seriesOkay, so the jingle has a fundamental frequency ( f_0 = 440 ) Hz. The harmonics are given by ( f_n = n cdot f_0 ) for ( n = 1, 2, 3, ldots ). The waveform ( x(t) ) is a sum of these harmonics, each with an amplitude of ( frac{1}{n^2} ). The period ( T ) is ( frac{1}{f_0} ), which would be ( frac{1}{440} ) seconds. That makes sense because the period is the reciprocal of the frequency.So, I remember that a Fourier series represents a periodic function as a sum of sines and cosines. Since the problem mentions harmonics, which are integer multiples of the fundamental frequency, it's definitely a Fourier series scenario.But wait, the problem says "express the jingle's waveform ( x(t) ) as a sum of its harmonics over one period ( T )." Hmm. So, is it a Fourier series in terms of sine and cosine terms or just a sum of complex exponentials? The problem doesn't specify, but since it's about harmonics, it might be a sum of sine terms because in music, harmonics are often associated with sine waves.But actually, in Fourier series, both sine and cosine terms are used, but if the waveform is symmetric or has certain properties, it might only have sine or cosine terms. However, since the problem doesn't specify, maybe it's safer to assume it's a sum of sine terms because it's talking about harmonics, which are typically sine waves in music.Wait, but in reality, a Fourier series for a general periodic function includes both sine and cosine terms unless the function is odd or even. Since the problem doesn't specify, maybe I should consider both? Hmm, but the problem says "the amplitude of each harmonic is ( frac{1}{n^2} )." So, each harmonic is a sine wave with amplitude ( frac{1}{n^2} ).Wait, but in Fourier series, each term is a sine or cosine with coefficients. So, if all the harmonics are sine terms with amplitude ( frac{1}{n^2} ), then the Fourier series would be a sum of sine terms with coefficients ( frac{1}{n^2} ).But hold on, in a Fourier series, the coefficients are usually given by integrals over the period, but here it's given directly. So, maybe the waveform is expressed as:( x(t) = sum_{n=1}^{infty} frac{1}{n^2} sin(2pi n f_0 t) )But let me check the units. The argument of the sine function should be in radians. Since ( f_0 ) is in Hz, which is 1/s, multiplying by t (in seconds) gives a dimensionless quantity, which is correct.Alternatively, since the period ( T = frac{1}{f_0} ), the angular frequency ( omega_0 = 2pi f_0 ), so each harmonic has angular frequency ( n omega_0 ). So, the sine terms would be ( sin(n omega_0 t) ).So, putting it together, the waveform is:( x(t) = sum_{n=1}^{infty} frac{1}{n^2} sin(n omega_0 t) )where ( omega_0 = 2pi f_0 = 2pi times 440 ) rad/s.Is that right? Let me think. Each harmonic is a sine wave with frequency ( n f_0 ), amplitude ( frac{1}{n^2} ), so yes, that seems correct.But wait, in Fourier series, we usually have both sine and cosine terms unless the function is odd. Since the problem doesn't specify, maybe it's just a sum of sine terms? Or perhaps it's a full Fourier series with both sine and cosine, but the coefficients are given as ( frac{1}{n^2} ). Hmm.Wait, the problem says "the amplitude of each harmonic is ( frac{1}{n^2} )." In Fourier series, the amplitude of each harmonic is related to the coefficients. For a sine term, the coefficient is the amplitude, but for a cosine term, it's similar. However, if the waveform is expressed purely as a sum of sine terms, then each coefficient is the amplitude. So, if the problem states that each harmonic has amplitude ( frac{1}{n^2} ), then it's likely a sum of sine terms with those amplitudes.Therefore, I think the correct expression is:( x(t) = sum_{n=1}^{infty} frac{1}{n^2} sin(2pi n f_0 t) )Alternatively, using angular frequency:( x(t) = sum_{n=1}^{infty} frac{1}{n^2} sin(n omega_0 t) )Either way is correct, but since ( omega_0 = 2pi f_0 ), both expressions are equivalent.So, I think that's the answer for part 1.Problem 2: Calculating the average energy per beatOkay, now the jingle writer wants to determine the average energy per beat over the 60-second duration. The instantaneous power is given by ( P(t) = x(t)^2 ). We need to calculate the average power ( langle P rangle ) over the period ( T ) and then determine the total energy per beat, noting that each beat lasts ( frac{1}{4} ) seconds.First, let's recall that power is energy per unit time, so average power over a period multiplied by the duration gives the energy. But here, we need the average power over one period ( T ), and then since each beat is ( frac{1}{4} ) seconds, we can find the energy per beat.Wait, but the jingle is 60 seconds long, and each beat is ( frac{1}{4} ) seconds, so there are ( 60 times 4 = 240 ) beats in total. But maybe that's not directly needed here.Let me break it down step by step.First, find the average power ( langle P rangle ) over one period ( T ). Since ( P(t) = x(t)^2 ), the average power is:( langle P rangle = frac{1}{T} int_{0}^{T} P(t) dt = frac{1}{T} int_{0}^{T} x(t)^2 dt )So, we need to compute this integral.Given that ( x(t) ) is a sum of sine terms, squaring it will result in cross terms. But due to the orthogonality of sine functions, the cross terms will integrate