🧠:<think>Alright, so I've got these two questions about an interactive art installation with a spherical dome and an AI system that generates light patterns based on visitors' movements. Let me try to unpack each part step by step.Starting with the first question: We need to determine the conditions under which the intensity function ( I(theta, phi) ) is maximized for a given movement trajectory ( mathbf{r}(t) ). The intensity is given by the integral ( I(theta, phi) = int_{0}^{T} e^{-|mathbf{r}(t) - mathbf{s}(theta, phi)|^2} , dt ). Hmm, okay. So, ( I(theta, phi) ) is integrating over time the exponential of the negative squared distance between the visitor's position ( mathbf{r}(t) ) and a point ( mathbf{s}(theta, phi) ) on the dome's surface. The exponential function here is a Gaussian, which peaks when the distance is zero and decays as the distance increases. So, the intensity at a point on the dome depends on how often and how close the visitor is to that point over time.To maximize ( I(theta, phi) ), we need to maximize the integral. Since the exponential is always positive, the integral will be larger when the integrand is larger over a significant portion of the interval [0, T]. The integrand is largest when ( mathbf{r}(t) ) is close to ( mathbf{s}(theta, phi) ). So, intuitively, the intensity is maximized at points on the dome where the visitor spends the most time being close to them.But let's think more formally. The function inside the integral is ( e^{-|mathbf{r}(t) - mathbf{s}|^2} ). To maximize the integral, we need to maximize the sum of these exponentials over time. Since the exponential is a decreasing function of the squared distance, the maximum occurs when the distance ( |mathbf{r}(t) - mathbf{s}| ) is minimized as much as possible over time.Therefore, the intensity ( I(theta, phi) ) is maximized when the point ( mathbf{s}(theta, phi) ) is as close as possible to the trajectory ( mathbf{r}(t) ) over the interval [0, T]. In other words, the maximum intensity occurs at the point(s) on the sphere that are closest to the path traced by the visitor.Wait, but is it just the closest point? Or is it the point where the visitor spends the most time nearby? Because even if a point is not the closest, if the visitor lingers near it for a longer duration, the integral could be larger.So, perhaps the maximum occurs where the visitor spends the most time in the vicinity of ( mathbf{s}(theta, phi) ). That is, if the visitor moves slowly near a certain area, that area would have a higher intensity because the integrand contributes more over a longer period.Alternatively, if the visitor moves quickly past a point, even if they get very close, the contribution to the integral might be less because the time spent near that point is shorter.So, to formalize this, the intensity ( I(theta, phi) ) is the integral over time of the exponential of the negative squared distance. Therefore, the maximum intensity occurs where the visitor spends the most time being close to that point on the sphere.But how do we express this mathematically? Maybe we can think about the trajectory ( mathbf{r}(t) ) and find points ( mathbf{s} ) such that the integral is maximized. This might involve calculus of variations or optimization techniques.Alternatively, we can consider that for each point ( mathbf{s} ), the integral is a measure of how much time the visitor spends near ( mathbf{s} ). So, the maximum intensity occurs at the point where the visitor spends the most time in its vicinity.But perhaps more precisely, the intensity is a kind of convolution between the visitor's trajectory and a Gaussian kernel on the sphere. So, the maximum intensity would be where the visitor's path has the highest density when convolved with this kernel.Wait, maybe another approach: Since the exponential function is being integrated over time, the intensity is essentially a weighted sum of the visitor's proximity to each point on the sphere. So, the point that is most frequently and closely approached by the visitor will have the highest intensity.Therefore, the condition for ( I(theta, phi) ) to be maximized is that the point ( mathbf{s}(theta, phi) ) is such that the trajectory ( mathbf{r}(t) ) spends the most time in the vicinity of ( mathbf{s} ). In other words, the point where the trajectory is densest or where the visitor lingers the longest.Alternatively, if we think about the function ( e^{-|mathbf{r}(t) - mathbf{s}|^2} ), it's maximized when ( mathbf{r}(t) = mathbf{s} ). So, if the visitor spends time at a particular point, that point will have a higher intensity. But if the visitor moves continuously, the intensity will be higher where the visitor's path is closest to that point and spends more time in that vicinity.So, perhaps the maximum occurs where the distance ( |mathbf{r}(t) - mathbf{s}| ) is minimized over time, but also considering the duration. It's a combination of proximity and duration.Wait, maybe we can think of it as the point ( mathbf{s} ) that minimizes the integral ( int_{0}^{T} |mathbf{r}(t) - mathbf{s}|^2 dt ). Because the exponential of the negative of that would be maximized when the integral is minimized.But actually, the integral in the intensity is ( int_{0}^{T} e^{-|mathbf{r}(t) - mathbf{s}|^2} dt ). So, it's not the integral of the squared distance, but the integral of the exponential of the negative squared distance.So, to maximize ( I(theta, phi) ), we need to maximize ( int_{0}^{T} e^{-|mathbf{r}(t) - mathbf{s}|^2} dt ). This integral is larger when the visitor is closer to ( mathbf{s} ) for a longer time.Therefore, the maximum occurs at the point ( mathbf{s} ) where the trajectory ( mathbf{r}(t) ) is closest to ( mathbf{s} ) for the longest duration.Alternatively, if the visitor spends a lot of time near ( mathbf{s} ), even if not the closest point overall, the integral could be larger.So, perhaps the maximum intensity occurs at the point ( mathbf{s} ) that is the most frequently visited or where the visitor lingers the longest.But how do we express this condition mathematically? Maybe we can take the derivative of ( I(theta, phi) ) with respect to ( theta ) and ( phi ) and set it to zero to find the maximum.But ( I(theta, phi) ) is a function on the sphere, so we can consider it as a function on a compact manifold, and by continuity, it must attain a maximum.To find where it's maximized, we can consider the points where the gradient of ( I ) is zero. That is, the points where the directional derivatives in all directions are zero.But computing the gradient might be complicated because ( I ) is defined as an integral over time of a function involving the distance between ( mathbf{r}(t) ) and ( mathbf{s}(theta, phi) ).Alternatively, perhaps we can think about the function ( I ) as a kind of heat kernel or something similar, where the intensity diffuses from the visitor's