🧠:<think>Alright, so I have this problem where a researcher is studying the impact of digital and transportation technology on urbanization. They've set up a partial differential equation (PDE) to model the population density over time. The equation is:[frac{partial P(t, x, y)}{partial t} = alpha nabla^2 P(t, x, y) + beta D(t, x, y) + gamma T(t, x, y)]They want me to first find the general solution for ( P(t, x, y) ) given the initial conditions and then, in part 2, consider specific forms for ( D(t, x, y) ) and ( T(t, x, y) ) and analyze the long-term behavior.Okay, let's start with part 1. The PDE is a linear parabolic equation because it has a Laplacian term and some source terms. The general form of such an equation is:[frac{partial P}{partial t} = alpha nabla^2 P + f(t, x, y)]where ( f(t, x, y) = beta D(t, x, y) + gamma T(t, x, y) ).To solve this, I remember that for linear PDEs, the solution can be expressed as the sum of the homogeneous solution and a particular solution. The homogeneous equation is:[frac{partial P_h}{partial t} = alpha nabla^2 P_h]And the particular solution ( P_p ) satisfies:[frac{partial P_p}{partial t} = alpha nabla^2 P_p + f(t, x, y)]So, the general solution is ( P = P_h + P_p ).For the homogeneous solution, assuming the domain is such that we can use separation of variables, the solution would typically involve eigenfunctions of the Laplacian. However, without specific boundary conditions, it's hard to write down the exact form. But generally, the homogeneous solution would decay or spread out depending on the initial conditions.For the particular solution, since ( f(t, x, y) ) is given, we can use methods like Duhamel's principle or Green's functions. The Green's function for the heat equation (which this is similar to) is well-known. The particular solution can be written as a convolution of the Green's function with the source term ( f(t, x, y) ).So, the general solution would be:[P(t, x, y) = int_{Omega} G(t, x, y; 0, x', y') P(0, x', y') dx' dy' + int_{0}^{t} int_{Omega} G(t - t', x, y; t', x', y') [beta D(t', x', y') + gamma T(t', x', y')] dx' dy' dt']Where ( G ) is the Green's function for the heat equation, which in two dimensions is:[G(t, x, y; t', x', y') = frac{1}{4 pi alpha (t - t')} expleft( -frac{(x - x')^2 + (y - y')^2}{4 alpha (t - t')} right)]But this is assuming infinite space, which might not be the case here. The researcher is looking at a metropolitan area, so maybe the domain is bounded. Without knowing the exact boundary conditions, it's tricky, but perhaps we can proceed with the Green's function approach assuming the domain is large enough that boundary effects are negligible, or that we can use periodic boundary conditions.Alternatively, if the domain is a rectangle, we can express the Green's function as a series solution involving sine and cosine terms. But since the problem doesn't specify, maybe it's safer to present the solution in terms of the Green's function as above.So, summarizing, the general solution is the sum of the initial population density convolved with the Green's function and the integral over time of the source terms convolved with the Green's function.Moving on to part 2. Now, ( D(t, x, y) = D_0 e^{delta t} ) and ( T(t, x, y) = T_0 + eta t ). So, substituting these into the particular solution part.First, let's write the particular solution:[P_p(t, x, y) = int_{0}^{t} int_{Omega} G(t - t', x, y; t', x', y') [beta D_0 e^{delta t'} + gamma (T_0 + eta t')] dx' dy' dt']This integral can be split into two parts:1. The term involving ( D(t', x', y') = D_0 e^{delta t'} )2. The term involving ( T(t', x', y') = T_0 + eta t' )So, let's handle them separately.First, the digital infrastructure term:[P_{p1}(t, x, y) = beta D_0 int_{0}^{t} e^{delta t'} int_{Omega} G(t - t', x, y; t', x', y') dx' dy' dt']But notice that the integral over ( Omega ) of the Green's function at time ( t - t' ) is 1, because the Green's function is a probability density function (it's the fundamental solution to the diffusion equation). So, integrating over all space gives 1. Therefore, this simplifies to:[P_{p1}(t, x, y) = beta D_0 int_{0}^{t} e^{delta t'} dt' = beta D_0 left( frac{e^{delta t} - 1}{delta} right)]Similarly, for the transportation network term:[P_{p2}(t, x, y) = gamma int_{0}^{t} int_{Omega} G(t - t', x, y; t', x', y') (T_0 + eta t') dx' dy' dt']Again, the integral over ( Omega ) of the Green's function is 1, so:[P_{p2}(t, x, y) = gamma int_{0}^{t} (T_0 + eta t') dt' = gamma left( T_0 t + frac{eta t^2}{2} right)]Therefore, the particular solution is:[P_p(t, x, y) = beta D_0 left( frac{e^{delta t} - 1}{delta} right) + gamma left( T_0 t + frac{eta t^2}{2} right)]Now, the homogeneous solution ( P_h(t, x, y) ) is the solution to the heat equation with initial condition ( P(0, x, y) ). As time progresses, this term will spread out and its amplitude will decrease unless there's some source term. However, since we have the particular solution accounting for the sources, the homogeneous part will tend to zero as ( t to infty ) if the domain is unbounded or if the boundary conditions allow dissipation.Therefore, the long-term behavior of ( P(t, x, y) ) will be dominated by the particular solution.Looking at the particular solution:- The term from digital infrastructure grows exponentially as ( e^{delta t} ).- The term from transportation infrastructure grows quadratically as ( t^2 ).So, as ( t to infty ), the exponential term will dominate over the quadratic term if ( delta > 0 ). Therefore, the population density will grow exponentially in the long run.However, we need to consider the constants ( beta ) and ( gamma ). If ( beta ) is positive, the digital infrastructure term contributes positively to population density, and vice versa. Similarly for ( gamma ). Assuming both ( beta ) and ( gamma ) are positive (which makes sense if better infrastructure attracts more population), then both terms contribute to growth.But the exponential growth from digital infrastructure will outpace the quadratic growth from transportation. So, the dominant term as ( t to infty ) is ( beta D_0 frac{e^{delta t}}{delta} ).Therefore, the long-term behavior is exponential growth of population density.But wait, is this realistic? Exponential growth in population density might lead to unbounded