🧠:<think>Okay, so I have this problem about modeling the decline of bee populations due to pesticides. The environmentalist has given me a differential equation to work with, and I need to solve it and analyze its long-term behavior. Let me try to break this down step by step.First, the differential equation is:[ frac{dB(t)}{dt} = -k B(t) P(t) ]And the pesticide usage ( P(t) ) is given by:[ P(t) = P_0 sin(omega t) + P_1 ]So, the rate at which the bee population is declining is proportional to both the current population and the amount of pesticide used. That makes sense because more bees would mean more affected by pesticides, and more pesticides would have a bigger impact.The first part is to solve this differential equation given the initial condition ( B(0) = B_0 ). Hmm, okay, so this is a first-order linear ordinary differential equation, but it's actually separable because I can write it in terms of ( B(t) ) and ( P(t) ).Let me rewrite the equation:[ frac{dB}{dt} = -k B(t) (P_0 sin(omega t) + P_1) ]This looks like a separable equation. So, I can separate the variables ( B ) and ( t ):[ frac{dB}{B} = -k (P_0 sin(omega t) + P_1) dt ]Now, I need to integrate both sides. On the left side, integrating ( frac{1}{B} dB ) gives ( ln|B| ). On the right side, I have to integrate ( -k (P_0 sin(omega t) + P_1) dt ).Let me compute the integral on the right side:First, split the integral into two parts:[ -k int P_0 sin(omega t) dt - k int P_1 dt ]Compute each integral separately.For the first integral:[ int P_0 sin(omega t) dt ]Let me recall that the integral of ( sin(ax) ) is ( -frac{1}{a} cos(ax) ). So, applying that here:[ P_0 int sin(omega t) dt = P_0 left( -frac{1}{omega} cos(omega t) right) + C = -frac{P_0}{omega} cos(omega t) + C ]For the second integral:[ int P_1 dt = P_1 t + C ]Putting it all together, the right side integral is:[ -k left( -frac{P_0}{omega} cos(omega t) + P_1 t right) + C ]Simplify that:[ frac{k P_0}{omega} cos(omega t) - k P_1 t + C ]So, now, going back to the left side, the integral of ( frac{1}{B} dB ) is ( ln|B| + C ). So, equating both sides:[ ln|B| = frac{k P_0}{omega} cos(omega t) - k P_1 t + C ]To solve for ( B ), exponentiate both sides:[ B(t) = e^{frac{k P_0}{omega} cos(omega t) - k P_1 t + C} ]We can write this as:[ B(t) = e^C cdot e^{frac{k P_0}{omega} cos(omega t)} cdot e^{-k P_1 t} ]Let me denote ( e^C ) as a constant, say ( C' ), which will be determined by the initial condition.So,[ B(t) = C' e^{frac{k P_0}{omega} cos(omega t)} e^{-k P_1 t} ]Now, let's apply the initial condition ( B(0) = B_0 ).At ( t = 0 ):[ B(0) = C' e^{frac{k P_0}{omega} cos(0)} e^{-k P_1 cdot 0} ]Simplify:[ B_0 = C' e^{frac{k P_0}{omega} cdot 1} cdot 1 ]So,[ C' = B_0 e^{-frac{k P_0}{omega}} ]Therefore, substituting back into the expression for ( B(t) ):[ B(t) = B_0 e^{-frac{k P_0}{omega}} e^{frac{k P_0}{omega} cos(omega t)} e^{-k P_1 t} ]Simplify the exponents:Combine the first two exponentials:[ e^{-frac{k P_0}{omega}} e^{frac{k P_0}{omega} cos(omega t)} = e^{frac{k P_0}{omega} (cos(omega t) - 1)} ]So, the expression becomes:[ B(t) = B_0 e^{frac{k P_0}{omega} (cos(omega t) - 1)} e^{-k P_1 t} ]Alternatively, we can write this as:[ B(t) = B_0 e^{-k P_1 t} e^{frac{k P_0}{omega} (cos(omega t) - 1)} ]That's the solution to the differential equation. Let me just check if this makes sense.At ( t = 0 ), we have:[ B(0) = B_0 e^{0} e^{0} = B_0 ]Which is correct.Also, the exponential terms make sense because the cosine term oscillates between -1 and 1, so ( cos(omega t) - 1 ) oscillates between -2 and 0. Therefore, the exponent ( frac{k P_0}{omega} (cos(omega t) - 1) ) oscillates between ( -frac{2 k P_0}{omega} ) and 0. So, the term ( e^{frac{k P_0}{omega} (cos(omega t) - 1)} ) oscillates between ( e^{-frac{2 k P_0}{omega}} ) and 1.Therefore, the bee population ( B(t) ) is oscillating due to the periodic pesticide usage, but also decaying exponentially due to the ( e^{-k P_1 t} ) term.Okay, so that seems reasonable.Now, moving on to the second part: determining the long-term behavior of ( B(t) ) as ( t ) approaches infinity.So, as ( t to infty ), what happens to ( B(t) )?Looking at the expression:[ B(t) = B_0 e^{-k P_1 t} e^{frac{k P_0}{omega} (cos(omega t) - 1)} ]We can analyze each factor:1. The term ( e^{-k P_1 t} ): This is an exponential decay term. If ( P_1 > 0 ), this term goes to zero as ( t to infty ). If ( P_1 = 0 ), this term is 1 for all ( t ). If ( P_1 < 0 ), this term would grow exponentially, but since ( P_1 ) is the baseline pesticide usage, it's unlikely to be negative. So, we can assume ( P_1 geq 0 ).2. The term ( e^{frac{k P_0}{omega} (cos(omega t) - 1)} ): As ( t to infty ), ( cos(omega t) ) oscillates between -1 and 1. Therefore, ( cos(omega t) - 1 ) oscillates between -2 and 0, so the exponent oscillates between ( -frac{2 k P_0}{omega} ) and 0. Thus, this term oscillates between ( e^{-frac{2 k P_0}{omega}} ) and 1. So, it doesn't settle to a single value but keeps oscillating.Putting these together, the dominant term as ( t to infty ) is ( e^{-k P_1 t} ), which tends to zero if ( P_1 > 0 ). The oscillating term doesn't affect the exponential decay; it just modulates the amplitude of the decay.Therefore, if ( P_1 > 0 ), the bee population ( B(t) ) will decay to zero as ( t to infty ), regardless of the oscillations in ( P(t) ). The oscillating term only causes the decay to be somewhat "wiggly," but the overall trend is towards extinction.If ( P_1 = 0 ), then the term ( e^{-k P_1 t} ) becomes 1, and the population becomes:[ B(t) = B_0 e^{frac{k P_0}{omega} (cos(omega t) - 1)} ]Which oscillates between ( B_0 e^{-frac{2 k P_0}{omega}} ) and ( B_0 ). So, the population doesn't decay to zero but oscillates between these two values. Whether the population stabilizes or not depends on the parameters.Wait, but actually, if ( P_1 = 0 ), the model becomes:[ frac{dB}{dt} = -k B(t) P_0 sin(omega t) ]Which is a differential equation with periodic forcing. The solution