trajectory onto the sphere.But maybe another approach: For each point ( mathbf{s} ), the intensity is the integral over time of the Gaussian centered at ( mathbf{r}(t) ). So, it's like accumulating a Gaussian bump at each position of the visitor over time.Therefore, the intensity function ( I ) is the sum (integral) of all these Gaussian bumps. The maximum of this sum would occur where the visitor's trajectory has the highest density, i.e., where the visitor spends the most time or where the trajectory is most concentrated.So, the condition for ( I(theta, phi) ) to be maximized is that ( mathbf{s}(theta, phi) ) is the point where the trajectory ( mathbf{r}(t) ) is most concentrated or where the visitor spends the most time in the vicinity.Alternatively, if we consider the function ( I ) as a convolution of the trajectory with a Gaussian kernel, the maximum occurs where the convolution is highest, which is typically where the trajectory is densest.But perhaps more formally, we can think about the intensity as a measure of how much time the visitor spends near each point on the sphere. So, the point with the highest intensity is where the visitor spends the most time being close to it.Therefore, the condition is that ( mathbf{s}(theta, phi) ) is the point on the sphere that is closest to the trajectory ( mathbf{r}(t) ) for the longest duration, or where the visitor's trajectory accumulates the most time in its vicinity.Alternatively, if we think about the integral ( I(theta, phi) ), it's essentially the sum over time of the proximity to ( mathbf{s} ). So, the maximum occurs where this sum is largest, which is where the visitor is closest for the longest time.But perhaps another way to look at it is to consider the function inside the integral, ( e^{-|mathbf{r}(t) - mathbf{s}|^2} ). For each ( t ), this is a function peaked at ( mathbf{r}(t) ). So, integrating over ( t ) from 0 to T, the intensity at ( mathbf{s} ) is the sum of these peaks. Therefore, the intensity is highest where these peaks overlap the most, i.e., where the visitor's trajectory is most concentrated.So, in summary, the intensity ( I(theta, phi) ) is maximized at the point(s) on the sphere where the visitor's trajectory is most concentrated, either because the visitor spends a lot of time near that point or because the visitor passes by it many times, each time contributing to the integral.Now, moving on to the second question: The artist wants to optimize the AI algorithm such that the total light energy over the surface of the dome is constant over time, despite fluctuations in visitor movement. The total energy is defined as ( E(t) = int_{0}^{2pi} int_{0}^{pi} I(theta, phi, t) sin theta , dtheta , dphi ), and we need to formulate the necessary condition for ( I(theta, phi, t) ) to satisfy ( E(t) = E_0 ) for all time ( t ).So, ( E(t) ) is the integral of ( I ) over the sphere, weighted by ( sin theta ) (which is the standard measure for spherical coordinates). The artist wants this integral to be constant, regardless of how the visitors move, which affects ( I ).Given that ( I(theta, phi, t) = int_{0}^{T} e^{-|mathbf{r}(t') - mathbf{s}(theta, phi)|^2} dt' ), but wait, actually, in the first part, ( I ) was defined as an integral over time from 0 to T, but in the second part, ( I ) is now a function of ( t ), so perhaps the definition has changed slightly.Wait, in the first part, ( I(theta, phi) ) was an integral over time from 0 to T, but in the second part, ( I(theta, phi, t) ) is defined, and ( E(t) ) is the integral over the sphere at time ( t ). So, perhaps ( I(theta, phi, t) ) is now a function that depends on time ( t ), and ( E(t) ) is the integral over the sphere at each time ( t ).Wait, but in the first part, ( I ) was an integral over time, but in the second part, it's a function of ( t ). Maybe the artist wants to adjust the algorithm so that for each time ( t ), the integral over the sphere is constant.So, perhaps the original definition was ( I(theta, phi) = int_{0}^{T} e^{-|mathbf{r}(t) - mathbf{s}|^2} dt ), but now, to make the total energy constant over time, they need to adjust ( I ) such that for each ( t ), the integral over the sphere is ( E_0 ).But wait, the problem says "the total light energy over the surface of the dome is constant over time, despite fluctuations in visitor movement." So, ( E(t) = int_{sphere} I(theta, phi, t) dA = E_0 ) for all ( t ).Given that ( I(theta, phi, t) ) is defined as ( int_{0}^{T} e^{-|mathbf{r}(t') - mathbf{s}|^2} dt' ), but now it's a function of ( t ). Wait, no, in the first part, ( I ) was a function of ( theta ) and ( phi ), integrating over time. But in the second part, ( I ) is a function of ( theta ), ( phi ), and ( t ), and ( E(t) ) is the integral over the sphere at each ( t ).Wait, perhaps the artist wants to adjust the algorithm so that for each time ( t ), the integral of ( I(theta, phi, t) ) over the sphere is constant. But in the original definition, ( I ) was an integral over time, so perhaps the artist is now considering a different formulation where ( I ) is a function that depends on time, and they want the integral over the sphere at each time to be constant.Alternatively, maybe the artist wants to modify the algorithm so that the total energy ( E(t) ) remains constant, even as the visitor's movement changes over time. So, perhaps they need to scale ( I(theta, phi, t) ) by some factor that depends on time to ensure that the integral remains ( E_0 ).So, if we denote the original intensity as ( I_0(theta, phi, t) = int_{0}^{T} e^{-|mathbf{r}(t') - mathbf{s}|^2} dt' ), then the total energy is ( E(t) = int_{sphere} I_0(theta, phi, t) dA ). To make this constant, we can scale ( I_0 ) by a factor ( alpha(t) ) such that ( E(t) = int_{sphere} alpha(t) I_0(theta, phi, t) dA = alpha(t) int_{sphere} I_0(theta, phi, t) dA = alpha(t) E_0' = E_0 ). Therefore, ( alpha(t) = E_0 / E_0' ), but this would require knowing ( E_0' ) at each time, which depends on the visitor's movement.But perhaps a better approach is to adjust the intensity function ( I ) such that its integral over the sphere is always ( E_0 ). So, we can write ( int_{sphere} I(theta, phi, t) sin theta dtheta dphi = E_0 ) for all ( t ).Given that ( I(theta, phi, t) ) is defined as ( int_{0}^{T} e^{-|mathbf{r}(t') - mathbf{s}|^2} dt' ), but now it's a function of ( t ), perhaps the artist wants to adjust the parameters of the algorithm so that this integral remains constant.Alternatively, maybe the artist needs to normalize ( I ) such that its integral is always ( E_0 ). So, if the original ( I ) has an integral ( E(t) ), then the normalized ( I ) would be ( I_{text{norm}} = (E_0 / E(t)) I ). But this would require knowing ( E(t) ) at each time, which depends on the visitor's movement.But the problem says "formulate the necessary condition for ( I(theta, phi, t) ) to satisfy the constraint ( E(t) = E_0 ) for all time ( t )." So, we need to find a condition on ( I ) such that its integral over the sphere is always ( E_0 ).One way to ensure this is to have ( I(theta, phi, t) ) scaled by a factor that depends on time, such that the integral is always ( E_0 ). So, if we let ( I(theta, phi, t) = k(t) I_0(theta, phi, t) ), where ( I_0 ) is the original intensity function, then ( E(t) = k(t) int_{sphere} I_0 dA = k(t) E_0' = E_0 ). Therefore, ( k(t) = E_0 / E_0' ), where ( E_0' = int_{sphere} I_0 dA ).But this requires knowing ( E_0' ) at each time, which depends on the visitor's movement. Alternatively, perhaps the artist can adjust the parameters of the algorithm, such as the exponent in the Gaussian, to control the total energy.Wait, the intensity function is ( I(theta, phi) = int_{0}^{T} e^{-|mathbf{r}(t) - mathbf{s}|^2} dt ). If we change the exponent, say ( e^{-k |mathbf{r}(t) - mathbf{s}|^2} ), then the total energy would change. So, perhaps by adjusting ( k ), we can control the total energy.But the problem is about keeping the total energy constant regardless of visitor movement. So, perhaps the artist needs to adjust ( k ) dynamically based on the current total energy.Alternatively, maybe the artist can use a different kernel function that automatically normalizes the total energy. For example, if the kernel is a probability density function, then the integral over the sphere would be 1, and scaling it by ( E_0 ) would make the total energy constant.But in the given intensity function, the kernel is ( e^{-|mathbf{r}(t) - mathbf{s}|^2} ), which is not necessarily a probability density. So, perhaps the artist needs to normalize this kernel such that its integral over the sphere is constant.Wait, but the intensity is the integral over time of the kernel. So, perhaps the artist can adjust the kernel's width or amplitude so that the total energy remains constant.Alternatively, perhaps the artist can adjust the exponent's coefficient. Let me think.Suppose the intensity is ( I(theta, phi) = int_{0}^{T} e^{-k(t) |mathbf{r}(t) - mathbf{s}|^2} dt ). Then, the total energy ( E(t) = int_{sphere} I dA = int_{sphere} int_{0}^{T} e^{-k(t) |mathbf{r}(t) - mathbf{s}|^2} dt dA ).If we want ( E(t) = E_0 ) for all ( t ), we need to choose ( k(t) ) such that this holds. However, this seems complicated because ( k(t) ) would have to adjust dynamically based on the visitor's movement.Alternatively, perhaps the artist can use a different form of the intensity function that inherently has a constant total energy. For example, if the intensity is a probability density function, then its integral over the sphere would be 1, and scaling it by ( E_0 ) would make the total energy constant.But in the given problem, the intensity is defined as an integral over time of a Gaussian. So, perhaps the artist needs to adjust the Gaussian's width or the integration time to ensure the total energy remains constant.Wait, another approach: If we consider that the total energy ( E(t) ) is the integral over the sphere of ( I(theta, phi, t) ), and we want this to be constant, then we can write ( frac{dE}{dt} = 0 ). So, the time derivative of ( E(t) ) must be zero.But ( E(t) = int_{sphere} I(theta, phi, t) dA ), so ( frac{dE}{dt} = int_{sphere} frac{partial I}{partial t} dA = 0 ).Given that ( I(theta, phi, t) = int_{0}^{T} e^{-|mathbf{r}(t') - mathbf{s}|^2} dt' ), but now it's a function of ( t ), perhaps ( T ) is a function of ( t ), or maybe ( T ) is fixed, and ( t ) is a parameter.Wait, I'm getting confused. Let me clarify.In the first part, ( I(theta, phi) ) was an integral over time from 0 to T, so it's a function of ( theta ) and ( phi ), not of time. But in the second part, ( I(theta, phi, t) ) is defined, and ( E(t) ) is the integral over the sphere at each time ( t ). So, perhaps the artist wants to redefine ( I ) such that it's a function of ( t ), and the integral over the sphere at each ( t ) is constant.Alternatively, maybe the artist wants to adjust the algorithm so that for each time ( t ), the intensity function ( I(theta, phi, t) ) is scaled such that its integral is ( E_0 ).So, if we denote ( I_0(theta, phi, t) = int_{0}^{T} e^{-|mathbf{r}(t') - mathbf{s}|^2} dt' ), then the total energy is ( E(t) = int_{sphere} I_0 dA ). To make this constant, we can define ( I = c(t) I_0 ), where ( c(t) ) is a scaling factor such that ( int_{sphere} c(t) I_0 dA = E_0 ). Therefore, ( c(t) = E_0 / int_{sphere} I_0 dA ).But this requires knowing ( int_{sphere} I_0 dA ) at each time ( t ), which depends on the visitor's movement. So, the necessary condition is that ( I(theta, phi, t) = frac{E_0}{int_{sphere} I_0(theta, phi, t) dA} I_0(theta, phi, t) ).Alternatively, perhaps the artist can adjust the exponent in the Gaussian to control the total energy. For example, if the exponent is ( -k(t) |mathbf{r}(t) - mathbf{s}|^2 ), then the total energy would be ( E(t) = int_{sphere} int_{0}^{T} e^{-k(t) |mathbf{r}(t') - mathbf{s}|^2} dt' dA ). To make ( E(t) ) constant, we can adjust ( k(t) ) such that this integral equals ( E_0 ).But this seems complicated because ( k(t) ) would have to be a function that depends on the visitor's movement. Alternatively, perhaps the artist can use a different kernel function that inherently has a constant integral over the sphere, regardless of the visitor's position.Wait, another idea: If the kernel is a delta function, then the integral over the sphere would be the time spent at each point, but that's not the case here. Alternatively, if the kernel is a probability density function, then its integral over the sphere would be 1, and scaling it by ( E_0 ) would make the total energy constant.But in the given problem, the kernel is ( e^{-|mathbf{r}(t) - mathbf{s}|^2} ), which is not normalized. So, perhaps the artist needs to normalize this kernel such that its integral over the sphere is constant.Wait, but the intensity is the integral over time of the kernel. So, perhaps the artist can adjust the kernel's width or the integration time to ensure that the total energy remains constant.Alternatively, maybe the artist can use a different exponent, such as ( e^{-k |mathbf{r}(t) - mathbf{s}|^2} ), and choose ( k ) such that the total energy is constant.But I think the key is that the artist needs to ensure that the integral of ( I ) over the sphere is constant. So, the necessary condition is that ( int_{sphere} I(theta, phi, t) sin theta dtheta dphi = E_0 ) for all ( t ).Therefore, the condition is that ( I(theta, phi, t) ) must be scaled such that its integral over the sphere is always ( E_0 ). This can be achieved by defining ( I(theta, phi, t) = frac{E_0}{int_{sphere} I_0(theta, phi, t) sin theta dtheta dphi} I_0(theta, phi, t) ), where ( I_0 ) is the original intensity function.Alternatively, if we consider that the artist can adjust the parameters of the algorithm, such as the exponent in the Gaussian, to control the total energy, then the necessary condition is that the integral of ( I ) over the sphere is constant, which can be expressed as ( int_{0}^{2pi} int_{0}^{pi} I(theta, phi, t) sin theta dtheta dphi = E_0 ) for all ( t ).So, in summary, for the first part, the intensity is maximized where the visitor spends the most time near that point on the sphere. For the second part, the necessary condition is that the integral of ( I ) over the sphere is always ( E_0 ), which can be achieved by appropriately scaling ( I ) or adjusting the kernel parameters.</think>