growth, which isn't sustainable in reality. However, the model doesn't include any limiting factors like carrying capacity or saturation effects. So, within the scope of this model, yes, it's exponential.Alternatively, if ( delta ) is negative, the exponential term would decay, but that would imply digital infrastructure is decreasing, which might not be the case over 10 years.So, assuming ( delta > 0 ), the population density grows exponentially in the long run.Putting it all together, the specific form of ( P(t, x, y) ) is the sum of the homogeneous solution (which decays over time) and the particular solution which has the exponential and quadratic terms.But since the homogeneous solution tends to zero, the specific form simplifies to the particular solution in the long run.So, summarizing:1. The general solution is a combination of the initial population diffusing over time and the cumulative effect of the digital and transportation infrastructure.2. With the given forms of ( D ) and ( T ), the population density grows exponentially due to the digital infrastructure and quadratically due to transportation, with the exponential term dominating as time increases.I think that's the gist of it. I should probably write this up more formally, but that's the thought process.Final Answer1. The general solution for ( P(t, x, y) ) is given by the convolution of the initial condition with the Green's function plus the integral of the source terms convolved with the Green's function. This can be expressed as: [ P(t, x, y) = int_{Omega} G(t, x, y; 0, x', y') P(0, x', y') dx' dy' + int_{0}^{t} int_{Omega} G(t - t', x, y; t', x', y') [beta D(t', x', y') + gamma T(t', x', y')] dx' dy' dt' ] where ( G ) is the Green's function for the heat equation.2. Substituting the given forms of ( D(t, x, y) ) and ( T(t, x, y) ), the specific solution is: [ P(t, x, y) = beta D_0 left( frac{e^{delta t} - 1}{delta} right) + gamma left( T_0 t + frac{eta t^2}{2} right) + text{(homogeneous solution)} ] As ( t to infty ), the population density grows exponentially due to the term ( beta D_0 frac{e^{delta t}}{delta} ).The long-term behavior of the population density is dominated by exponential growth.[boxed{P(t, x, y) = beta D_0 left( frac{e^{delta t} - 1}{delta} right) + gamma left( T_0 t + frac{eta t^2}{2} right) + text{(diffusion term)}}]</think>

🧠:<think>Okay, so I have this problem about modeling the spread of cultural artifacts using a partial differential equation. It's divided into two parts. Let me try to tackle them step by step.Starting with the first part: solving the PDE. The equation given is[ frac{partial u}{partial t} = D left( frac{partial^2 u}{partial r^2} + frac{1}{r} frac{partial u}{partial r} right). ]Hmm, this looks like a radial diffusion equation in two dimensions because of the ( frac{1}{r} frac{partial u}{partial r} ) term. I remember that in cylindrical coordinates, the Laplacian in radial direction has that extra term. So, yeah, this is a 2D diffusion equation.The initial condition is ( u(r,0) = u_0 e^{-alpha r^2} ). So, it's a Gaussian distribution centered at the origin. That makes sense because it's a common initial condition for diffusion problems.I think I need to solve this PDE using separation of variables or maybe Fourier transforms. But since it's in radial coordinates, separation of variables might be more straightforward.Let me recall how to solve the heat equation in polar coordinates. The general solution can be expressed as a series expansion using Bessel functions because of the angular dependence, but since the problem is radially symmetric, the solution should only depend on ( r ) and ( t ), not on the angle ( theta ). So, maybe I can use a Fourier transform in radial coordinates, which is related to the Hankel transform.Wait, Hankel transforms are used for solving PDEs in cylindrical coordinates when the solution is radially symmetric. The Hankel transform of order zero might be applicable here because the equation doesn't involve any angular derivatives.Let me try to apply the Hankel transform to the PDE. The Hankel transform of order ( nu ) is defined as:[ mathcal{H}_nu { f(r) } = int_0^infty f(r) J_nu(kr) r , dr, ]where ( J_nu ) is the Bessel function of the first kind of order ( nu ). Since our equation is radially symmetric, we can use ( nu = 0 ).Taking the Hankel transform of both sides of the PDE. Let me denote the Hankel transform of ( u(r,t) ) as ( tilde{u}(k,t) ).Applying the Hankel transform to the PDE:[ mathcal{H}_0 left{ frac{partial u}{partial t} right} = mathcal{H}_0 left{ D left( frac{partial^2 u}{partial r^2} + frac{1}{r} frac{partial u}{partial r} right) right}. ]Since the Hankel transform is linear, this becomes:[ frac{partial tilde{u}}{partial t} = D mathcal{H}_0 left{ frac{partial^2 u}{partial r^2} + frac{1}{r} frac{partial u}{partial r} right}. ]Now, I need to compute the Hankel transform of the Laplacian term. I remember that the Hankel transform of the derivative of a function can be expressed in terms of the Hankel transform of the function itself. Specifically, for the second derivative:[ mathcal{H}_0 left{ frac{partial^2 u}{partial r^2} right} = -k^2 tilde{u}(k,t) - frac{1}{k} frac{partial}{partial k} left( k tilde{u}(k,t) right). ]Wait, is that right? Let me double-check. I think the Hankel transform of the second derivative involves the second derivative of the transform. Alternatively, maybe I should look up the properties of the Hankel transform.Alternatively, I recall that for the Laplacian in 2D, the Hankel transform converts the Laplacian into a multiplication by ( -k^2 ). So, perhaps:[ mathcal{H}_0 left{ nabla^2 u right} = -k^2 tilde{u}(k,t). ]But in our case, the equation is:[ frac{partial u}{partial t} = D left( frac{partial^2 u}{partial r^2} + frac{1}{r} frac{partial u}{partial r} right). ]Which is exactly the Laplacian in 2D polar coordinates. So, the equation is the heat equation in 2D, and its solution can be found using the Hankel transform.Therefore, taking the Hankel transform of both sides:[ frac{partial tilde{u}}{partial t} = -D k^2 tilde{u}(k,t). ]This is an ordinary differential equation (ODE) in ( t ) with variable ( k ). The solution to this ODE is:[ tilde{u}(k,t) = tilde{u}(k,0) e^{-D k^2 t}. ]Now, I need to find the Hankel transform of the initial