we found shows that the population oscillates without a trend. So, in this case, the population doesn't go extinct; it just fluctuates.But wait, is that the case? Let me think.If ( P_1 = 0 ), the equation is:[ frac{dB}{dt} = -k B(t) P_0 sin(omega t) ]This is a linear differential equation with a periodic coefficient. The solution we found is:[ B(t) = B_0 e^{frac{k P_0}{omega} (cos(omega t) - 1)} ]Which is indeed oscillating between ( B_0 e^{-frac{2 k P_0}{omega}} ) and ( B_0 ). So, the population doesn't go extinct; it just oscillates. Therefore, in this case, the bee population stabilizes in the sense that it doesn't tend to zero or infinity, but fluctuates periodically.However, if ( P_1 > 0 ), then the exponential decay term dominates, and the population tends to zero.But wait, what if ( P_1 < 0 )? Is that possible? Since ( P_1 ) is the baseline pesticide usage, it's probably non-negative. So, we can assume ( P_1 geq 0 ).Therefore, summarizing:- If ( P_1 > 0 ): The bee population ( B(t) ) tends to zero as ( t to infty ). The population declines to extinction.- If ( P_1 = 0 ): The bee population oscillates between ( B_0 e^{-frac{2 k P_0}{omega}} ) and ( B_0 ). So, it doesn't go extinct; it stabilizes in the sense that it fluctuates periodically.But wait, actually, even when ( P_1 = 0 ), the population doesn't necessarily stabilize at a fixed number; it keeps oscillating. So, maybe "stabilizes" isn't the right word. It might be more accurate to say that the population fluctuates without a trend.But in the context of the question, they ask about the long-term behavior, whether it stabilizes, grows, or declines to extinction.So, if ( P_1 > 0 ), it declines to extinction.If ( P_1 = 0 ), it oscillates between two values, so it doesn't stabilize or grow; it just fluctuates.But wait, let's think again. The equation when ( P_1 = 0 ) is:[ frac{dB}{dt} = -k B(t) P_0 sin(omega t) ]This is a Bernoulli equation, and the solution oscillates. So, the population doesn't go extinct, but it doesn't grow either; it just oscillates.But is that the case? Let me consider the integral of the growth rate.The growth rate is ( -k P_0 sin(omega t) ). The average of ( sin(omega t) ) over a period is zero. So, the average growth rate is zero. Therefore, over time, the population doesn't have a net growth or decay; it just oscillates.Therefore, in the case ( P_1 = 0 ), the population doesn't go extinct, doesn't grow indefinitely, but oscillates periodically.But wait, in our solution, when ( P_1 = 0 ), the population is:[ B(t) = B_0 e^{frac{k P_0}{omega} (cos(omega t) - 1)} ]Which is:[ B(t) = B_0 e^{-frac{k P_0}{omega}} e^{frac{k P_0}{omega} cos(omega t)} ]So, it's oscillating between ( B_0 e^{-frac{2 k P_0}{omega}} ) and ( B_0 ). So, the amplitude of oscillation is fixed.Therefore, the population doesn't stabilize in the sense of approaching a fixed number, but it does not go extinct. It just keeps oscillating between two fixed bounds.So, in terms of the question: the long-term behavior is either extinction, oscillation without trend, or... Wait, can the population grow?Wait, let's see. If ( P_1 ) is negative, which is not realistic, but mathematically, if ( P_1 < 0 ), then ( e^{-k P_1 t} ) would be ( e^{|k P_1| t} ), which grows exponentially. So, in that case, the population would grow exponentially, despite the oscillations.But since ( P_1 ) is the baseline pesticide usage, it's non-negative. So, in reality, ( P_1 geq 0 ). So, the population either goes extinct or oscillates without trend.But wait, let's think about the case when ( P_1 = 0 ) and ( P_0 = 0 ). Then, ( P(t) = 0 ), so the differential equation becomes ( frac{dB}{dt} = 0 ), so the population remains constant. So, in that case, the population stabilizes.But in the general case, when ( P_1 = 0 ) and ( P_0 > 0 ), the population oscillates.So, to sum up:- If ( P_1 > 0 ): The population tends to zero (extinction).- If ( P_1 = 0 ): - If ( P_0 = 0 ): Population remains constant (stabilizes). - If ( P_0 > 0 ): Population oscillates between two values (doesn't stabilize, doesn't grow, doesn't go extinct).But in the context of the question, they might consider oscillation as a form of stabilization, but technically, it's fluctuating.But perhaps, in the long term, if ( P_1 = 0 ), the population doesn't go extinct, but it doesn't grow either; it just oscillates. So, it's neither growing nor declining to extinction.But wait, let's think about the integral of the growth rate. The growth rate is ( -k P(t) B(t) ). If ( P(t) ) is periodic, the average growth rate over a period is:[ frac{1}{T} int_0^T -k P(t) B(t) dt ]But since ( P(t) ) is periodic and ( B(t) ) is also periodic (when ( P_1 = 0 )), the integral might not necessarily be zero. Wait, no, because ( B(t) ) is also a function of ( P(t) ). Hmm, this might be more complicated.Alternatively, perhaps it's better to think in terms of the solution we have. When ( P_1 = 0 ), the population oscillates between two fixed values, so it doesn't have a trend towards zero or infinity. Therefore, in the long term, it doesn't go extinct, doesn't grow, but oscillates.Therefore, the conditions are:- If ( P_1 > 0 ): Bee population declines to extinction.- If ( P_1 = 0 ): - If ( P_0 = 0 ): Population remains constant (stabilizes). - If ( P_0 > 0 ): Population oscillates without trend (doesn't stabilize, doesn't grow, doesn't go extinct).But the question asks for the conditions under which the population stabilizes, grows, or declines to extinction.So, in terms of the parameters:- Decline to extinction: ( P_1 > 0 ).- Stabilizes: ( P_1 = 0 ) and ( P_0 = 0 ).- Oscillates (no growth or decline): ( P_1 = 0 ) and ( P_0 > 0 ).But the question might consider oscillation as a form of stabilization, but technically, it's fluctuating. So, perhaps the answer is:- If ( P_1 > 0 ): Decline to extinction.