🧠:<think>Alright, so I've got this problem about cubic Bézier curves and approximating them with Chebyshev polynomials. Let me try to figure this out step by step. First, part 1 asks for the parametric equations of a cubic Bézier curve. I remember that Bézier curves are defined using control points and a parameter t. For a cubic one, there are four control points: P0, P1, P2, P3. The general formula involves Bernstein polynomials, right? So, the parametric equations for x and y coordinates should be a combination of these control points multiplied by the Bernstein basis polynomials. The Bernstein polynomial for degree 3 is given by B_{i,3}(t) = C(3,i) * t^i * (1-t)^{3-i}, where C is the combination function. Therefore, for each coordinate, x(t) and y(t), it should be the sum from i=0 to 3 of P_i multiplied by B_{i,3}(t). Let me write that out:x(t) = P0_x * (1-t)^3 + 3*P1_x * t*(1-t)^2 + 3*P2_x * t^2*(1-t) + P3_x * t^3Similarly for y(t):y(t) = P0_y * (1-t)^3 + 3*P1_y * t*(1-t)^2 + 3*P2_y * t^2*(1-t) + P3_y * t^3I think that's correct. It's a weighted sum of the control points with the Bernstein polynomials as weights.Now, moving on to part 2. The programmer wants to approximate the cubic Bézier curve with a polynomial of degree 3 using Chebyshev polynomials. Hmm, Chebyshev polynomials are known for their minimal maximum error property, so they might be good for approximation.Chebyshev nodes are the points where the Chebyshev polynomial of a certain degree reaches its extrema. For approximation, using these nodes can minimize the error. But how does this apply to approximating a Bézier curve?Since the Bézier curve is already a cubic polynomial, approximating it with another cubic polynomial might seem redundant, but perhaps the goal is to represent it in a different basis, maybe to reduce the number of control points or for more efficient computation.Wait, the problem says they are exploring the use of Chebyshev polynomials for this purpose. So, maybe instead of using the Bernstein basis, they want to express the curve in terms of Chebyshev polynomials. Chebyshev polynomials of the first kind are defined as T_n(t) = cos(n * arccos(t)), but they can also be expressed recursively or via explicit formulas. For degree 3, T_3(t) = 4t^3 - 3t.But how do we approximate the Bézier curve with a Chebyshev polynomial? I think the idea is to express the parametric equations x(t) and y(t) as a linear combination of Chebyshev polynomials up to degree 3.So, x(t) ≈ a0*T0(t) + a1*T1(t) + a2*T2(t) + a3*T3(t)Similarly for y(t). But since T0(t) = 1, T1(t) = t, T2(t) = 2t^2 -1, T3(t) =4t^3 -3t.But wait, the original Bézier curve is a cubic polynomial in t, so it can be expressed exactly as a combination of Chebyshev polynomials up to degree 3. So, maybe the approximation is exact in this case?But the problem says "approximate", so perhaps they are using a different approach, like interpolation at Chebyshev nodes.Chebyshev nodes for degree 3 would be t_k = cos((2k-1)*π/(2*3)) for k=1,2,3,4. Wait, no, for n points, the nodes are cos((2k-1)π/(2n)) for k=1 to n. So for degree 3, we have 4 nodes.But since the curve is already a cubic, interpolating at 4 points would give an exact representation. So maybe the idea is to represent the curve in terms of Chebyshev polynomials evaluated at these nodes.Alternatively, perhaps they are using the Chebyshev polynomial basis to represent the curve, which might have better numerical properties or allow for more efficient computation.But the question is about approximating the cubic Bézier curve with a polynomial of degree 3 using Chebyshev nodes. So, maybe it's about expressing the curve in the Chebyshev basis.Let me recall that any cubic polynomial can be expressed as a linear combination of Chebyshev polynomials up to degree 3. So, if we have x(t) as a cubic, we can write it as a combination of T0, T1, T2, T3.To find the coefficients, we can use the inner product with respect to the Chebyshev weight function. The coefficients a_n are given by (2/π) * integral from -1 to 1 of x(t) * T_n(t) / sqrt(1 - t^2) dt, but since our t is from 0 to 1, maybe we need to adjust the interval.Alternatively, since the curve is defined on [0,1], we might need to use a shifted Chebyshev polynomial basis. Shifted Chebyshev polynomials are defined on the interval [0,1] instead of [-1,1]. The shifted T_n(t) can be expressed as T_n(2t -1).So, for the shifted Chebyshev polynomials, T0(t) =1, T1(t)=2t -1, T2(t)=4t^2 -4t +1, T3(t)=8t^3 -12t^2 +6t -1.Then, any cubic polynomial on [0,1] can be expressed as a combination of these shifted Chebyshev polynomials.So, to approximate x(t) and y(t), which are cubic polynomials, we can express them exactly in terms of the shifted Chebyshev polynomials up to degree 3.Therefore, the approximation would be exact, and the coefficients can be found by solving the system of equations or using the orthogonality properties.But since the original curve is already a cubic, using Chebyshev approximation might not reduce the number of control points, unless they are using a lower degree approximation, but the problem says degree 3.Wait, maybe the idea is to represent the curve in terms of Chebyshev polynomials, which might have fewer coefficients or allow for more efficient storage. But since it's still degree 3, the number of coefficients is the same.Alternatively, perhaps the programmer is considering using fewer points by leveraging the properties of Chebyshev polynomials, but I'm not sure.Wait, the problem says "approximate the given cubic Bézier curve with a polynomial of degree 3". So, it's still a cubic, but expressed in a different basis.So, the explicit form would be expressing x(t) and y(t) as a combination of T0, T1, T2, T3 with appropriate coefficients.But to find those coefficients, we can use the fact that the Bernstein basis and Chebyshev basis are both bases for the space of cubic polynomials. So, we can convert the coefficients from Bernstein to Chebyshev.Alternatively, we can use the inner product to find the coefficients. For each coordinate, x(t) is a cubic polynomial, so we can write:x(t) = a0*T0(t) + a1*T1(t) + a2*T2(t) + a3*T3(t)Similarly for y(t).To find a0, a1, a2, a3, we can use the orthogonality of Chebyshev polynomials with respect to the weight