condition ( u(r,0) = u_0 e^{-alpha r^2} ).So, compute ( tilde{u}(k,0) = mathcal{H}_0 { u_0 e^{-alpha r^2} } ).I think the Hankel transform of a Gaussian function ( e^{-a r^2} ) is another Gaussian function. Let me recall the formula.The Hankel transform of order ( nu ) of ( r^nu e^{-a r^2} ) is ( frac{sqrt{pi}}{2 a^{3/2}}} k^nu e^{-k^2 / (4a)} ). Wait, I might be mixing up some constants.Alternatively, I can look up the Hankel transform of ( e^{-a r^2} ). Let me think.The Hankel transform of order 0 of ( e^{-a r^2} ) is:[ int_0^infty e^{-a r^2} J_0(k r) r , dr. ]I think this integral is known. Let me recall that:[ int_0^infty e^{-a r^2} J_0(k r) r , dr = frac{1}{2 a} e^{-k^2 / (4a)}. ]Yes, that seems familiar. So, substituting ( a = alpha ), we have:[ tilde{u}(k,0) = u_0 cdot frac{1}{2 alpha} e^{-k^2 / (4 alpha)}. ]Therefore, the Hankel transform of the initial condition is:[ tilde{u}(k,0) = frac{u_0}{2 alpha} e^{-k^2 / (4 alpha)}. ]So, plugging this into the solution of the ODE:[ tilde{u}(k,t) = frac{u_0}{2 alpha} e^{-k^2 / (4 alpha)} e^{-D k^2 t}. ]Simplify the exponents:[ tilde{u}(k,t) = frac{u_0}{2 alpha} e^{-k^2 (1/(4 alpha) + D t)}. ]Let me write the exponent as:[ -k^2 left( frac{1}{4 alpha} + D t right) = -k^2 left( frac{1 + 4 alpha D t}{4 alpha} right). ]So,[ tilde{u}(k,t) = frac{u_0}{2 alpha} e^{-k^2 (1 + 4 alpha D t)/(4 alpha)}. ]Now, to find ( u(r,t) ), we need to take the inverse Hankel transform of ( tilde{u}(k,t) ). The inverse Hankel transform is given by:[ u(r,t) = int_0^infty tilde{u}(k,t) J_0(k r) k , dk. ]Substituting ( tilde{u}(k,t) ):[ u(r,t) = frac{u_0}{2 alpha} int_0^infty e^{-k^2 (1 + 4 alpha D t)/(4 alpha)} J_0(k r) k , dk. ]Hmm, this integral looks similar to the Hankel transform we did earlier. In fact, if I let ( a = (1 + 4 alpha D t)/(4 alpha) ), then the integral becomes:[ int_0^infty e^{-a k^2} J_0(k r) k , dk. ]From earlier, we know that:[ int_0^infty e^{-a k^2} J_0(k r) k , dk = frac{1}{2 a} e^{-r^2 / (4a)}. ]So, substituting back ( a = (1 + 4 alpha D t)/(4 alpha) ), we have:[ int_0^infty e^{-a k^2} J_0(k r) k , dk = frac{1}{2 a} e^{-r^2 / (4a)} = frac{1}{2 cdot frac{1 + 4 alpha D t}{4 alpha}} e^{-r^2 / (4 cdot frac{1 + 4 alpha D t}{4 alpha})}. ]Simplify the denominator:[ 2 cdot frac{1 + 4 alpha D t}{4 alpha} = frac{1 + 4 alpha D t}{2 alpha}. ]And the exponent:[ -r^2 / left( frac{1 + 4 alpha D t}{alpha} right) = -frac{alpha r^2}{1 + 4 alpha D t}. ]So, putting it all together:[ int_0^infty e^{-a k^2} J_0(k r) k , dk = frac{2 alpha}{1 + 4 alpha D t} e^{- frac{alpha r^2}{1 + 4 alpha D t}}. ]Therefore, substituting back into ( u(r,t) ):[ u(r,t) = frac{u_0}{2 alpha} cdot frac{2 alpha}{1 + 4 alpha D t} e^{- frac{alpha r^2}{1 + 4 alpha D t}}. ]Simplify the constants:[ frac{u_0}{2 alpha} cdot frac{2 alpha}{1 + 4 alpha D t} = frac{u_0}{1 + 4 alpha D t}. ]So, the solution is:[ u(r,t) = frac{u_0}{1 + 4 alpha D t} e^{- frac{alpha r^2}{1 + 4 alpha D t}}. ]Let me check the dimensions to see if this makes sense. The exponent must be dimensionless. ( alpha ) has units of inverse length squared, ( r^2 ) is length squared, so ( alpha r^2 ) is dimensionless. The denominator ( 1 + 4 alpha D t ) must also be dimensionless. ( D ) has units of length squared over time, ( t ) is time, so ( D t ) is length squared. ( alpha D t ) is (inverse length squared)(length squared) = dimensionless. So, yes, the denominator is dimensionless, and the exponent is dimensionless. The coefficient ( u_0 ) is density, so the entire expression is density, which matches.Also, at ( t = 0 ), we get ( u(r,0) = u_0 e^{-alpha r^2} ), which matches the initial condition. As ( t ) increases, the denominator grows, so the amplitude decreases, and the exponent's coefficient decreases, meaning the Gaussian broadens. That makes sense for diffusion.Okay, so I think that's the solution for the first part.Moving on to the second part: determining the mean squared displacement ( langle r^2(t) rangle ).The mean squared displacement is defined as:[ langle r^2(t) rangle = int_0^infty r^2 u(r,t) cdot 2 pi r , dr. ]Wait, no. Wait, actually, in 2D polar coordinates, the area element is ( 2 pi r , dr ). But since ( u(r,t) ) is the density, the total number of artifacts is ( int_0^infty u(r,t) cdot 2 pi r , dr ). But for the mean squared displacement, it's the expectation value of ( r^2 ), so:[ langle r^2(t) rangle = frac{1}{N(t)} int_0^infty r^2 u(r,t) cdot 2 pi r , dr, ]where ( N(t) = int_0^infty u(r,t) cdot 2 pi r , dr ) is the total number of artifacts at time ( t ).But wait, in our case, the initial condition is ( u(r,0) = u_0 e^{-alpha r^2} ), which is a Gaussian, so the total number is:[ N(0) = int_0^infty u_0 e^{-alpha r^2} cdot 2 pi r , dr. ]Let me compute that:Let ( x = r^2 ), then ( dx = 2 r dr ), so ( r dr = dx/2 ).Thus,[ N(0) = 2 pi u_0 int_0^infty e^{-alpha x} cdot frac{dx}{2} = pi u_0 int_0^infty e^{-alpha x} dx = pi u_0 cdot frac{1}{alpha}. ]So, ( N(0) = frac{pi u_0}{alpha} ).Now, looking at the solution ( u(r,t) = frac{u_0}{1 + 4 alpha D t} e^{- frac{alpha r^2}{1 + 4 alpha D t}} ), let's compute ( N(t) ):[ N(t) = int_0^infty u(r,t) cdot 2 pi r , dr = 2 pi int_0^infty frac{u_0}{1 + 4 alpha D t} e^{- frac{alpha r^2}{1 + 4 alpha D t}} r , dr. ]Again, let me make a substitution. Let ( y = r^2 ), so ( dy = 2 r dr ), ( r dr = dy/2 ).Thus,[ N(t) = 2 pi cdot frac{u_0}{1 + 4 alpha D t} cdot frac{1}{2} int_0^infty e^{- frac{alpha y}{1 + 4 alpha D t}} dy. ]The integral is:[ int_0^infty e^{-b y} dy = frac{1}{b}, ]where ( b = frac{alpha}{1 + 4 alpha D t} ).So,[ N(t) = pi cdot frac{u_0}{1 + 4 alpha D t} cdot frac{1 + 4 alpha D t}{alpha} = frac{pi u_0}{alpha}. ]So, ( N(t) = N(0) ), which makes sense because the total number of artifacts should be conserved in the diffusion process. Good, that checks out.Now, compute ( langle r^2(t) rangle ):[ langle r^2(t) rangle = frac{1}{N(t)} int_0^infty r^2 u(r,t) cdot 2 pi r , dr. ]Substituting ( u(r,t) ):[ langle r^2(t) rangle = frac{1}{N(0)} cdot 2 pi int_0^infty r^3 cdot frac{u_0}{1 + 4 alpha D t} e^{- frac{alpha r^2}{1 + 4 alpha D t}} dr. ]Simplify:[ langle r^2(t) rangle = frac{2 pi u_0}{N(0) (1 + 4 alpha D t)} int_0^infty r^3 e^{- frac{alpha r^2}{1 + 4 alpha D t}} dr. ]Again, let me make a substitution. Let ( y = r^2 ), so ( dy = 2 r dr ), ( r^3 dr = y cdot r dr = y cdot frac{dy}{2} ).Thus, the integral becomes:[ int_0^infty r^3 e^{- frac{alpha r^2}{1 + 4 alpha D t}} dr = frac{1}{2} int_0^infty y e^{- frac{alpha y}{1 + 4 alpha D t}} dy. ]Compute this integral:Let ( b = frac{alpha}{1 + 4 alpha D t} ), then:[ int_0^infty y e^{-b y} dy = frac{1}{b^2}. ]So,[ int_0^infty r^3 e^{- frac{alpha r^2}{1 + 4 alpha D t}} dr = frac{1}{2} cdot frac{1}{b^2} = frac{1}{2} cdot left( frac{1 + 4 alpha D t}{alpha} right)^2. ]Substituting back into ( langle r^2(t) rangle ):[ langle r^2(t) rangle = frac{2 pi u_0}{N(0) (1 + 4 alpha D t)} cdot frac{1}{2} left( frac{1 + 4 alpha D t}{alpha} right)^2. ]Simplify:The 2 and 1/2 cancel out.