- If ( P_1 = 0 ): - If ( P_0 = 0 ): Stabilizes. - If ( P_0 > 0 ): Oscillates, neither growing nor declining.But the question is about long-term behavior, so in the case of oscillation, it doesn't stabilize in the sense of approaching a limit, but it doesn't go extinct either.Alternatively, maybe the question is expecting to consider the average effect. Since the average of ( P(t) ) when ( P_1 = 0 ) is ( P_1 = 0 ), but the oscillating term has an average of zero. So, the average growth rate is zero, leading to no net change. But in reality, the population is oscillating.But in our solution, when ( P_1 = 0 ), the population doesn't go extinct, but it also doesn't grow. It just oscillates.Therefore, the conditions are:- If ( P_1 > 0 ): The bee population will decline to extinction as ( t to infty ).- If ( P_1 = 0 ): - If ( P_0 = 0 ): The bee population remains constant (stabilizes). - If ( P_0 > 0 ): The bee population oscillates periodically without a trend (neither grows nor declines to extinction).But the question is phrased as "discuss the conditions under which the bee population stabilizes, grows, or declines to extinction."So, in terms of the answer:- The population stabilizes if ( P_1 = 0 ) and ( P_0 = 0 ).- The population grows if... Wait, in our model, can the population grow? Let's see.Looking back at the differential equation:[ frac{dB}{dt} = -k B(t) P(t) ]Since ( k ) is positive, and ( P(t) ) is given as ( P_0 sin(omega t) + P_1 ). If ( P(t) ) is positive, then ( frac{dB}{dt} ) is negative, meaning the population is decreasing. If ( P(t) ) is negative, then ( frac{dB}{dt} ) is positive, meaning the population is increasing.But in the context of the problem, ( P(t) ) represents pesticide usage, which is a non-negative quantity. So, ( P(t) geq 0 ). Therefore, ( P(t) ) cannot be negative, so ( frac{dB}{dt} leq 0 ). Therefore, the population cannot grow; it can only stay the same or decrease.Wait, that's an important point. Since ( P(t) ) is non-negative, the growth rate ( frac{dB}{dt} ) is non-positive. Therefore, the population cannot grow; it can only decline or stay the same.Therefore, in the case when ( P_1 = 0 ) and ( P_0 = 0 ), ( P(t) = 0 ), so ( frac{dB}{dt} = 0 ), so the population remains constant.When ( P_1 = 0 ) and ( P_0 > 0 ), ( P(t) ) oscillates between ( -P_0 ) and ( P_0 ). But wait, no, ( P(t) = P_0 sin(omega t) + P_1 ). If ( P_1 = 0 ), then ( P(t) = P_0 sin(omega t) ), which oscillates between ( -P_0 ) and ( P_0 ). But since ( P(t) ) is pesticide usage, it can't be negative. So, perhaps the model assumes ( P(t) geq 0 ), which would require ( P_1 geq P_0 ), so that ( P(t) = P_0 sin(omega t) + P_1 geq 0 ) for all ( t ).Wait, that's a good point. If ( P(t) ) is a pesticide usage, it can't be negative. So, the model must have ( P(t) geq 0 ) for all ( t ). Therefore, ( P_0 sin(omega t) + P_1 geq 0 ) for all ( t ).The minimum value of ( P(t) ) is ( P_1 - P_0 ). Therefore, to ensure ( P(t) geq 0 ), we must have ( P_1 - P_0 geq 0 ), so ( P_1 geq P_0 ).Therefore, in the case when ( P_1 = 0 ), ( P(t) ) would be ( P_0 sin(omega t) ), which is negative for half the period, which is not physically meaningful. Therefore, perhaps the model assumes ( P_1 geq P_0 ), so that ( P(t) geq 0 ) for all ( t ).Therefore, in the case when ( P_1 = 0 ), it's actually not a valid scenario because ( P(t) ) would become negative. So, we can assume ( P_1 geq P_0 ).Therefore, in the case when ( P_1 = 0 ), it's invalid because ( P(t) ) would be negative. So, the only valid cases are when ( P_1 geq P_0 ), ensuring ( P(t) geq 0 ).Therefore, in the context of the problem, ( P_1 geq P_0 ), so ( P(t) ) is always non-negative.Therefore, going back to the long-term behavior:- If ( P_1 > 0 ): The population tends to zero (extinction).- If ( P_1 = 0 ): Not a valid case because ( P(t) ) would be negative.Wait, but if ( P_1 = 0 ), and ( P_0 = 0 ), then ( P(t) = 0 ), which is valid. So, in that case, the population remains constant.But if ( P_1 = 0 ) and ( P_0 > 0 ), ( P(t) ) becomes negative, which is invalid. Therefore, the only valid cases are when ( P_1 geq P_0 ), ensuring ( P(t) geq 0 ).Therefore, in the valid cases:- If ( P_1 > 0 ): Population tends to zero.- If ( P_1 = 0 ) and ( P_0 = 0 ): Population remains constant.But since ( P_1 geq P_0 ), when ( P_1 = 0 ), ( P_0 ) must also be zero.Therefore, the only valid cases are:- If ( P_1 > 0 ): Population declines to extinction.- If ( P_1 = 0 ) and ( P_0 = 0 ): Population remains constant.Therefore, in the context of the problem, the population either remains constant or declines to extinction, depending on whether ( P_1 ) is zero or positive.Wait, but in the case when ( P_1 > 0 ), even if ( P_0 ) is positive, the population still tends to zero because of the exponential decay term ( e^{-k P_1 t} ). The oscillating term only modulates the decay but doesn't prevent it.Therefore, the conclusion is:- If ( P_1 > 0 ): Bee population declines to extinction.- If ( P_1 = 0 ) and ( P_0 = 0 ): Bee population remains constant.But in the problem statement, the environmentalist is campaigning for stricter pesticide regulations because of their harmful effects. So, she is likely concerned about the case when ( P_1 > 0 ), leading to extinction.But to answer the question as posed:1. Solve the differential equation.2. Determine the long-term behavior.So, for part 2, the long-term behavior is:- If ( P_1 > 0 ): ( B(t) to 0 ) as ( t to infty ). The bee population declines to extinction.- If ( P_1 = 0 ): - If ( P_0 = 0 ): ( B(t) ) remains constant at ( B_0 ). - If ( P_0 > 0 ): ( B(t) ) oscillates between ( B_0 e^{-frac{2 k P_0}{omega}} ) and ( B_0 ). However, since ( P(t) ) must be non-negative, ( P_1 geq P_0 ), so ( P_1 = 0 ) and ( P_0 > 0 ) is invalid. Therefore, the only valid case when ( P_1 = 0 ) is ( P_0 = 0 ), leading to a constant population.Therefore, in the valid scenarios:- If ( P_1 > 0 ): Extinction.