to zero over one period. So, the integral simplifies to the sum of the squares of the amplitudes of each harmonic.Wait, let me recall. If ( x(t) = sum_{n=1}^{infty} a_n sin(n omega_0 t) ), then ( x(t)^2 = sum_{n=1}^{infty} a_n^2 sin^2(n omega_0 t) + 2 sum_{n=1}^{infty} sum_{m > n} a_n a_m sin(n omega_0 t) sin(m omega_0 t) )When we integrate ( x(t)^2 ) over one period ( T ), the cross terms (the double sum) will integrate to zero because the product of sine functions with different frequencies are orthogonal over the period. So, we're left with:( int_{0}^{T} x(t)^2 dt = sum_{n=1}^{infty} a_n^2 int_{0}^{T} sin^2(n omega_0 t) dt )Now, the integral of ( sin^2(k omega_0 t) ) over one period ( T ) is ( frac{T}{2} ), because ( sin^2 theta ) has an average value of ( frac{1}{2} ) over a full period.Therefore,( int_{0}^{T} x(t)^2 dt = sum_{n=1}^{infty} a_n^2 cdot frac{T}{2} )So, the average power ( langle P rangle ) is:( langle P rangle = frac{1}{T} cdot sum_{n=1}^{infty} a_n^2 cdot frac{T}{2} = frac{1}{2} sum_{n=1}^{infty} a_n^2 )Given that each ( a_n = frac{1}{n^2} ), so ( a_n^2 = frac{1}{n^4} ). Therefore,( langle P rangle = frac{1}{2} sum_{n=1}^{infty} frac{1}{n^4} )I remember that the sum ( sum_{n=1}^{infty} frac{1}{n^4} ) is known and equals ( frac{pi^4}{90} ). Let me confirm that. Yes, the Riemann zeta function ( zeta(4) = frac{pi^4}{90} ).So,( langle P rangle = frac{1}{2} cdot frac{pi^4}{90} = frac{pi^4}{180} )So, that's the average power over one period ( T ).Now, we need to determine the total energy per beat. Each beat lasts ( frac{1}{4} ) seconds. Since power is energy per unit time, energy is power multiplied by time. Therefore, the energy per beat ( E_{beat} ) is:( E_{beat} = langle P rangle times text{duration of one beat} = langle P rangle times frac{1}{4} )Substituting the value of ( langle P rangle ):( E_{beat} = frac{pi^4}{180} times frac{1}{4} = frac{pi^4}{720} )But wait, let me think again. The average power is over one period ( T ), which is ( frac{1}{440} ) seconds. But each beat is ( frac{1}{4} ) seconds. So, how many periods are in one beat?Number of periods in one beat: ( frac{frac{1}{4}}{T} = frac{frac{1}{4}}{frac{1}{440}} = frac{440}{4} = 110 ). So, each beat consists of 110 periods.Therefore, the energy per beat would be the average power multiplied by the duration of one beat, which is ( langle P rangle times frac{1}{4} ). But wait, since the average power is already the average over one period, multiplying by the duration of one beat gives the total energy in that beat.Alternatively, since each beat has 110 periods, the energy per beat is 110 times the energy per period.Energy per period ( E_{period} = langle P rangle times T = frac{pi^4}{180} times frac{1}{440} )Then, energy per beat ( E_{beat} = 110 times E_{period} = 110 times frac{pi^4}{180} times frac{1}{440} )Simplify:( 110 times frac{1}{440} = frac{1}{4} ), so( E_{beat} = frac{pi^4}{180} times frac{1}{4} = frac{pi^4}{720} )So, same result as before. Therefore, the total energy per beat is ( frac{pi^4}{720} ).But wait, let me make sure. The average power is ( frac{pi^4}{180} ) watts (assuming the units are consistent). Then, energy is power multiplied by time, so ( frac{pi^4}{180} times frac{1}{4} ) seconds gives energy in joules.Yes, that makes sense.Alternatively, if we think in terms of the entire 60-second jingle, the total energy would be ( langle P rangle times 60 ). But the question specifically asks for the total energy per beat, so we just need ( frac{pi^4}{720} ) joules per beat.Wait, but let me check the units again. The problem doesn't specify the units for the waveform ( x(t) ). If ( x(t) ) is in volts, then power would be in watts (volts squared over ohms, but since impedance isn't given, maybe it's just in terms of voltage squared). But since the problem doesn't specify, maybe we can just leave it in terms of the squared waveform.But in any case, the average power is ( frac{pi^4}{180} ), and energy per beat is ( frac{pi^4}{720} ).So, summarizing:1. The waveform is ( x(t) = sum_{n=1}^{infty} frac{1}{n^2} sin(n omega_0 t) ), where ( omega_0 = 2pi f_0 = 880pi ) rad/s.2. The average power is ( frac{pi^4}{180} ), and the energy per beat is ( frac{pi^4}{720} ).But wait, let me make sure I didn't make a mistake in calculating the sum. The sum ( sum_{n=1}^{infty} frac{1}{n^4} ) is indeed ( frac{pi^4}{90} ). Yes, that's correct. So, the average power is half of that, which is ( frac{pi^4}{180} ).And then, energy per beat is average power times ( frac{1}{4} ) seconds, so ( frac{pi^4}{720} ) joules per beat.I think that's it. Let me just recap:- Expressed the waveform as a Fourier series with sine terms, each with amplitude ( frac{1}{n^2} ).- Calculated the average power by integrating the square of the waveform over one period, using orthogonality to simplify the integral to a sum of squares of amplitudes.- Recognized the sum as a known Riemann zeta function value.- Then, calculated the energy per beat by multiplying the average power by the duration of one beat.Yes, that seems solid.Final Answer1. The waveform is expressed as ( boxed{x(t) = sum_{n=1}^{infty} frac{1}{n^2} sin(2pi n f_0 t)} ).2. The total energy per beat is ( boxed{dfrac{pi^4}{720}} ) joules.</think>