[ langle r^2(t) rangle = frac{pi u_0}{N(0)} cdot frac{(1 + 4 alpha D t)^2}{alpha^2 (1 + 4 alpha D t)} = frac{pi u_0}{N(0)} cdot frac{1 + 4 alpha D t}{alpha^2}. ]But ( N(0) = frac{pi u_0}{alpha} ), so:[ langle r^2(t) rangle = frac{pi u_0}{frac{pi u_0}{alpha}} cdot frac{1 + 4 alpha D t}{alpha^2} = alpha cdot frac{1 + 4 alpha D t}{alpha^2} = frac{1 + 4 alpha D t}{alpha}. ]Simplify:[ langle r^2(t) rangle = frac{1}{alpha} + 4 D t. ]Wait, that seems interesting. The mean squared displacement has two terms: a constant term ( 1/alpha ) and a term linear in time ( 4 D t ).But let's think about the initial condition. At ( t = 0 ), ( langle r^2(0) rangle = 1/alpha ). Let me check what the variance of the initial Gaussian is.The initial distribution is ( u(r,0) = u_0 e^{-alpha r^2} ). In 2D, the variance ( sigma^2 ) of a Gaussian ( e^{-a r^2} ) is ( 1/(4a) ). Wait, is that right?Wait, in 1D, the variance of ( e^{-a x^2} ) is ( 1/(2a) ). In 2D polar coordinates, the variance in ( r ) is different. Let me recall.Actually, for a 2D Gaussian distribution ( e^{-a r^2} ), the radial variance is ( langle r^2 rangle = 1/(2a) ). Wait, but in our case, the initial distribution is ( u(r,0) = u_0 e^{-alpha r^2} ). So, the variance should be ( langle r^2 rangle = 1/(2 alpha) ). But according to our calculation, at ( t = 0 ), ( langle r^2(0) rangle = 1/alpha ). Hmm, that's twice the expected variance.Wait, maybe I made a mistake in the calculation. Let me go back.When I computed ( langle r^2(t) rangle ), I had:[ langle r^2(t) rangle = frac{1}{N(t)} int_0^infty r^2 u(r,t) cdot 2 pi r , dr. ]Wait, but in 2D, the mean squared displacement is actually ( langle r^2 rangle ), which is the expectation of ( r^2 ). However, in our case, the integral is ( int_0^infty r^2 u(r,t) cdot 2 pi r , dr ). Wait, no, that's not correct.Wait, actually, the correct expression for ( langle r^2 rangle ) is:[ langle r^2 rangle = frac{1}{N(t)} int_0^infty r^2 u(r,t) cdot 2 pi r , dr. ]Because ( u(r,t) ) is the density, and the probability density in 2D is ( u(r,t) cdot 2 pi r , dr ). So, to compute the expectation of ( r^2 ), it's:[ langle r^2 rangle = frac{int_0^infty r^2 cdot u(r,t) cdot 2 pi r , dr}{int_0^infty u(r,t) cdot 2 pi r , dr}. ]So, in our case, the numerator is:[ int_0^infty r^3 u(r,t) cdot 2 pi , dr ]Wait, no, wait. Let me clarify.In 2D polar coordinates, the probability of finding an artifact in the annulus between ( r ) and ( r + dr ) is ( u(r,t) cdot 2 pi r , dr ). Therefore, the expectation value ( langle r^2 rangle ) is:[ langle r^2 rangle = frac{int_0^infty r^2 cdot u(r,t) cdot 2 pi r , dr}{int_0^infty u(r,t) cdot 2 pi r , dr}. ]So, the numerator is ( int_0^infty r^3 u(r,t) cdot 2 pi , dr ), and the denominator is ( N(t) = int_0^infty u(r,t) cdot 2 pi r , dr ).Wait, but in our calculation above, we had:[ langle r^2(t) rangle = frac{1}{N(t)} cdot 2 pi int_0^infty r^3 u(r,t) , dr. ]But actually, it should be:[ langle r^2(t) rangle = frac{int_0^infty r^2 cdot u(r,t) cdot 2 pi r , dr}{int_0^infty u(r,t) cdot 2 pi r , dr} = frac{int_0^infty r^3 u(r,t) cdot 2 pi , dr}{N(t)}. ]Wait, no, hold on. Let me think carefully.The expectation value ( langle r^2 rangle ) is:[ langle r^2 rangle = frac{int_0^infty r^2 cdot [u(r,t) cdot 2 pi r , dr]}{int_0^infty u(r,t) cdot 2 pi r , dr} = frac{int_0^infty r^3 u(r,t) cdot 2 pi , dr}{N(t)}. ]Yes, that's correct. So, in our calculation, we had:[ langle r^2(t) rangle = frac{2 pi u_0}{N(0) (1 + 4 alpha D t)} int_0^infty r^3 e^{- frac{alpha r^2}{1 + 4 alpha D t}} dr. ]But when we computed the integral, we substituted ( y = r^2 ), so ( r^3 dr = y cdot frac{dy}{2} ). Therefore, the integral became:[ int_0^infty r^3 e^{-b r^2} dr = frac{1}{2} int_0^infty y e^{-b y} dy = frac{1}{2} cdot frac{1}{b^2}. ]Which is correct. Then, substituting back, we had:[ langle r^2(t) rangle = frac{2 pi u_0}{N(0) (1 + 4 alpha D t)} cdot frac{1}{2} cdot frac{(1 + 4 alpha D t)^2}{alpha^2} = frac{pi u_0}{N(0)} cdot frac{1 + 4 alpha D t}{alpha^2}. ]Since ( N(0) = frac{pi u_0}{alpha} ), substituting:[ langle r^2(t) rangle = frac{pi u_0}{frac{pi u_0}{alpha}} cdot frac{1 + 4 alpha D t}{alpha^2} = alpha cdot frac{1 + 4 alpha D t}{alpha^2} = frac{1 + 4 alpha D t}{alpha} = frac{1}{alpha} + 4 D t. ]So, the mean squared displacement is ( langle r^2(t) rangle = frac{1}{alpha} + 4 D t ).Wait, but earlier I thought the variance of the initial Gaussian should be ( 1/(2 alpha) ), but according to this, ( langle r^2(0) rangle = 1/alpha ). Let me check that.In 2D, the variance of a Gaussian distribution ( e^{-a r^2} ) is ( sigma^2 = 1/(2a) ). So, if ( a = alpha ), then ( sigma^2 = 1/(2 alpha) ). But according to our calculation, ( langle r^2(0) rangle = 1/alpha ). That suggests that our calculation is giving twice the variance. Hmm, that's a problem.Wait, perhaps I made a mistake in the definition of ( u(r,t) ). Let me think.The initial condition is ( u(r,0) = u_0 e^{-alpha r^2} ). The total number is ( N(0) = int_0^infty u(r,0) cdot 2 pi r , dr = frac{pi u_0}{alpha} ).But in 2D, the variance of ( r^2 ) is ( langle r^2 rangle - langle r rangle^2 ). However, for a radially symmetric Gaussian, ( langle r rangle ) is not zero, but the variance is ( langle r^2 rangle - langle r rangle^2 ). Wait, but in our case, the distribution is ( u(r,t) ), which is the density. So, the expectation ( langle r^2 rangle ) is indeed ( int_0^infty r^2 u(r,t) cdot 2 pi r , dr / N(t) ).Wait, but for a 2D Gaussian distribution ( e^{-a r^2} ), the expectation ( langle r^2 rangle ) is ( 1/(2a) ). So, if our initial condition is ( u(r,0) = u_0 e^{-alpha r^2} ), then ( langle r^2(0) rangle = 1/(2 alpha) ). But according to our calculation, it's ( 1/alpha ). So, there must be a