- If ( P_1 = 0 ) and ( P_0 = 0 ): Constant population.So, the bee population stabilizes only if both ( P_1 = 0 ) and ( P_0 = 0 ). Otherwise, if ( P_1 > 0 ), it declines to extinction.But wait, in the case when ( P_1 > 0 ), even if ( P_0 ) is positive, the population still tends to zero. So, the key factor is ( P_1 ). If ( P_1 > 0 ), extinction; if ( P_1 = 0 ) and ( P_0 = 0 ), constant population.Therefore, the conditions are:- The bee population stabilizes (remains constant) if and only if ( P_1 = 0 ) and ( P_0 = 0 ).- The bee population declines to extinction if ( P_1 > 0 ).- If ( P_1 = 0 ) and ( P_0 > 0 ), the model is invalid because ( P(t) ) becomes negative.Therefore, in the context of the problem, the environmentalist is concerned about ( P_1 > 0 ), leading to extinction.So, to wrap up:1. The solution to the differential equation is:[ B(t) = B_0 e^{-k P_1 t} e^{frac{k P_0}{omega} (cos(omega t) - 1)} ]2. The long-term behavior is:- If ( P_1 > 0 ): The bee population declines to extinction.- If ( P_1 = 0 ) and ( P_0 = 0 ): The bee population remains constant.- If ( P_1 = 0 ) and ( P_0 > 0 ): The model is invalid as ( P(t) ) becomes negative.Therefore, under valid conditions, the population either stabilizes (if no pesticides are used) or declines to extinction (if pesticides are used with a positive baseline ( P_1 )).</think>

🧠:<think>Alright, so I'm trying to help this college student with their family tree software. They're using a directed acyclic graph (DAG) where each node is a person, and edges go from parents to children. Each node has a weight, which is a unique integer representing significance. The first task is to find the longest ancestral line, which is a path from a root node (someone with no parents) to a leaf node (someone with no children), and the weight is the sum of all nodes in that path. They want an algorithm to compute this.Hmm, okay. So, thinking about this, it's similar to finding the longest path in a DAG. I remember that DAGs have a topological order, which is an ordering of the nodes where every directed edge goes from an earlier node to a later node. If we can process the nodes in topological order, we can compute the longest path efficiently.So, the plan is to perform a topological sort on the DAG. Once we have that order, we can process each node in that order and relax all its outgoing edges. Relaxing an edge here means checking if going through the current node to a child gives a longer path than what's already recorded for that child.Let me outline the steps:1. Topological Sort: First, we need to topologically sort the DAG. This can be done using Kahn's algorithm or DFS-based approach. Since the graph is a family tree, which is a DAG, it should be straightforward.2. Dynamic Programming Setup: For each node, we'll keep track of the maximum weight path ending at that node. Let's call this `max_weight[node]`. Initially, each node's `max_weight` is just its own weight because the path could start at itself.3. Processing in Topological Order: For each node `u` in the topological order, we look at all its children `v`. For each child, we check if `max_weight[u] + weight[v]` is greater than `max_weight[v]`. If it is, we update `max_weight[v]` to this new value. This way, each node's maximum path is built upon the maximum paths of its ancestors.4. Finding the Longest Path: After processing all nodes, the longest path will be the maximum value in `max_weight`. The corresponding path can be reconstructed by backtracking from the node with the maximum weight.Now, considering the time complexity. Topological sorting takes O(V + E) time, where V is the number of nodes and E is the number of edges. Then, processing each node and its edges is also O(V + E). So overall, the algorithm runs in O(V + E) time, which is efficient for this problem.Moving on to the second part. The student wants to handle updates efficiently, especially adding a new individual or altering relationships. They're using an adjacency matrix, which is a 2D array where `matrix[i][j]` is true if there's an edge from node i to node j.When adding a new individual, we need to add a new row and column to the adjacency matrix. If the matrix is stored as a list of lists, adding a new row is O(V) time because we have to add a new list of size V. Similarly, adding a new column requires iterating through each row and adding a new element, which is also O(V) time. So, the total time complexity for adding a new node is O(V).But if the graph is large, say with thousands of nodes, this could be slow because each addition would require O(V) operations. To minimize the computational cost, maybe using a different data structure would be better. An adjacency list uses less space and allows for more efficient updates. When adding a new node, you just append a new entry to the list, which is O(1) time. Adding edges would be O(1) per edge as well.However, the student is currently using an adjacency matrix. If they want to stick with it for some reason, perhaps they can use a more dynamic data structure, like a dictionary of dictionaries, where each node points to its children. This way, adding a new node doesn't require resizing the entire matrix, but rather just adding new keys to the dictionaries. This would make adding nodes more efficient, with O(1) time for the node itself and O(1) per edge added.Alternatively, if they need to keep the adjacency matrix for certain operations, they could use a sparse matrix representation, which only stores the edges that exist. This would save space and potentially make updates faster, especially if the graph isn't too dense.In summary, while the adjacency matrix has O(V) time complexity for adding a node, switching to an adjacency list or a more dynamic structure like a dictionary could significantly reduce the time complexity for updates, making the software more efficient when handling frequent changes.</think>