function 1/sqrt(1 - t^2) on [-1,1]. But since our t is in [0,1], we need to use the shifted Chebyshev polynomials.The formula for the coefficients in the shifted case is:a_n = (2/(π)) * integral from 0 to1 of x(t) * T_n(2t -1) / sqrt(t(1 - t)) dtBut this integral might be complicated. Alternatively, we can use the fact that the Bernstein polynomials can be expressed in terms of Chebyshev polynomials.But maybe a simpler approach is to evaluate the Bézier curve at the Chebyshev nodes and then perform interpolation.Wait, Chebyshev nodes for degree 3 would be t_k = (1 + cos((2k-1)π/(2*3)))/2 for k=1,2,3,4. So, four points in [0,1].Then, we can evaluate x(t) and y(t) at these four points and set up a system of equations to solve for the coefficients a0, a1, a2, a3 in the Chebyshev polynomial.Since we have four equations (from four nodes) and four unknowns, we can solve for the coefficients exactly.So, the explicit form would be:x(t) ≈ a0*T0(t) + a1*T1(t) + a2*T2(t) + a3*T3(t)Similarly for y(t), where a0, a1, a2, a3 are determined by solving the system obtained by evaluating x(t) and y(t) at the four Chebyshev nodes.But since the original curve is a cubic, this approximation will be exact, meaning that the coefficients will be such that the Chebyshev polynomial representation matches the Bézier curve exactly.Therefore, the polynomial approximation is exact and can be written as a combination of the shifted Chebyshev polynomials with coefficients determined by interpolation at the Chebyshev nodes.So, putting it all together, the parametric equations for the Bézier curve are as I wrote in part 1, and the approximation using Chebyshev polynomials involves expressing x(t) and y(t) as a combination of T0, T1, T2, T3 with coefficients found by evaluating the curve at the Chebyshev nodes and solving the resulting system.I think that's the gist of it. Let me just make sure I didn't miss anything. The key points are:1. Bézier curve parametric equations using Bernstein polynomials.2. Approximating with Chebyshev polynomials by expressing the cubic as a combination of shifted Chebyshev polynomials, which can be done exactly since both are cubic. The coefficients are found via interpolation at Chebyshev nodes, ensuring minimal error and maintaining continuity and differentiability since it's an exact representation.Yeah, that seems right.</think>

🧠:<think>Alright, so I have this problem about a transgender woman who is a theater critic. She evaluates 20 performances over a month, and her scores follow a sinusoidal pattern. The function given is ( S(t) = 5sinleft(frac{pi}{15}t + phiright) + 5 ). She notices that her peak mood is at ( t = 15 ) days, which gives the highest scores around that day. First, I need to figure out the phase shift ( phi ). Since the peak occurs at ( t = 15 ), that should correspond to the maximum of the sine function. The sine function reaches its maximum at ( frac{pi}{2} ) radians. So, plugging ( t = 15 ) into the argument of the sine function should give ( frac{pi}{2} ).Let me write that out:( frac{pi}{15} times 15 + phi = frac{pi}{2} )Simplifying, ( pi + phi = frac{pi}{2} ), so ( phi = frac{pi}{2} - pi = -frac{pi}{2} ).Wait, that seems a bit off. Let me double-check. The general sine function is ( sin(Bt + C) ), and its maximum occurs when ( Bt + C = frac{pi}{2} + 2pi k ) for integer ( k ). Since we're dealing with a single period here, we can ignore the ( 2pi k ) part.So, ( frac{pi}{15} times 15 + phi = frac{pi}{2} )Which simplifies to ( pi + phi = frac{pi}{2} ), so ( phi = -frac{pi}{2} ). Yeah, that seems correct. So the function becomes:( S(t) = 5sinleft(frac{pi}{15}t - frac{pi}{2}right) + 5 )Alternatively, using the sine identity, ( sin(theta - frac{pi}{2}) = -cos(theta) ). So, substituting that in:( S(t) = 5(-cos(frac{pi}{15}t)) + 5 = -5cosleft(frac{pi}{15}tright) + 5 )Hmm, that's another way to write it. So, the function is a cosine function shifted and scaled. Since cosine has a maximum at 0, and we have a negative cosine, the maximum will occur at ( t = 0 ), but wait, that contradicts the peak at ( t = 15 ). Wait, maybe I made a mistake.Wait, no. Let's think again. The original function is ( S(t) = 5sinleft(frac{pi}{15}t + phiright) + 5 ). When ( t = 15 ), the argument is ( frac{pi}{15} times 15 + phi = pi + phi ). We set this equal to ( frac{pi}{2} ) because that's where sine reaches its maximum. So, ( pi + phi = frac{pi}{2} ) implies ( phi = -frac{pi}{2} ). So, substituting back, we get:( S(t) = 5sinleft(frac{pi}{15}t - frac{pi}{2}right) + 5 )Using the identity ( sin(alpha - frac{pi}{2}) = -cos(alpha) ), so:( S(t) = -5cosleft(frac{pi}{15}tright) + 5 )So, the function is a cosine function flipped vertically and shifted up by 5. The maximum of this function occurs when ( cosleft(frac{pi}{15}tright) ) is minimized, which is -1. So, ( S(t) = -5(-1) + 5 = 5 + 5 = 10 ). The minimum occurs when ( cosleft(frac{pi}{15}tright) = 1 ), so ( S(t) = -5(1) + 5 = 0 ). Wait, but the original function was supposed to be between 1 and 10. Hmm, that's a problem.Wait, no. The original function is ( 5sin(...) + 5 ). The sine function ranges from -1 to 1, so ( 5sin(...) ) ranges from -5 to 5, and adding 5 gives a range from 0 to 10. So, the scores go from 0 to 10, but the problem says she assigns scores from 1 to 10. Hmm, that's a discrepancy. Maybe the function is actually ( 5sin(...) + 5 ), which gives 0 to 10, but she only assigns 1 to 10. So perhaps the minimum score is 1, not 0. Maybe I need to adjust the function accordingly.Wait, the problem says the scores are on a scale from 1 to 10, but the function given is ( S(t) = 5sin(...) + 5 ), which ranges from 0 to 10. So, perhaps the scores are actually from 0 to 10, but she only assigns 1 to 10, meaning 0 is not used. Or maybe the function is adjusted. Hmm, perhaps I should proceed with the given function, assuming that the scores can go down to 0, even though she assigns from 1 to 10. Maybe it's a typo or oversight in the problem. I'll proceed with the given function, which ranges from 0 to 10.So, moving on. She wants to normalize her scores to a scale from 0 to 1 using the function ( N(S) = frac{S(t) - min(S(t))}{max(S(t)) - min(S(t))} ).First, I need to find the minimum and maximum of ( S(t) ) over the interval ( t in [0, 30] ).Given ( S(t) = 5sinleft(frac{pi}{15}t - frac{pi}{2}right) + 5 ), which simplifies to ( S(t) = -5cosleft(frac{pi}{15}tright) + 5 ).The cosine function ( cosleft(frac{pi}{15}tright) ) has a period of ( frac{2pi}{pi/15} } = 30 ) days, which makes sense since the interval is 30 days. So, over 30 days, the function completes one full cycle.The maximum of ( S(t) ) occurs when ( cosleft(frac{pi}{15}tright) ) is minimized, which is -1. So, ( S(t) = -5(-1) + 5 = 5 + 5 = 10 ).The minimum occurs when ( cosleft(frac{pi}{15}tright) = 1 ), so ( S(t) = -5(1) + 5 = 0 ).Therefore, ( min(S(t)) = 0 ) and ( max(S(t)) = 