mistake.Wait, let's recompute ( langle r^2(0) rangle ) using our solution.At ( t = 0 ), ( u(r,0) = u_0 e^{-alpha r^2} ). So,[ langle r^2(0) rangle = frac{int_0^infty r^2 u(r,0) cdot 2 pi r , dr}{N(0)} = frac{int_0^infty r^3 u_0 e^{-alpha r^2} cdot 2 pi , dr}{frac{pi u_0}{alpha}}. ]Compute the numerator:Let ( y = r^2 ), ( dy = 2 r dr ), so ( r^3 dr = y cdot frac{dy}{2} ).Thus,[ int_0^infty r^3 u_0 e^{-alpha r^2} cdot 2 pi , dr = 2 pi u_0 cdot frac{1}{2} int_0^infty y e^{-alpha y} dy = pi u_0 cdot frac{1}{alpha^2}. ]Therefore,[ langle r^2(0) rangle = frac{pi u_0 / alpha^2}{pi u_0 / alpha} = frac{1}{alpha}. ]But according to the properties of a 2D Gaussian, it should be ( 1/(2 alpha) ). So, where is the discrepancy?Wait, perhaps the issue is that in 2D, the variance is ( sigma^2 = langle r^2 rangle - langle r rangle^2 ). But for a Gaussian centered at the origin, ( langle r rangle = 0 ), so ( sigma^2 = langle r^2 rangle ). But in our case, ( langle r^2 rangle = 1/alpha ), which is twice the expected variance.Wait, no, actually, in 2D, the variance in Cartesian coordinates would be ( sigma_x^2 = sigma_y^2 = 1/(2 alpha) ), so the variance in radial coordinates is ( sigma_r^2 = sigma_x^2 + sigma_y^2 = 1/alpha ). So, actually, ( langle r^2 rangle = sigma_r^2 = 1/alpha ). So, our calculation is correct. The variance in radial coordinates is indeed ( 1/alpha ), not ( 1/(2 alpha) ). I was confusing the variance in Cartesian coordinates with the radial variance.So, that resolves the confusion. Therefore, our calculation is correct, and ( langle r^2(t) rangle = frac{1}{alpha} + 4 D t ).So, the mean squared displacement depends linearly on time with a coefficient ( 4 D ), and it also has a constant term ( 1/alpha ) which represents the initial variance of the distribution.Therefore, the dependence is:- Linear on ( D ): The coefficient of ( t ) is ( 4 D ).- The constant term depends on ( alpha ) as ( 1/alpha ).- ( u_0 ) does not affect the mean squared displacement because it cancels out in the calculation.So, summarizing:The solution to the PDE is:[ u(r,t) = frac{u_0}{1 + 4 alpha D t} e^{- frac{alpha r^2}{1 + 4 alpha D t}}. ]And the mean squared displacement is:[ langle r^2(t) rangle = frac{1}{alpha} + 4 D t. ]This shows that the cultural artifacts spread out with a mean squared displacement that increases linearly with time, with a rate determined by the diffusion coefficient ( D ), and an initial contribution from the variance of the initial Gaussian distribution.Final Answer1. The solution to the PDE is ( boxed{u(r, t) = dfrac{u_0}{1 + 4 alpha D t} e^{- dfrac{alpha r^2}{1 + 4 alpha D t}}} ).2. The mean squared displacement is ( boxed{langle r^2(t) rangle = dfrac{1}{alpha} + 4 D t} ), which depends linearly on time with a coefficient proportional to ( D ) and includes a constant term dependent on ( alpha ).</think>

🧠:<think>Alright, so I have this problem about analyzing actors' performance trajectories using quadratic functions. I need to figure out when each actor reached their peak number of performances and then predict their cumulative performances over the next five years. Let me take this step by step.First, let's tackle part 1. Each actor's performance is modeled by a quadratic function ( P(t) = at^2 + bt + c ). I remember that quadratic functions graph as parabolas, and depending on the coefficient ( a ), they either open upwards (if ( a > 0 )) or downwards (if ( a < 0 )). The peak or trough of the parabola occurs at the vertex.For each actor, I need to find the vertex of their quadratic function because that will give me the year ( t ) when they reached their peak performances. The formula for the vertex (specifically the ( t )-coordinate) is ( t = -frac{b}{2a} ). Once I have that ( t ), I can plug it back into the original equation to find the peak performance value ( P(t) ).Let's start with Actor A: ( P_A(t) = 2t^2 + 3t + 5 ). Here, ( a = 2 ), ( b = 3 ). So, the vertex occurs at ( t = -frac{3}{2*2} = -frac{3}{4} ). Wait, that gives me a negative time, which doesn't make sense because ( t ) represents years since debut, so it can't be negative. Hmm, that must mean that the parabola opens upwards (since ( a = 2 > 0 )), so the vertex is the minimum point, not the maximum. Therefore, Actor A's performances are increasing over time because the parabola is opening upwards. So, does that mean Actor A doesn't have a peak? Or maybe the peak is at the last year we have data for? Wait, the data is over the past 10 years, so ( t ) ranges from 0 to 10. Since the vertex is at ( t = -0.75 ), which is before the actor's debut, the minimum performance is at ( t = 0 ), and performances increase from there. So, in this case, the peak performance would be at ( t = 10 ), the latest year we have data for. Let me verify that.Calculating ( P_A(10) ): ( 2*(10)^2 + 3*(10) + 5 = 2*100 + 30 + 5 = 200 + 30 + 5 = 235 ). So, yes, Actor A's peak is at ( t = 10 ) with 235 performances.Moving on to Actor B: ( P_B(t) = -t^2 + 6t + 2 ). Here, ( a = -1 ), so the parabola opens downward, meaning the vertex is the maximum point. So, the peak occurs at ( t = -frac{b}{2a} = -frac{6}{2*(-1)} = -frac{6}{-2} = 3 ). So, the peak is at ( t = 3 ). Let me calculate the peak performance: ( P_B(3) = -(3)^2 + 6*(3) + 2 = -9 + 18 + 2 = 11 ). So, Actor B peaked at year 3 with 11 performances.Now, Actor C: ( P_C(t) = 4t^2 - 2t + 1 ). Here, ( a = 4 ), which is positive, so the parabola opens upwards, meaning the vertex is a minimum. So, similar to Actor A, the peak would be at the latest year, ( t = 10 ). Let me compute ( P_C(10) ): ( 4*(10)^2 - 2*(10) + 1 = 4*100 - 20 + 1 = 400 - 20 + 1 = 381 ). So, Actor C's peak is at ( t = 10 ) with 381 performances.Wait a second, let me just make sure I didn't make a mistake with Actor A and C. Since their parabolas open upwards, their performance numbers are increasing over time, so their peaks are indeed at the highest ( t ) value we have, which is 10. So, that seems correct.So, summarizing part 1:- Actor A: Peak at ( t = 10 ), 235 performances.- Actor B: Peak at ( t = 3 ), 11 performances.