🧠:<think>Okay, so I have this problem about Liam, a retired farmer from Northern Ireland, who used to participate in ploughing contests. The field he had to plough was rectangular, 300 meters by 200 meters. He had to plough the entire field in straight parallel lines, each spanning the width of the field, which is 200 meters, and spaced 1 meter apart. There are two parts to the problem. The first part is about calculating the total additional length of ploughed lines due to sinusoidal deviations. The second part is about calculating the additional time required because of these deviations, given that it takes him 5 minutes to plough 1 meter of line.Let me start with the first part.So, the field is 300 meters long and 200 meters wide. He's ploughing parallel lines that span the width, so each line is 200 meters long. These lines are spaced 1 meter apart. So, how many lines does he have to plough? Well, the field is 300 meters long, and each line is spaced 1 meter apart. So, the number of lines would be 300 meters divided by 1 meter spacing, which is 300 lines. But wait, actually, if you have a field of length L and you're spacing lines S meters apart, the number of lines is L / S. But since the field is 300 meters, and each line is 1 meter apart, starting from 0, the number of lines would actually be 301? Hmm, wait, no. Because if you have a field of length 300 meters, and you start at 0, then the next line is at 1 meter, and so on, up to 300 meters. So, actually, it's 301 lines. Hmm, but wait, no, because each line is 200 meters long, so the starting point is at 0 meters, and the next line is at 1 meter, and so on, until 299 meters, because 300 meters would be the end. So, actually, the number of lines is 300. Because from 0 to 299 meters, that's 300 lines spaced 1 meter apart. Yeah, that makes sense.So, 300 lines, each 200 meters long. So, the total straight-line length he would have ploughed is 300 lines * 200 meters = 60,000 meters.But because of the sinusoidal deviations, each line isn't straight. The deviation is described by the function y = A sin(kx + φ). Given that A is 0.5 meters, k is π/100, and φ is 0. So, the equation simplifies to y = 0.5 sin(πx / 100).So, each ploughed line isn't straight; it's wavy. The question is, how much longer is each wavy line compared to the straight line? Then, we can multiply that additional length per line by the number of lines to get the total additional length.So, the key is to find the length of one sinusoidal curve over the interval of the field's length, which is 300 meters, and then subtract the straight-line distance to find the additional length. Then, multiply that by the number of lines.Wait, actually, each line is 200 meters long, but the sinusoidal function is in terms of x, which is the length along the field. So, each line is 200 meters in width, but the sinusoidal deviation is along the length of the field, which is 300 meters. Hmm, maybe I need to clarify the coordinate system.Wait, perhaps the field is 300 meters in length and 200 meters in width. So, when ploughing, each line is along the width, which is 200 meters, and the lines are spaced 1 meter apart along the length, which is 300 meters. So, each line is 200 meters long, and the spacing between lines is 1 meter along the 300-meter length.So, the sinusoidal deviation is along the direction of ploughing, which is along the width. Wait, no, the function is y = A sin(kx + φ). So, is x the direction along the length or the width?Hmm, perhaps I need to think about this. The field is 300 meters by 200 meters. So, if he's ploughing along the width, which is 200 meters, then each line is 200 meters long. The sinusoidal deviation would be in the direction perpendicular to the ploughing direction, which would be along the length of the field.Wait, so the ploughing is along the width, 200 meters, and the sinusoidal deviation is along the length, 300 meters. So, the function y = 0.5 sin(πx / 100) would describe the deviation along the length as a function of position along the width.Wait, that might be a bit confusing. Let me try to visualize it.Imagine the field as a rectangle, 300 meters long (let's say along the x-axis) and 200 meters wide (along the y-axis). Liam is ploughing along the width, so each ploughing line is from one end to the other along the y-axis, 200 meters. But due to the sinusoidal deviation, instead of ploughing a straight line along y, he deviates sinusoidally along the x-axis.So, for each line, which is supposed to be along y from 0 to 200 meters, he instead ploughs a curve where his x-coordinate varies sinusoidally as a function of y. So, the function would be x = A sin(ky + φ). Wait, but the given function is y = A sin(kx + φ). Hmm, maybe I need to clarify.Wait, perhaps the coordinate system is such that x is along the length of the field (300 meters) and y is along the width (200 meters). So, each ploughing line is along the y-axis, from y=0 to y=200, but the deviation is along the x-axis, so the position along x is given by y = A sin(kx + φ). Wait, that seems conflicting.Wait, maybe the function is given as y = A sin(kx + φ), where x is the position along the ploughing direction. So, if he's ploughing along the y-axis, then x would be the position along the y-axis. So, for each line, as he ploughs from y=0 to y=200, his x-coordinate deviates sinusoidally as y increases.So, the function is y = 0.5 sin(πx / 100), but x here is the position along the ploughing direction, which is the y-axis. So, actually, maybe it's better to denote the ploughing direction as s, from 0 to 200 meters, and the deviation is x(s) = 0.5 sin(πs / 100). So, for each ploughing line, as he moves along s from 0 to 200 meters, his x-coordinate deviates sinusoidally.Therefore, the length of each ploughed line is the arc length of the curve x(s) = 0.5 sin(πs / 100) from s=0 to s=200.So, the arc length of a curve given by x(s) is the integral from s=0 to s=200 of sqrt(1 + (dx/ds)^2) ds.So, let's compute that.First, find dx/ds. Given x(s) = 0.5 sin(πs / 100), so dx/ds = 0.5 * (π / 100) cos(πs / 100) = (π / 200) cos(πs / 100).So, (dx/ds)^2 = (π^2 / 40000) cos^2(πs / 100).Therefore, the integrand