10 ).So, the normalization function ( N(S) ) becomes:( N(S) = frac{S(t) - 0}{10 - 0} = frac{S(t)}{10} )Therefore, the normalized score distribution function is simply ( N(S) = frac{S(t)}{10} ).But wait, let me think again. The function ( S(t) ) is given as ( 5sin(...) + 5 ), which ranges from 0 to 10. So, the normalization is straightforward: subtract the minimum (0) and divide by the range (10). So yes, ( N(S) = S(t)/10 ).So, the normalized function is ( N(t) = frac{S(t)}{10} = frac{-5cosleft(frac{pi}{15}tright) + 5}{10} = frac{-5cosleft(frac{pi}{15}tright)}{10} + frac{5}{10} = -frac{1}{2}cosleft(frac{pi}{15}tright) + frac{1}{2} ).Alternatively, simplifying:( N(t) = frac{1}{2} - frac{1}{2}cosleft(frac{pi}{15}tright) )That's the normalized score distribution function.Now, she wants to find the average normalized score over the 30-day period. The average value of a function over an interval [a, b] is given by:( text{Average} = frac{1}{b - a} int_{a}^{b} N(t) dt )Here, ( a = 0 ), ( b = 30 ), so:( text{Average} = frac{1}{30} int_{0}^{30} left( frac{1}{2} - frac{1}{2}cosleft(frac{pi}{15}tright) right) dt )Let me compute this integral step by step.First, split the integral into two parts:( int_{0}^{30} frac{1}{2} dt - frac{1}{2} int_{0}^{30} cosleft(frac{pi}{15}tright) dt )Compute the first integral:( frac{1}{2} int_{0}^{30} dt = frac{1}{2} [t]_{0}^{30} = frac{1}{2} (30 - 0) = 15 )Now, compute the second integral:( frac{1}{2} int_{0}^{30} cosleft(frac{pi}{15}tright) dt )Let me make a substitution to solve this integral. Let ( u = frac{pi}{15}t ), so ( du = frac{pi}{15} dt ), which implies ( dt = frac{15}{pi} du ).When ( t = 0 ), ( u = 0 ). When ( t = 30 ), ( u = frac{pi}{15} times 30 = 2pi ).So, the integral becomes:( frac{1}{2} times frac{15}{pi} int_{0}^{2pi} cos(u) du )Compute the integral of ( cos(u) ):( int cos(u) du = sin(u) + C )So, evaluating from 0 to ( 2pi ):( sin(2pi) - sin(0) = 0 - 0 = 0 )Therefore, the second integral is:( frac{1}{2} times frac{15}{pi} times 0 = 0 )Putting it all together, the average is:( frac{1}{30} (15 - 0) = frac{15}{30} = frac{1}{2} )So, the average normalized score over the month is 0.5.Wait, let me double-check the integral calculations.First integral: ( frac{1}{2} times 30 = 15 ). Correct.Second integral: The integral of ( cos ) over a full period (0 to ( 2pi )) is zero. So, the second term is zero. Therefore, the average is ( 15/30 = 0.5 ). That makes sense because the normalized function is symmetric around 0.5, oscillating between 0 and 1. So, the average should be the midpoint, which is 0.5.Therefore, the normalized score distribution function is ( N(t) = frac{1}{2} - frac{1}{2}cosleft(frac{pi}{15}tright) ), and the average normalized score is 0.5.I think that's it. Let me summarize:1. The phase shift ( phi ) is ( -frac{pi}{2} ), so the function becomes ( S(t) = -5cosleft(frac{pi}{15}tright) + 5 ).2. The minimum score is 0 and the maximum is 10, so normalization is ( N(S) = S(t)/10 ), which simplifies to ( N(t) = frac{1}{2} - frac{1}{2}cosleft(frac{pi}{15}tright) ).3. The average normalized score is 0.5.Yeah, that seems correct.</think>

🧠:<think>Okay, so I have this problem about a photographer using a drone to capture panoramic images over a desert area. The area is a rectangle with vertices at (0,0), (a,0), (a,b), and (0,b). There are two parts to the problem. Let me tackle them one by one.Problem 1: Calculating the total distance the drone travels along the sinusoidal path.The drone's path is given by the equation ( y = frac{b}{2} sinleft(frac{2pi x}{a}right) + frac{b}{2} ) from ( x = 0 ) to ( x = a ). I need to find the total distance traveled along this curve. Hmm, I remember that the formula for the length of a curve ( y = f(x) ) from ( x = a ) to ( x = b ) is given by the integral:[L = int_{a}^{b} sqrt{1 + left( frac{dy}{dx} right)^2} dx]So, first, I need to find the derivative of ( y ) with respect to ( x ).Given ( y = frac{b}{2} sinleft(frac{2pi x}{a}right) + frac{b}{2} ), let's compute ( frac{dy}{dx} ).The derivative of ( sin(u) ) is ( cos(u) cdot u' ), so:[frac{dy}{dx} = frac{b}{2} cdot cosleft(frac{2pi x}{a}right) cdot frac{2pi}{a}]Simplify that:[frac{dy}{dx} = frac{b}{2} cdot frac{2pi}{a} cosleft(frac{2pi x}{a}right) = frac{bpi}{a} cosleft(frac{2pi x}{a}right)]Okay, so now we have ( frac{dy}{dx} = frac{bpi}{a} cosleft(frac{2pi x}{a}right) ). Next, we need to compute ( left( frac{dy}{dx} right)^2 ):[left( frac{dy}{dx} right)^2 = left( frac{bpi}{a} cosleft(frac{2pi x}{a}right) right)^2 = left( frac{b^2 pi^2}{a^2} right) cos^2left(frac{2pi x}{a}right)]So, the integrand becomes:[sqrt{1 + frac{b^2 pi^2}{a^2} cos^2left(frac{2pi x}{a}right)}]Therefore, the total distance ( L ) is:[L = int_{0}^{a} sqrt{1 + frac{b^2 pi^2}{a^2} cos^2left(frac{2pi x}{a}right)} dx]Hmm, this integral looks a bit complicated. I wonder if there's a substitution that can simplify it. Let me consider substituting ( u = frac{2pi x}{a} ). Then, ( du = frac{2pi}{a} dx ), so ( dx = frac{a}{2pi} du ). Changing the limits of integration: when ( x = 0 ), ( u = 0 ); when ( x = a ), ( u = 2pi ). Substituting into the integral:[L = int_{0}^{2pi} sqrt{1 + frac{b^2 pi^2}{a^2} cos^2(u)} cdot frac{a}{2pi} du]Simplify the constants:[L = frac{a}{2pi} int_{0}^{2pi} sqrt{1 + frac{b^2 pi^2}{a^2} cos^2(u)} du]Let me denote ( k = frac{b pi}{a} ), so ( k^2 = frac{b^2 pi^2}{a^2} ). Then, the integral becomes:[L = frac{a}{2pi} int_{0}^{2pi} sqrt{1 + k^2 cos^2(u)} du]Hmm, this integral is still not straightforward. I recall that integrals of the form ( int sqrt{1 + k^2 cos^2(u)} du ) don't have elementary antiderivatives. They are related to elliptic integrals. Wait, but maybe I can express it in terms of an elliptic integral. 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, we have ( sqrt{1 + k^2 cos^2(u)} ). Let me see if I can manipulate it to match the form of an elliptic integral.Note that ( cos^2(u) = 1 - sin^2(u) ), so:[sqrt{1 + k^2 cos^2(u)} = sqrt{1 + k^2 (1 - sin^2(u))} = sqrt{1 + k^2 - k^2 sin^2(u)} = sqrt{(1 + k^2) - k^2 sin^2(u)}]Let me factor out ( sqrt{1 + k^2} ):[sqrt{1 + k^2} cdot sqrt{1 - frac{k^2}{1 + k^2} sin^2(u)}]So, the integral becomes:[sqrt{1 + k^2} int_{0}^{2pi} sqrt{1 - left( frac{k^2}{1 + k^2} right) sin^2(u)} du]But the integral over ( 0 ) to ( 2pi ) can be expressed as 4 times the integral from ( 0 ) to ( frac{pi}{2} ) because the integrand is periodic with period ( pi ) and symmetric.So,[int_{0}^{2pi} sqrt{1 - m sin^2(u)} du = 4 int_{0}^{frac{pi}{2}} sqrt{1 - m sin^2(u)} du = 4 E(m)]Where ( m = frac{k^2}{1 + k^2} ).Therefore, our integral becomes:[sqrt{1 + k^2} cdot 4 Eleft( frac{k^2}{1 + k^2} right)]So, putting it all together:[L = frac{a}{2pi} cdot sqrt{1 + k^2} cdot 4 Eleft( frac{k^2}{1 + k^2} right)]Simplify:[L = frac{2a}{pi} sqrt{1 + k^2} Eleft( frac{k^2}{1 + k^2} right)]But ( k = frac{b pi}{a} ), so ( k^2 = frac{b^2 pi^2}{a^2} ). Therefore,[sqrt{1 + k^2} = sqrt{1 + frac{b^2 pi^2}{a^2}} = sqrt{frac{a^2 + b^2 pi^2}{a^2}}} = frac{sqrt{a^2 + b^2 pi^2}}{a}]So, substituting back:[L = frac{2a}{pi} cdot frac{sqrt{a^2 + b^2 pi^2}}{a} cdot Eleft( frac{frac{b^2 pi^2}{a^2}}{1 + frac{b^2 pi^2}{a^2}} right)]Simplify the constants:[L = frac{2}{pi} sqrt{a^2 + b^2 pi^2} cdot Eleft( frac{b^2 pi^2}{a^2 + b^2 pi^2} right)]Hmm, this is getting a bit messy, but I think this is as far as I can go analytically. So, the total distance is expressed in terms of the complete elliptic integral of the second kind. Alternatively, if I didn't know about elliptic integrals, I might have to approximate the integral numerically, but since the problem asks for an integral calculus solution, expressing it in terms of elliptic integrals is acceptable.Wait, let me check if I made any mistakes in substitution or simplification.Starting from:[L = frac{a}{2pi} int_{0}^{2pi} sqrt{1 + k^2 cos^2(u)} du]Then, I expressed it as:[sqrt{1 + k^2} cdot sqrt{1 - frac{k^2}{1 + k^2} sin^2(u)}]Which is correct because:[1 + k^2 cos^2(u) = 1 + k^2 (1 - sin^2(u)) = 1 + k^2 - k^2 sin^2(u) = (1 + k^2)(1 - frac{k^2}{1 + k^2} sin^2(u))]Yes, that's correct. Then, the integral over 0 to 2π becomes 4 times the integral from 0 to π/2, which is 4E(m). So, that seems right.Therefore, the total distance is:[L = frac{2}{pi} sqrt{a^2 + b^2 pi^2} cdot Eleft( frac{b^2 pi^2}{a^2 + b^2 pi^2} right)]I think that's the answer for part 1.Problem 2: Determining the average intensity of sunlight along the sinusoidal path.The intensity at any point (x, y) is given by:[I(x, y) = I_0 e^{-frac{(x - frac{a}{2})^2 + (y - frac{b}{2})^2}{2sigma^2}}]We need to find the average intensity along the path. The average value of a function along a curve is given by:[text{Average Intensity} = frac{1}{L} int_{C} I(x, y) ds]Where ( L ) is the total length of the curve (which we found in part 1), and ( ds ) is the differential arc length along the curve.But since we already have ( ds = sqrt{1 + left( frac{dy}{dx} right)^2} dx ), we can write:[text{Average Intensity} = frac{1}{L} int_{0}^{a} I(x, y(x)) sqrt{1 + left( frac{dy}{dx} right)^2} dx]Wait, but in part 1, we already have ( L = int_{0}^{a} sqrt{1 + left( frac{dy}{dx} right)^2} dx ). So, the average intensity is:[text{Average Intensity} = frac{1}{L} int_{0}^{a} I(x, y(x)) sqrt{1 + left( frac{dy}{dx} right)^2} dx]But since ( L ) is the integral of the same square root term, this is essentially the weighted average of ( I(x, y(x)) ) along the path.So, let me write out the integral:[text{Average Intensity} = frac{1}{L} int_{0}^{a} I_0 e^{-frac{(x - frac{a}{2})^2 + (y - frac{b}{2})^2}{2sigma^2}} sqrt{1 + left( frac{dy}{dx} right)^2} dx]We already have ( y = frac{b}{2} sinleft( frac{2pi x}{a} right) + frac{b}{2} ), so ( y - frac{b}{2} = frac{b}{2} sinleft( frac{2pi x}{a} right) ). Therefore, ( (y - frac{b}{2})^2 = frac{b^2}{4} sin^2left( frac{2pi x}{a} right) ).So, the exponent in the intensity function becomes:[-frac{(x - frac{a}{2})^2 + frac{b^2}{4} sin^2left( frac{2pi x}{a} right)}{2sigma^2}]Therefore, the integral becomes:[text{Average Intensity} = frac{I_0}{L} int_{0}^{a} e^{-frac{(x - frac{a}{2})^2 + frac{b^2}{4} sin^2left( frac{2pi x}{a} right)}{2sigma^2}} sqrt{1 + left( frac{bpi}{a} cosleft( frac{2pi x}{a} right) right)^2} dx]This integral looks quite complicated. I don't think it has an elementary antiderivative either. Maybe we can express it in terms of some special functions or approximate it numerically, but since the problem asks for an integral calculus solution, perhaps we can leave it in integral form or see if there's a way to simplify it.Alternatively, maybe we can make a substitution similar to part 1. Let me try substituting ( u = frac{2pi x}{a} ), so ( x = frac{a u}{2pi} ), and ( dx = frac{a}{2pi} du ). The limits of integration become ( u = 0 ) to ( u = 2pi ).Let me rewrite the integral:First, express ( x - frac{a}{2} ) in terms of ( u ):[x - frac{a}{2} = frac{a u}{2pi} - frac{a}{2} = frac{a}{2} left( frac{u}{pi} - 1 right)]So, ( (x - frac{a}{2})^2 = frac{a^2}{4} left( frac{u}{pi} - 1 right)^2 )Also, ( sin^2left( frac{2pi x}{a} right) = sin^2(u) )So, the exponent becomes:[-frac{frac{a^2}{4} left( frac{u}{pi} - 1 right)^2 + frac{b^2}{4} sin^2(u)}{2sigma^2} = -frac{a^2}{8sigma^2} left( frac{u}{pi} - 1 right)^2 - frac{b^2}{8sigma^2} sin^2(u)]So, the integral becomes:[text{Average Intensity} = frac{I_0}{L} cdot frac{a}{2pi} int_{0}^{2pi} e^{-frac{a^2}{8sigma^2} left( frac{u}{pi} - 1 right)^2 - frac{b^2}{8sigma^2} sin^2(u)} sqrt{1 + left( frac{bpi}{a} cos(u) right)^2} du]Hmm, this still seems quite complex. I don't see an obvious way to simplify this further. Maybe we can consider expanding the exponent or using some approximation, but without more information on the parameters ( a, b, sigma ), it's hard to proceed.Alternatively, perhaps we can write the average intensity as:[text{Average Intensity} = frac{I_0}{L} int_{0}^{a} e^{-frac{(x - frac{a}{2})^2 + frac{b^2}{4} sin^2left( frac{2pi x}{a} right)}{2sigma^2}} sqrt{1 + left( frac{bpi}{a} cosleft( frac{2pi x}{a} right) right)^2} dx]Which is the same as:[text{Average Intensity} = frac{I_0}{L} int_{0}^{a} e^{-frac{(x - frac{a}{2})^2}{2sigma^2} - frac{b^2}{8sigma^2} sin^2left( frac{2pi x}{a} right)} sqrt{1 + left( frac{bpi}{a} cosleft( frac{2pi x}{a} right) right)^2} dx]But I don't think this helps much. Maybe we can separate the exponent into two terms:[e^{-frac{(x - frac{a}{2})^2}{2sigma^2}} cdot e^{-frac{b^2}{8sigma^2} sin^2left( frac{2pi x}{a} right)}]So, the integral becomes:[text{Average Intensity} = frac{I_0}{L} int_{0}^{a} e^{-frac{(x - frac{a}{2})^2}{2sigma^2}} e^{-frac{b^2}{8sigma^2} sin^2left( frac{2pi x}{a} right)} sqrt{1 + left( frac{bpi}{a} cosleft( frac{2pi x}{a} right) right)^2} dx]This might allow us to consider the integral as a product of functions, but I don't see a straightforward way to evaluate it. Perhaps using series expansions or numerical integration, but since the problem is theoretical, maybe we can leave it in this form.Alternatively, if ( sigma ) is large compared to the variations in ( x ) and ( y ), the exponential term might be approximately constant, but without knowing the relationship between ( sigma ), ( a ), and ( b ), it's hard to say.Wait, maybe we can consider the average intensity as the integral over the path, which is a sinusoidal curve, of the intensity function. Given that the intensity is a Gaussian centered at ( (a/2, b/2) ), the average intensity would depend on how the path is distributed around this center.But I don't think that gives us a way to compute the integral analytically. So, perhaps the best we can do is express the average intensity as the integral given above.Alternatively, maybe we can make a substitution similar to part 1, but I don't see it immediately.Wait, let me recall that in part 1, we had:[L = frac{2}{pi} sqrt{a^2 + b^2 