- Actor C: Peak at ( t = 10 ), 381 performances.Okay, that seems solid.Now, moving on to part 2: predicting the cumulative number of performances from ( t = 11 ) to ( t = 15 ) for each actor. So, we need to calculate the sum of ( P(t) ) for each year from 11 to 15 inclusive.I can approach this by calculating ( P(11) ), ( P(12) ), ( P(13) ), ( P(14) ), ( P(15) ) for each actor and then adding them up.Let's start with Actor A: ( P_A(t) = 2t^2 + 3t + 5 ).Compute each year:- ( P_A(11) = 2*(11)^2 + 3*(11) + 5 = 2*121 + 33 + 5 = 242 + 33 + 5 = 280 )- ( P_A(12) = 2*(12)^2 + 3*(12) + 5 = 2*144 + 36 + 5 = 288 + 36 + 5 = 329 )- ( P_A(13) = 2*(13)^2 + 3*(13) + 5 = 2*169 + 39 + 5 = 338 + 39 + 5 = 382 )- ( P_A(14) = 2*(14)^2 + 3*(14) + 5 = 2*196 + 42 + 5 = 392 + 42 + 5 = 439 )- ( P_A(15) = 2*(15)^2 + 3*(15) + 5 = 2*225 + 45 + 5 = 450 + 45 + 5 = 499 + 5 = 500 ) Wait, 450 + 45 is 495, plus 5 is 500. Yes.Now, sum these up: 280 + 329 + 382 + 439 + 500.Let me add them step by step:280 + 329 = 609609 + 382 = 991991 + 439 = 14301430 + 500 = 1930So, Actor A's cumulative performances from t=11 to t=15 are 1930.Now, Actor B: ( P_B(t) = -t^2 + 6t + 2 ).Compute each year:- ( P_B(11) = -(11)^2 + 6*(11) + 2 = -121 + 66 + 2 = (-121 + 66) + 2 = (-55) + 2 = -53 ). Wait, negative performances? That doesn't make sense. Maybe I made a mistake.Wait, let me recalculate ( P_B(11) ):( P_B(11) = -(11)^2 + 6*11 + 2 = -121 + 66 + 2 ). So, -121 + 66 is -55, plus 2 is -53. Hmm, negative performances? That can't be right. Maybe the quadratic model isn't valid beyond a certain point? Or perhaps the model is only accurate for the first 10 years, and beyond that, it's not reliable? The problem says we have data for the past 10 years, so t=0 to t=10. So, t=11 to t=15 are predictions beyond the data. So, perhaps the model can still be used, but we might get negative values, which would imply that the actor has stopped performing or maybe retired.But in the context of the problem, we're supposed to calculate the cumulative number of performances, so even if it's negative, we have to include it? Or maybe we should take it as zero? The problem doesn't specify, so perhaps we just proceed with the calculation as is.But let me check if I did the calculation correctly:( P_B(11) = -(11)^2 + 6*11 + 2 = -121 + 66 + 2 = (-121 + 66) = -55; -55 + 2 = -53 ). Yes, that's correct.Similarly, let's compute the rest:- ( P_B(12) = -(12)^2 + 6*12 + 2 = -144 + 72 + 2 = (-144 + 72) = -72; -72 + 2 = -70 )- ( P_B(13) = -(13)^2 + 6*13 + 2 = -169 + 78 + 2 = (-169 + 78) = -91; -91 + 2 = -89 )- ( P_B(14) = -(14)^2 + 6*14 + 2 = -196 + 84 + 2 = (-196 + 84) = -112; -112 + 2 = -110 )- ( P_B(15) = -(15)^2 + 6*15 + 2 = -225 + 90 + 2 = (-225 + 90) = -135; -135 + 2 = -133 )So, all these are negative. That seems odd, but perhaps the model is correct. So, the cumulative performances would be the sum of these negative numbers. But that would result in a negative cumulative, which doesn't make sense in real life. Maybe the model is only valid up to a certain point, and beyond that, the actor isn't performing anymore, so performances would be zero. But the problem doesn't specify, so I think we have to proceed with the model's predictions, even if they result in negative numbers.So, let's sum them:-53 + (-70) + (-89) + (-110) + (-133)Let me add them step by step:-53 -70 = -123-123 -89 = -212-212 -110 = -322-322 -133 = -455So, cumulative performances for Actor B from t=11 to t=15 is -455. But since negative performances don't make sense, perhaps we should interpret this as zero? Or maybe the model isn't suitable beyond t=10. Hmm, the problem says "predict the cumulative number of performances", so maybe we should just go with the model's output, even if it's negative. Alternatively, maybe the model is only valid up to t=10, and beyond that, we can assume zero performances. But the problem doesn't specify, so I think we have to use the quadratic model as given.So, I'll proceed with -455, but note that in reality, this wouldn't make sense.Now, moving on to Actor C: ( P_C(t) = 4t^2 - 2t + 1 ).Compute each year:- ( P_C(11) = 4*(11)^2 - 2*(11) + 1 = 4*121 - 22 + 1 = 484 - 22 + 1 = 463 )- ( P_C(12) = 4*(12)^2 - 2*(12) + 1 = 4*144 - 24 + 1 = 576 - 24 + 1 = 553 )- ( P_C(13) = 4*(13)^2 - 2*(13) + 1 = 4*169 - 26 + 1 = 676 - 26 + 1 = 651 )- ( P_C(14) = 4*(14)^2 - 2*(14) + 1 = 4*196 - 28 + 1 = 784 - 28 + 1 = 757 )- ( P_C(15) = 4*(15)^2 - 2*(15) + 1 = 4*225 - 30 + 1 = 900 - 30 + 1 = 871 )Now, sum these up: 463 + 553 + 651 + 757 + 871.Let me add them step by step:463 + 553 = 10161016 + 651 = 16671667 + 757 = 24242424 + 871 = 3295So, Actor C's cumulative performances from t=11 to t=15 are 3295.Wait, let me double-check the calculations for each year to make sure I didn't make any arithmetic errors.For Actor C:- ( P_C(11) = 4*121 = 484; 484 -22 = 462; 462 +1 = 463. Correct.- ( P_C(12) = 4*144 = 576; 576 -24 = 552; 552 +1 = 553. Correct.- ( P_C(13) = 4*169 = 676; 676 -26 = 650; 650 +1 = 651. Correct.- ( P_C(14) = 4*196 = 784; 784 -28 = 756; 756 +1 = 757. Correct.- ( P_C(15) = 4*225 = 900; 900 -30 = 870; 870 +1 = 871. Correct.Sum: 463 + 553 = 1016; 1016 + 651 = 1667; 1667 + 757 = 2424; 2424 + 871 = 3295. Correct.So, that seems right.Now, just to recap:- Actor A: 1930- Actor B: -455 (but in reality, probably zero)- Actor C: 3295But since the problem asks for the cumulative number, and the model gives negative for Actor B, I think we have to report it as -455. Alternatively, maybe the model is only valid up to t=10, so beyond that, performances are zero. But the problem doesn't specify, so I think we have to go with the model's output.Alternatively, perhaps I made a mistake in interpreting the vertex for Actor B. Wait, Actor B's peak was at t=3, so after that, their performances decrease. So, beyond t=10, it's possible that the model predicts negative performances, which might indicate that the actor has stopped performing. So, perhaps for the purpose of cumulative performances, we should consider only up to the point where performances are positive, and beyond that, it's zero. But the problem says to calculate from t=11 to t=15, so we have to include those years, even if the model gives negative numbers.Alternatively, maybe the model is only valid for t=0 to t=10, and beyond that, we can't predict. But the problem says "predict the cumulative number of performances for each actor over the next 5 years", so t=11 to t=15. So, we have to use the model as given.So, I think the answer is as calculated: Actor A: 1930, Actor B: -455, Actor C: 3295.But just to make sure, let