becomes sqrt(1 + (π^2 / 40000) cos^2(πs / 100)).So, the arc length L is the integral from 0 to 200 of sqrt(1 + (π^2 / 40000) cos^2(πs / 100)) ds.Hmm, that integral doesn't look straightforward. Maybe we can approximate it or find a way to compute it.Alternatively, perhaps we can use the formula for the length of a sine wave. The general formula for the length of one period of a sine wave y = A sin(kx) is 4A sqrt(k^2 + (1/(2π))^2). Wait, no, that might not be correct.Wait, actually, the length of one period of a sine wave y = A sin(kx) over a period T = 2π/k is given by the integral from 0 to T of sqrt(1 + (A k cos(kx))^2) dx.Which is 4A sqrt(k^2 + (1/(2π))^2). Wait, no, that doesn't seem right. Let me think.Wait, actually, the exact integral is 4A sqrt(k^2 + (1/(2π))^2) multiplied by something? Hmm, maybe I need to recall the formula.Alternatively, for small amplitudes, the length can be approximated by the straight-line distance plus a term proportional to the square of the amplitude. But in this case, the amplitude is 0.5 meters, and the wavelength is 2π/k = 2π/(π/100) = 200 meters. So, the wavelength is 200 meters, which is the same as the length of each ploughing line.Wait, that's interesting. So, each ploughing line is 200 meters long, and the sinusoidal deviation has a wavelength of 200 meters. So, each line has exactly one wavelength of deviation.Therefore, the curve is a sine wave with one complete cycle over the length of the ploughing line.So, the length of one wavelength of a sine wave is given by 4A sqrt(1 + (π/(2A))^2). Wait, no, that might not be correct.Wait, let me think again. The length of a sine wave over one period is given by the integral from 0 to T of sqrt(1 + (dy/dx)^2) dx, where y = A sin(kx), and T = 2π/k.So, dy/dx = A k cos(kx), so (dy/dx)^2 = A^2 k^2 cos^2(kx).Therefore, the integrand is sqrt(1 + A^2 k^2 cos^2(kx)).So, the integral becomes ∫₀^T sqrt(1 + A^2 k^2 cos^2(kx)) dx.This integral doesn't have a closed-form solution, but it can be expressed in terms of elliptic integrals. Alternatively, for small amplitudes, we can approximate it.But in this case, A = 0.5 meters, and k = π/100, so A k = 0.5 * π/100 ≈ 0.0157. So, A k is small, which suggests that the amplitude is small relative to the wavelength, so we can approximate the integral using a Taylor expansion.The expansion for sqrt(1 + ε cos^2(kx)) is approximately 1 + (ε/2) cos^2(kx) - (ε^2)/8 cos^4(kx) + ..., where ε = A^2 k^2.So, since ε is small, we can approximate the integral as:∫₀^T [1 + (A^2 k^2 / 2) cos^2(kx)] dx.Therefore, the length L ≈ ∫₀^T 1 dx + (A^2 k^2 / 2) ∫₀^T cos^2(kx) dx.Compute each integral:First integral: ∫₀^T 1 dx = T.Second integral: ∫₀^T cos^2(kx) dx.We know that cos^2(kx) = (1 + cos(2kx))/2, so:∫₀^T cos^2(kx) dx = ∫₀^T (1 + cos(2kx))/2 dx = (T/2) + (1/(4k)) sin(2kx) evaluated from 0 to T.But sin(2kT) = sin(2k*(2π/k)) = sin(4π) = 0, similarly sin(0) = 0. So, the second term is zero.Therefore, ∫₀^T cos^2(kx) dx = T/2.So, putting it all together:L ≈ T + (A^2 k^2 / 2) * (T / 2) = T + (A^2 k^2 T) / 4.So, substituting T = 2π/k:L ≈ (2π/k) + (A^2 k^2 * 2π/k) / 4 = (2π/k) + (A^2 k * 2π) / 4 = (2π/k) + (A^2 π k) / 2.So, the additional length per line is L - T = (A^2 π k)/2.Wait, that seems too simple. Let me check.Wait, no, actually, L ≈ T + (A^2 k^2 T)/4, so the additional length is (A^2 k^2 T)/4.But T = 2π/k, so substituting:Additional length ≈ (A^2 k^2 * 2π/k)/4 = (A^2 k * 2π)/4 = (A^2 π k)/2.Yes, that's correct.So, the additional length per line is (A^2 π k)/2.Given that A = 0.5 meters, k = π/100, so:Additional length ≈ (0.5^2 * π * (π/100)) / 2 = (0.25 * π^2 / 100) / 2 = (π^2 / 400) / 2 = π^2 / 800 ≈ (9.8696)/800 ≈ 0.012337 meters.So, approximately 0.01234 meters additional length per line.But wait, that seems really small. Is that correct?Wait, let me think again. The approximation was for small A k, which in this case, A k = 0.5 * π/100 ≈ 0.0157, which is indeed small, so the approximation should be reasonable.But let me check the exact integral for a sine wave over one period.The exact length of one period of y = A sin(kx) is 4A sqrt(k^2 + (1/(2π))^2). Wait, no, that doesn't seem right.Wait, actually, the exact length is given by 4A sqrt(1 + (k)^2) * E(k), where E is the complete elliptic integral of the second kind. Hmm, that's more complicated.Alternatively, maybe I can use a better approximation.Wait, another way to approximate the arc length is to use the formula:L ≈ T + (A^2 k^2 / 4) * T.Wait, that's similar to what I had before.But let me check with a different approach.The arc length of a sine wave over one period can be approximated as:L ≈ 4A sqrt(1 + (k)^2) * (1 - (1/2) sin^2(θ)) where θ is some angle. Hmm, maybe not.Alternatively, I remember that for small amplitudes, the arc length can be approximated as L ≈ T + (π A^2)/(2λ), where λ is the wavelength.Wait, let's see. The wavelength λ = 2π/k. So, substituting, we get L ≈ T + (π A^2)/(2*(2π/k)) ) = T + (π A^2 k)/(4π) ) = T + (A^2 k)/4.Wait, that's similar to what I had earlier.So, in our case, T = 200 meters, A = 0.5, k = π/100.So, L ≈ 200 + (0.5^2 * π/100)/4 = 200 + (0.25 * π / 100)/4 = 200 + (π / 400)/4 = 200 + π / 1600 ≈ 200 + 0.0019635 ≈ 200.0019635 meters.Wait, that's a very small additional length, about 0.002 meters per line.But earlier, I had a different approximation giving 0.01234 meters. Hmm, conflicting results.Wait, perhaps I made a mistake in the first approximation.Let me go back.The first approximation was:L ≈ T + (A^2 k^2 T)/4.Given T = 200, A = 0.5, k = π/100.So, A^2 k^2 = 0.25 * (π^2 / 10000) ≈ 0.25 * 0.00098696 ≈ 0.00024674.Then, (A^2 k^2 T)/4 ≈ 0.00024674 * 200 / 4 ≈ 0.00024674 * 50 ≈ 0.012337 meters.So, that's about 0.01234 meters additional length per line.But when I used another approach, I got L ≈ T + (A^2 k)/4 ≈ 200 + 0.0019635.Wait, that's inconsistent.I think the confusion arises from different approximations.Wait, perhaps I should use the exact formula for the arc length