pi^2} cdot Eleft( frac{b^2 pi^2}{a^2 + b^2 pi^2} right)]So, if we can express the integral for the average intensity in terms of similar elliptic integrals, that might be a way forward. But I don't see a direct connection.Alternatively, perhaps we can consider expanding the exponential term in a Taylor series around the center ( (a/2, b/2) ), but that might complicate things further.Alternatively, maybe we can change variables to center the coordinate system at ( (a/2, b/2) ). Let me try that.Let ( x' = x - frac{a}{2} ) and ( y' = y - frac{b}{2} ). Then, the intensity function becomes:[I(x', y') = I_0 e^{-frac{x'^2 + y'^2}{2sigma^2}}]And the path becomes:[y' = frac{b}{2} sinleft( frac{2pi (x' + frac{a}{2})}{a} right)]Simplify the argument of the sine:[frac{2pi (x' + frac{a}{2})}{a} = frac{2pi x'}{a} + pi]So,[y' = frac{b}{2} sinleft( frac{2pi x'}{a} + pi right) = frac{b}{2} cdot (-sinleft( frac{2pi x'}{a} right)) = -frac{b}{2} sinleft( frac{2pi x'}{a} right)]So, the path in terms of ( x' ) is:[y' = -frac{b}{2} sinleft( frac{2pi x'}{a} right)]But ( x' ) ranges from ( -frac{a}{2} ) to ( frac{a}{2} ) as ( x ) goes from 0 to ( a ).So, the intensity function is now:[I(x', y') = I_0 e^{-frac{x'^2 + y'^2}{2sigma^2}} = I_0 e^{-frac{x'^2 + frac{b^2}{4} sin^2left( frac{2pi x'}{a} right)}{2sigma^2}}]So, the integral for the average intensity becomes:[text{Average Intensity} = frac{I_0}{L} int_{-frac{a}{2}}^{frac{a}{2}} e^{-frac{x'^2 + frac{b^2}{4} sin^2left( frac{2pi x'}{a} right)}{2sigma^2}} sqrt{1 + left( frac{dy'}{dx'} right)^2} dx']But ( frac{dy'}{dx'} = -frac{b}{2} cdot frac{2pi}{a} cosleft( frac{2pi x'}{a} right) = -frac{bpi}{a} cosleft( frac{2pi x'}{a} right) ), so ( left( frac{dy'}{dx'} right)^2 = left( frac{bpi}{a} cosleft( frac{2pi x'}{a} right) right)^2 ), which is the same as in part 1.Therefore, the integral is symmetric around ( x' = 0 ), so we can write:[text{Average Intensity} = frac{2 I_0}{L} int_{0}^{frac{a}{2}} e^{-frac{x'^2 + frac{b^2}{4} sin^2left( frac{2pi x'}{a} right)}{2sigma^2}} sqrt{1 + left( frac{bpi}{a} cosleft( frac{2pi x'}{a} right) right)^2} dx']But I don't think this helps much either. It still seems like a complicated integral.Alternatively, maybe we can consider expanding the exponential term in a Fourier series or something, but that might be overcomplicating.Given that both integrals in part 1 and part 2 result in expressions involving elliptic integrals or complicated exponentials, perhaps the best way to present the answers is to leave them in integral form, acknowledging that they don't have elementary antiderivatives.But wait, let me think again. Maybe for part 2, we can express the average intensity in terms of the same elliptic integral as in part 1, but multiplied by some factor involving the Gaussian.But I don't see a direct way to relate the two integrals. The presence of the Gaussian complicates things.Alternatively, perhaps if we consider that the Gaussian is symmetric and the path is symmetric around the center, we might be able to make some approximations, but without more information, it's hard.Given all this, I think the answers are:1. The total distance is expressed in terms of the complete elliptic integral of the second kind as:[L = frac{2}{pi} sqrt{a^2 + b^2 pi^2} cdot Eleft( frac{b^2 pi^2}{a^2 + b^2 pi^2} right)]2. The average intensity is given by the integral:[text{Average Intensity} = frac{I_0}{L} int_{0}^{a} e^{-frac{(x - frac{a}{2})^2 + frac{b^2}{4} sin^2left( frac{2pi x}{a} right)}{2sigma^2}} sqrt{1 + left( frac{bpi}{a} cosleft( frac{2pi x}{a} right) right)^2} dx]Which can be written as:[text{Average Intensity} = frac{I_0}{L} int_{0}^{a} e^{-frac{(x - frac{a}{2})^2 + frac{b^2}{4} sin^2left( frac{2pi x}{a} right)}{2sigma^2}} sqrt{1 + left( frac{bpi}{a} cosleft( frac{2pi x}{a} right) right)^2} dx]But since ( L ) itself is an elliptic integral, perhaps we can write the average intensity as:[text{Average Intensity} = frac{I_0}{L} cdot text{[another integral]}]But without further simplification, I think this is as far as we can go.Wait a second, maybe I can factor out the Gaussian term and express the integral in terms of the same substitution as in part 1. Let me try that.In part 1, we had:[L = frac{a}{2pi} int_{0}^{2pi} sqrt{1 + k^2 cos^2(u)} du]Where ( k = frac{bpi}{a} ).In part 2, after substitution, we have:[text{Average Intensity} = frac{I_0}{L} cdot frac{a}{2pi} int_{0}^{2pi} e^{-frac{frac{a^2}{4} left( frac{u}{pi} - 1 right)^2 + frac{b^2}{4} sin^2(u)}{2sigma^2}} sqrt{1 + k^2 cos^2(u)} du]Wait, that's:[text{Average Intensity} = frac{I_0}{L} cdot frac{a}{2pi} int_{0}^{2pi} e^{-frac{a^2}{8sigma^2} left( frac{u}{pi} - 1 right)^2 - frac{b^2}{8sigma^2} sin^2(u)} sqrt{1 + k^2 cos^2(u)} du]But since ( L = frac{a}{2pi} int_{0}^{2pi} sqrt{1 + k^2 cos^2(u)} du ), we can write:[text{Average Intensity} = frac{I_0}{L} cdot frac{a}{2pi} int_{0}^{2pi} e^{-frac{a^2}{8sigma^2} left( frac{u}{pi} - 1 right)^2 - frac{b^2}{8sigma^2} sin^2(u)} sqrt{1 + k^2 cos^2(u)} du]Which simplifies to:[text{Average Intensity} = I_0 cdot frac{int_{0}^{2pi} e^{-frac{a^2}{8sigma^2} left( frac{u}{pi} - 1 right)^2 - frac{b^2}{8sigma^2} sin^2(u)} sqrt{1 + k^2 cos^2(u)} du}{int_{0}^{2pi} sqrt{1 + k^2 cos^2(u)} du}]So, the average intensity is the ratio of two integrals, both over the same interval, with the numerator being the weighted integral by the Gaussian term.But without further simplification, this is as far as we can go analytically. Therefore, the average intensity is given by this ratio.Alternatively, if we denote the numerator integral as ( N ) and the denominator as ( D ), then:[text{Average Intensity} = I_0 cdot frac{N}{D}]Where:[N = int_{0}^{2pi} e^{-frac{a^2}{8sigma^2} left( frac{u}{pi} - 1 right)^2 - frac{b^2}{8sigma^2} sin^2(u)} sqrt{1 + k^2 cos^2(u)} du]And:[D = int_{0}^{2pi} sqrt{1 + k^2 cos^2(u)} du = frac{2pi}{a} L]But I don't think this helps much in terms of simplifying.Given all this, I think the answers are:1. The total distance is expressed using the complete elliptic integral of the second kind.2. The average intensity is expressed as an integral involving the Gaussian function and the same square root term as in part 1.So, summarizing:1. Total distance ( L ):[L = frac{2}{pi} sqrt{a^2 + b^2 pi^2} cdot Eleft( frac{b^2 pi^2}{a^2 + b^2 pi^2} right)]2. Average intensity ( bar{I} ):[bar{I} = frac{I_0}{L} int_{0}^{a} e^{-frac{(x - frac{a}{2})^2 + frac{b^2}{4} sin^2left( frac{2pi x}{a} right)}{2sigma^2}} sqrt{1 + left( frac{bpi}{a} cosleft( frac{2pi x}{a} right) right)^2} dx]Alternatively, expressed in terms of the substitution ( u ):[bar{I} = I_0 cdot frac{int_{0}^{2pi} e^{-frac{a^2}{8sigma^2} left( frac{u}{pi} - 1 right)^2 - frac{b^2}{8sigma^2} sin^2(u)} sqrt{1 + k^2 cos^2(u)} du}{int_{0}^{2pi} sqrt{1 + k^2 cos^2(u)} du}]Where ( k = frac{bpi}{a} ).I think that's as far as I can go without numerical methods or further special functions.</think>