me check if there's another way to compute the cumulative sum. Instead of calculating each year individually, maybe we can use the formula for the sum of a quadratic function over an interval. The sum from t=11 to t=15 of ( P(t) ) can be calculated using the formula for the sum of a quadratic sequence.The general formula for the sum of ( P(t) = at^2 + bt + c ) from t = m to t = n is:( sum_{t=m}^{n} P(t) = a sum_{t=m}^{n} t^2 + b sum_{t=m}^{n} t + c sum_{t=m}^{n} 1 )We can use the formulas for the sum of squares, sum of integers, and sum of constants.The sum of squares from m to n is ( frac{n(n+1)(2n+1)}{6} - frac{(m-1)m(2m-1)}{6} )The sum of integers from m to n is ( frac{n(n+1)}{2} - frac{(m-1)m}{2} )The sum of constants from m to n is ( c*(n - m + 1) )So, let's try this method for Actor A to see if we get the same result.Actor A: ( P_A(t) = 2t^2 + 3t + 5 )Sum from t=11 to t=15:Sum = 2*(sum of squares from 11 to 15) + 3*(sum of integers from 11 to 15) + 5*(number of terms)Number of terms = 15 - 11 + 1 = 5Sum of squares from 11 to 15:Using the formula:( frac{15*16*31}{6} - frac{10*11*19}{6} )Compute each part:First part: ( frac{15*16*31}{6} )15*16 = 240; 240*31 = 7440; 7440/6 = 1240Second part: ( frac{10*11*19}{6} )10*11 = 110; 110*19 = 2090; 2090/6 ≈ 348.333...So, sum of squares = 1240 - 348.333... ≈ 891.666...But let me compute it more accurately:15*16*31 = 15*16=240; 240*31=744010*11*19=10*11=110; 110*19=2090So, sum of squares = (7440 - 2090)/6 = 5350/6 ≈ 891.666...But actually, the formula is:Sum of squares from 1 to n is ( frac{n(n+1)(2n+1)}{6} ). So, sum from m to n is sum from 1 to n minus sum from 1 to m-1.So, sum from 11 to 15 is sum from 1 to 15 minus sum from 1 to 10.Sum from 1 to 15: ( frac{15*16*31}{6} = frac{7440}{6} = 1240 )Sum from 1 to 10: ( frac{10*11*21}{6} = frac{2310}{6} = 385 )So, sum of squares from 11 to 15: 1240 - 385 = 855Wait, that's different from my previous calculation. I think I made a mistake earlier. Let me recalculate:Sum from 1 to 15: 15*16*31/6 = (15*16)=240; 240*31=7440; 7440/6=1240Sum from 1 to 10: 10*11*21/6 = (10*11)=110; 110*21=2310; 2310/6=385So, sum from 11 to 15: 1240 - 385 = 855Similarly, sum of integers from 11 to 15:Sum from 1 to 15: ( frac{15*16}{2} = 120 )Sum from 1 to 10: ( frac{10*11}{2} = 55 )So, sum from 11 to 15: 120 - 55 = 65Sum of constants: 5*5 = 25Now, putting it all together:Sum = 2*(855) + 3*(65) + 5*(25) = 1710 + 195 + 125 = 1710 + 195 = 1905; 1905 + 125 = 2030Wait, but earlier when I calculated individually, I got 1930. There's a discrepancy here. Hmm, that means I must have made a mistake in one of the methods.Wait, let's recalculate the individual method:Actor A:- t=11: 280- t=12: 329- t=13: 382- t=14: 439- t=15: 500Sum: 280 + 329 = 609; 609 + 382 = 991; 991 + 439 = 1430; 1430 + 500 = 1930But using the formula, I got 2030. So, which one is correct?Wait, let's check the sum of squares from 11 to 15:Compute each square:11²=12112²=14413²=16914²=19615²=225Sum: 121 + 144 = 265; 265 + 169 = 434; 434 + 196 = 630; 630 + 225 = 855. So, that's correct.Sum of integers from 11 to 15:11 + 12 + 13 + 14 + 15 = 11+15=26; 12+14=26; 13=13; total=26+26+13=65. Correct.So, sum = 2*855 + 3*65 + 5*5 = 1710 + 195 + 25 = 1710 + 195 = 1905; 1905 +25=1930.Wait, that's different from my previous calculation. Wait, in the formula, I think I added 5*5=25, but in the individual calculation, the constants were 5 each year, so 5*5=25. So, 2*855=1710, 3*65=195, 5*5=25. 1710+195=1905; 1905+25=1930. So, that matches the individual calculation. Earlier, I must have miscalculated when I thought the formula gave 2030. I think I added 5*5 as 25, but maybe I miscounted. So, the correct sum is 1930, which matches the individual calculation.So, the formula method is correct. I must have made a mistake earlier when I thought the formula gave 2030. So, that's settled.Similarly, let's try the formula method for Actor C to verify.Actor C: ( P_C(t) = 4t^2 - 2t + 1 )Sum from t=11 to t=15:Sum = 4*(sum of squares from 11 to 15) - 2*(sum of integers from 11 to 15) + 1*(number of terms)We already know sum of squares from 11 to 15 is 855, sum of integers is 65, number of terms is 5.So, Sum = 4*855 - 2*65 + 1*5 = 3420 - 130 + 5 = 3420 - 130 = 3290; 3290 +5=3295. Which matches the individual calculation. So, that's correct.Now, for Actor B, let's try the formula method.Actor B: ( P_B(t) = -t^2 + 6t + 2 )Sum from t=11 to t=15:Sum = -1*(sum of squares from 11 to 15) + 6*(sum of integers from 11 to 15) + 2*(number of terms)Sum of squares: 855Sum of integers: 65Number of terms:5So, Sum = -855 + 6*65 + 2*5 = -855 + 390 + 10 = (-855 + 390) = -465; -465 +10 = -455. Which matches the individual calculation. So, that's correct.Therefore, the cumulative performances are:- Actor A: 1930- Actor B: -455- Actor C: 3295But as I thought earlier, negative performances don't make sense, so maybe we should consider only the positive contributions. However, since the problem doesn't specify, I think we have to report the values as calculated.So, to summarize:1. Peak performances:- Actor A: t=10, 235- Actor B: t=3, 11- Actor C: t=10, 3812. Cumulative performances from t=11 to t=15:- Actor A: 1930- Actor B: -455- Actor C: 3295But since negative cumulative doesn't make sense, perhaps for Actor B, we should consider zero. However, the problem didn't specify, so I think we have to go with the model's output.Alternatively, maybe the model is only valid up to t=10, and beyond that, the actor's performances are zero. So, for t=11 to t=15, Actor B's performances would be zero, resulting in a cumulative of zero. But the problem says "predict the cumulative number of performances", so it's ambiguous. Since the model gives negative, but in reality, it's zero, perhaps we should adjust it.But without specific instructions, I think we have to use the model's predictions. So, I'll proceed with the calculated values.</think>

🧠:<think>Alright, so I need to figure out the quadratic model for Mike Gartner's goals per season. The model is given by ( G_n = an^2 + bn + c ). I have three data points: in the 5th season, he scored 48 goals; in the 10th season, 41 goals; and in the 15th season, 35 goals. First, I think I can set up a system of equations using these points. Each season corresponds to a value of n, so plugging those into the equation should give me three equations with three unknowns: a, b, and c.Let me write down the equations:For the 5th season (n=5):( a(5)^2 + b(5) + c = 48 )Which simplifies to:( 25a + 5b + c = 48 ) ...(1)For the 10th season (n=10):( a(10)^2 + b(10) + c = 41 )Simplifies to:( 100a + 10b + c = 41 ) ...(2)For the 15th season (n=15):( a(15)^2 + b(15) + c = 35 )Which becomes:( 225a + 15b + c = 35 ) ...(3)Now I have three equations:1. ( 25a + 5b + c = 48 )2. ( 100a + 10b + c = 41 )3. ( 225a + 15b + c = 35 )I need to solve this system for a, b, and c. I can use elimination to solve for the variables.First, let's subtract equation (1) from equation (2):Equation (2) - Equation (1):( (100a - 25a) + (10b - 5b) + (c - c) = 41 - 48 )Simplifies to:( 75a + 5b = -7 ) ...(4)Similarly, subtract equation (2) from equation (3):Equation (3) - Equation (2):( (225a - 100a) + (15b - 10b) + (c - c) = 35 - 41 )Simplifies to:( 125a + 5b = -6 ) ...(5)Now, I have two equations:4. ( 75a + 5b = -7 )5. ( 125a + 5b = -6 )Subtract equation (4) from equation (5):( (125a - 75a) + (5b - 5b) = -6 - (-7) )Simplifies to:( 50a = 1 )So, ( a = 1/50 ) or 0.02.Now, plug a back into equation (4) to find b:( 75*(1/50) + 5b = -7 )Calculate 75*(1/50): 75 divided by 50 is 1.5, so:1.5 + 5b = -7Subtract 1.5 from both sides:5b = -8.5Divide by 5:b = -8.5 / 5 = -1.7So, b is -1.7.Now, plug a and b into equation (1) to find c:25a + 5b + c = 4825*(0.02) + 5*(-1.7) + c = 48Calculate each term:25*0.02 = 0.55*(-1.7) = -8.5So, 0.5 - 8.5 + c = 48Combine like terms:-8 + c = 48Add 8 to both sides:c = 56So, c is 56.Let me double-check these values with another equation to ensure consistency. Let's use equation (2):100a + 10b + c = 41100*(0.02) + 10*(-1.7) + 56 = ?100*0.02 = 210*(-1.7) = -17So, 2 -17 + 56 = 41Which is correct.Similarly, equation (3):225a + 15b + c = 35225*0.02 = 4.515*(-1.7) = -25.54.5 -25.5 +56 = 35Which is also correct.So, the constants are:a = 0.02b = -1.7c = 56So, the quadratic model is ( G_n = 0.02n^2 - 1.7n + 56 ).Now, moving on to part 2: calculating the total number of goals over a 20-season career.To find the total goals, I need to sum ( G_n ) from n=1 to n=20.Given that ( G_n = 0.02n^2 - 1.7n + 56 ), the total goals T is:( T = sum_{n=1}^{20} (0.02n^2 - 1.7n + 56) )I can split this sum into three separate sums:( T = 0.02sum_{n=1}^{20}n^2 - 1.7sum_{n=1}^{20}n + 56sum_{n=1}^{20}1 )I remember the formulas for these sums:1. Sum of squares: ( sum_{n=1}^{k}n^2 = frac{k(k+1)(2k+1)}{6} )2. Sum of first k natural numbers: ( sum_{n=1}^{k}n = frac{k(k+1)}{2} )3. Sum of 1 k times: ( sum_{n=1}^{k}1 = k )So, plugging in k=20:First, calculate each sum:1. ( sum_{n=1}^{20}n^2 = frac{20*21*41}{6} )Let me compute that:20*21 = 420420*41 = 17220Divide by 6: 17220 /6 = 28702. ( sum_{n=1}^{20}n = frac{20*21}{2} = 210 )3. ( sum_{n=1}^{20}1 = 20 )Now, plug these into T:( T = 0.02*2870 - 1.7*210 + 56*20 )Compute each term:0.02*2870 = 57.41.7*210 = 35756*20 = 1120So, T = 57.4 - 357 + 1120Compute step by step:57.4 - 357 = -299.6-299.6 + 1120 = 820.4So, the total goals are 820.4. Since goals can't be a fraction, but since we're dealing with a model, it's acceptable for the calculation. However, in reality, goals are whole numbers, but since we're using a quadratic model, it's okay.But the question says to calculate the total number of goals, so 820.4 is the result. But maybe we should round it? Or perhaps the model is precise. Hmm.But moving on, the referee's insights say that 5% of his calculated total goals were disallowed due to offside calls. So, we need to adjust the total by subtracting 5%.First, calculate 5% of 820.4:5% of 820.4 is 0.05*820.4 = 41.02So, the adjusted total goals would be 820.4 - 41.02 = 779.38Again, since goals are whole numbers, we might need to round this. Depending on the context, we could round to the nearest whole number, which would be 779 goals.But let me double-check my calculations to make sure I didn't make any errors.First, the sum of squares for n=1 to 20:20*21*41 /6 = (20/6)*21*41 = (10/3)*21*41 = 70*41 = 2870. Correct.Sum of n=1 to 20: 20*21/2 = 210. Correct.Sum of 1s: 20. Correct.Then, T = 0.02*2870 = 57.4; 1.7*210=357; 56*20=1120. Correct.So, 57.4 - 357 = -299.6; -299.6 + 1120 = 820.4. Correct.5% of 820.4 is 41.02; 820.4 - 41.02 = 779.38. Rounded to 779.Alternatively, if we keep it as 779.38, but since goals are whole numbers, 779 is appropriate.Alternatively, perhaps we should consider whether the model's total is 820.4, which is approximately 820 goals, and then 5% of 820 is 41, so 820 - 41 = 779. So, same result.Therefore, the adjusted total is 779 goals.Wait, but let me think again: the model gives 820.4, which is approximately 820, but if we use the exact value, 820.4, then 5% is 41.02, so 820.4 - 41.02 = 779.38, which is approximately 779.38. Since we can't have a fraction of a goal, we should round to the nearest whole number, which is 779.Alternatively, if we consider that the model's total is 820.4, which is 820 goals when rounded down, but 820.4 is closer to 820 than 821, so 820. Then, 5% of 820 is 41, so 820 - 41 = 779.Either way, the result is 779.So, the total goals over 20 seasons would be approximately 779 after adjusting for the disallowed goals.Wait, but let me think again: the model's total is 820.4, which is 820.4 goals. So, 5% of that is 41.02, so subtracting that gives 779.38, which is approximately 779.38, so 779 goals when rounded down, or 779.4, which is 779 when rounded to the nearest whole number.Alternatively, if we keep it as 779.38, but since we can't have a fraction, 779 is the correct number.Therefore, the adjusted total is 779 goals.So, summarizing:1. The constants are a=0.02, b=-1.7, c=56.2. The total goals over 20 seasons are approximately 820.4, which after adjusting for 5% disallowed goals, becomes approximately 779 goals.But let me just verify the quadratic model again with the given data points to make sure.For n=5: G_5 = 0.02*(25) -1.7*5 +56 = 0.5 -8.5 +56 = 48. Correct.n=10: 0.02*100 -1.7*10 +56 = 2 -17 +56 = 41. Correct.n=15: 0.02*225 -1.7*15 +56 = 4.5 -25.5 +56 = 35. Correct.So, the model is correct.Therefore, the calculations for the total goals are accurate.Final Answer1. The constants are ( a = boxed{0.02} ), ( b = boxed{-1.7} ), and ( c = boxed{56} ).2. The adjusted total number of goals is ( boxed{779} ).</think>