of a sine wave over one period.The exact arc length is given by:L = 4A * sqrt(k^2 + (1/(2π))^2) * E(k), where E(k) is the complete elliptic integral of the second kind.But that might be too complicated.Alternatively, I can use the series expansion for the arc length.The arc length of y = A sin(kx) over one period is:L = (2π/k) * sqrt(1 + (A k)^2) * [1 - (1/8)(A k)^2 + ...]Wait, that's another approximation.So, if we take the first two terms, we have:L ≈ (2π/k) * sqrt(1 + (A k)^2) ≈ (2π/k) * [1 + (1/2)(A k)^2].So, L ≈ (2π/k) + (2π/k) * (1/2)(A k)^2 = (2π/k) + (π A^2 k)/2.Which is the same as the first approximation.So, substituting the values:(2π/k) = 2π / (π/100) ) = 200 meters.And (π A^2 k)/2 = π * (0.5)^2 * (π/100) / 2 = π * 0.25 * π / 100 / 2 = (π^2 / 400) / 2 = π^2 / 800 ≈ 9.8696 / 800 ≈ 0.012337 meters.So, the total length is approximately 200 + 0.012337 ≈ 200.012337 meters.Therefore, the additional length per line is approximately 0.012337 meters.So, that seems consistent.Therefore, each line is approximately 0.012337 meters longer due to the sinusoidal deviation.Now, since there are 300 lines, the total additional length is 300 * 0.012337 ≈ 3.7011 meters.Wait, so approximately 3.7 meters of additional length.But let me verify this with another method.Alternatively, perhaps I can compute the exact integral numerically.The integral for the arc length is:L = ∫₀^200 sqrt(1 + (π/200 cos(π s / 100))^2) ds.Let me compute this integral numerically.First, let's note that the integrand is sqrt(1 + (π/200 cos(π s / 100))^2).Let me make a substitution: let θ = π s / 100, so s = 100θ / π, and ds = 100 dθ / π.When s = 0, θ = 0; when s = 200, θ = 2π.So, the integral becomes:L = ∫₀^{2π} sqrt(1 + (π/200 cos θ)^2) * (100 / π) dθ.Simplify:L = (100 / π) ∫₀^{2π} sqrt(1 + (π^2 / 40000) cos^2 θ) dθ.So, L = (100 / π) * ∫₀^{2π} sqrt(1 + (π^2 / 40000) cos^2 θ) dθ.This integral is symmetric over 0 to 2π, so we can compute it as 4 times the integral from 0 to π/2.But regardless, let's compute it numerically.Let me approximate the integral using Simpson's rule or another numerical method.But since I don't have a calculator here, I can use the fact that for small ε = π^2 / 40000 ≈ 0.00024674, the integral can be approximated as:∫₀^{2π} sqrt(1 + ε cos^2 θ) dθ ≈ 2π [1 + (ε/4)].Wait, that's another approximation for small ε.So, ∫₀^{2π} sqrt(1 + ε cos^2 θ) dθ ≈ 2π [1 + (ε/4)].Therefore, L ≈ (100 / π) * 2π [1 + (ε/4)] = 200 [1 + (ε/4)].Substituting ε = π^2 / 40000:L ≈ 200 [1 + (π^2 / 160000)] ≈ 200 + 200*(π^2 / 160000) ≈ 200 + (π^2 / 800) ≈ 200 + 0.012337 ≈ 200.012337 meters.So, same result as before.Therefore, the additional length per line is approximately 0.012337 meters.So, total additional length for 300 lines is 300 * 0.012337 ≈ 3.7011 meters.So, approximately 3.7 meters.Therefore, the answer to part 1 is approximately 3.7 meters.Now, moving on to part 2.Given that it takes Liam 5 minutes to plough 1 meter of line, we need to calculate the total additional time required due to the sinusoidal deviations.So, the additional length is approximately 3.7 meters, as calculated above.Therefore, the additional time is 3.7 meters * 5 minutes/meter = 18.5 minutes.So, approximately 18.5 minutes.But let me double-check the calculations.First, additional length per line: ~0.01234 meters.Number of lines: 300.Total additional length: 300 * 0.01234 ≈ 3.702 meters.Time per meter: 5 minutes.Total additional time: 3.702 * 5 ≈ 18.51 minutes.So, approximately 18.5 minutes.But let me think if there's another way to approach this.Alternatively, since the additional length per line is 0.01234 meters, and each line takes 5 minutes per meter, the additional time per line is 0.01234 * 5 ≈ 0.0617 minutes per line.Then, total additional time is 300 * 0.0617 ≈ 18.51 minutes.Same result.Therefore, the total additional time is approximately 18.5 minutes.But let me think if there's a more precise way to calculate the arc length without approximation.Alternatively, perhaps I can use the exact integral.The exact arc length is given by:L = ∫₀^200 sqrt(1 + (π/200 cos(π s / 100))^2) ds.Let me make a substitution: let u = π s / 100, so s = 100u / π, ds = 100 du / π.When s = 0, u = 0; s = 200, u = 2π.So, L = ∫₀^{2π} sqrt(1 + (π/200 cos u)^2) * (100 / π) du.Simplify:L = (100 / π) ∫₀^{2π} sqrt(1 + (π^2 / 40000) cos^2 u) du.This integral is known as an elliptic integral of the second kind.The complete elliptic integral of the second kind, E(m), is defined as:E(m) = ∫₀^{π/2} sqrt(1 - m sin^2 θ) dθ.But our integral is ∫₀^{2π} sqrt(1 + k cos^2 u) du, where k = π^2 / 40000.We can express this as 4 ∫₀^{π/2} sqrt(1 + k cos^2 u) du, due to symmetry.But to relate it to the standard elliptic integral, we can use the identity:∫₀^{2π} sqrt(1 + k cos^2 u) du = 4 ∫₀^{π/2} sqrt(1 + k cos^2 u) du.And, using the identity that ∫₀^{π/2} sqrt(1 + k cos^2 u) du = E(-k).Wait, actually, the standard form is E(m) = ∫₀^{π/2} sqrt(1 - m sin^2 θ) dθ.But we have sqrt(1 + k cos^2 u). Let's express this in terms of sin.Note that cos^2 u = 1 - sin^2 u, so:sqrt(1 + k cos^2 u) = sqrt(1 + k - k sin^2 u) = sqrt((1 + k) - k sin^2 u).So, we can write it as sqrt(1 + k) * sqrt(1 - (k / (1 + k)) sin^2 u).Therefore, ∫₀^{π/2} sqrt(1 + k cos^2 u) du = sqrt(1 + k) ∫₀^{π/2} sqrt(1 - (k / (1 + k)) sin^2 u) du = sqrt(1 + k) E(k / (1 + k)).Therefore, the integral ∫₀^{2π} sqrt(1 + k cos^2 u) du = 4 sqrt(1 + k) E(k / (1 + k)).So, substituting back into L:L = (100 / π) * 4 sqrt(1 + k) E(k / (1 + k)).Where k = π^2 / 40000 ≈ 0.00024674.So, let's compute this.First, compute sqrt(1 + k) ≈ sqrt(1 + 0.00024674) ≈ 1 + 0.00012337 ≈ 1.00012337.Next, compute m = k / (1 + k) ≈ 0.00024674 / 1.00024674 ≈ 0.00024667.Now, we need to compute E(m), the complete elliptic integral of the second kind for m ≈ 0.00024667.For small m, E(m) ≈ π/2 [1 - (1/4)m - (3/64)m^2 - ...].So, E(m) ≈ π/2 [1 - (1/4)m].Substituting m ≈ 0.00024667:E(m) ≈ π/2 [1 - 0.000061667] ≈ π/2 * 0.999938333 ≈ (1.570796327) * 0.999938333 ≈ 1.570756.Therefore, L ≈ (100 / π) * 4 * 1.00012337 * 1.570756.Compute step by step:First, 4 * 1.00012337 ≈ 4.00049348.Then, 4.00049348 * 1.570756 ≈ 6.283185307 (since 4 * π ≈ 12.56637, but here it's 4.00049348 * 1.570756 ≈ 6.283185307).Wait, actually, 4 * 1.570756 ≈ 6.283024, and 0.00049348 * 1.570756 ≈ 0.000773, so total ≈ 6.283024 + 0.000773 ≈ 6.283797.Then, L ≈ (100 / π) * 6.283797 ≈ (100 / 3.141592654) * 6.283797 ≈ 31.83098862 * 6.283797 ≈ 200.012337 meters.So, same result as before.Therefore, the exact calculation gives us the same approximate additional length per line as the approximation.Therefore, the total additional length is approximately 3.7011 meters, and the additional time is approximately 18.5055 minutes.So, rounding to a reasonable precision, perhaps 3.7 meters and 18.5 minutes.But let me check if the question expects an exact answer or if it's okay to approximate.The problem states that the sinusoidal function is given, and it's a math problem, so perhaps we can express the additional length in terms of π.From the earlier approximation, the additional length per line is (A^2 π k)/2.Given A = 0.5, k = π/100.So, additional length per line = (0.5^2 * π * π/100)/2 = (0.25 * π^2 / 100)/2 = π^2 / 800.Therefore, total additional length = 300 * π^2 / 800 = (300 / 800) π^2 = (3/8) π^2 ≈ (3/8)*9.8696 ≈ 3.7011 meters.So, exactly, it's (3/8) π^2 meters.Similarly, the additional time is (3/8) π^2 meters * 5 minutes/meter = (15/8) π^2 minutes ≈ 18.5055 minutes.So, perhaps expressing the answers in terms of π^2 is acceptable.Therefore, the total additional length is (3/8) π^2 meters, and the additional time is (15/8) π^2 minutes.But let me compute (3/8) π^2:π^2 ≈ 9.8696, so (3/8)*9.8696 ≈ 3.7011 meters.Similarly, (15/8)*9.8696 ≈ 18.5055 minutes.So, the answers are approximately 3.7011 meters and 18.5055 minutes.But perhaps the question expects an exact expression in terms of π.So, total additional length = (3/8) π^2 meters.Total additional time = (15/8) π^2 minutes.Alternatively, if we want to write it as a multiple of π^2, that's fine.But let me check if my initial assumption about the number of lines is correct.Earlier, I thought it's 300 lines because the field is 300 meters long, spaced 1 meter apart. So, starting at 0, the next line is at 1, ..., up to 299, making 300 lines.But actually, if the field is 300 meters long, and each line is spaced 1 meter apart, the number of lines is 301. Because from 0 to 300 meters, inclusive, you have 301 lines.Wait, that's a crucial point.So, if the field is 300 meters long, and the lines are spaced 1 meter apart, starting at 0, the next line is at 1, then 2, ..., up to 300 meters. So, that's 301 lines.Therefore, the number of lines is 301, not 300.So, that changes the total additional length and time.So, let me recalculate.Additional length per line: π^2 / 800 ≈ 0.012337 meters.Total additional length: 301 * 0.012337 ≈ 3.703 meters.Similarly, total additional time: 301 * 0.012337 * 5 ≈ 3.703 * 5 ≈ 18.515 minutes.But let's express it exactly.Total additional length = 301 * (π^2 / 800) = (301/800) π^2.Similarly, total additional time = 301 * (π^2 / 800) * 5 = (1505/800) π^2 = (301/160) π^2.But 301 is a prime number, so it can't be simplified.Alternatively, perhaps the question assumes 300 lines, given that the field is 300 meters long, and the lines are spaced 1 meter apart, so 300 lines.But actually, in reality, the number of lines would be 301, because you include both endpoints.So, perhaps the correct number is 301.But since the problem says "spaced 1 meter apart," it's a bit ambiguous whether it's 300 or 301 lines.But in agricultural terms, when you have a field of length L and you space lines S meters apart, the number of lines is typically L / S, but whether it's inclusive or exclusive depends on the context.But given that the field is 300 meters, and the lines are spaced 1 meter apart, starting at 0, the next line is at 1, ..., up to 299, making 300 lines, each 1 meter apart, covering 300 meters.Wait, no, because 300 lines spaced 1 meter apart would cover 299 meters, right? Because the distance between the first and last line is (n-1)*S.So, if you have n lines spaced S meters apart, the total length covered is (n-1)*S.Therefore, to cover 300 meters, you need (n-1)*1 = 300, so n = 301 lines.Therefore, the correct number of lines is 301.So, that changes the total additional length and time.Therefore, total additional length = 301 * (π^2 / 800) ≈ 301 * 0.012337 ≈ 3.703 meters.Total additional time = 3.703 * 5 ≈ 18.515 minutes.But let's compute it exactly:Total additional length = 301 * (π^2 / 800) = (301/800) π^2.Total additional time = (301/800) π^2 * 5 = (1505/800) π^2 = (301/160) π^2.But 301/160 is approximately 1.88125.So, 1.88125 * π^2 ≈ 1.88125 * 9.8696 ≈ 18.515 minutes.Therefore, the answers are approximately 3.703 meters and 18.515 minutes.But since the problem might expect an exact answer in terms of π, we can write:Total additional length = (301/800) π^2 meters.Total additional time = (301/160) π^2 minutes.Alternatively, if we consider that the number of lines is 300, then:Total additional length = 300 * (π^2 / 800) = (3/8) π^2 ≈ 3.7011 meters.Total additional time = (3/8) π^2 * 5 = (15/8) π^2 ≈ 18.5055 minutes.But given the field is 300 meters, and lines are spaced 1 meter apart, the correct number of lines is 301.Therefore, the answers should be based on 301 lines.But to be safe, perhaps the problem assumes 300 lines, as it's a math problem, and sometimes they simplify.But in reality, it's 301 lines.However, since the problem says "spaced 1 meter apart," it's more accurate to say 301 lines.But let me check the problem statement again."he had to plough the entire field in straight parallel lines, each line spanning the width of the field (200 meters) and spaced 1 meter apart."So, "spaced 1 meter apart" usually means the distance between adjacent lines is 1 meter. So, the number of lines is (length / spacing) + 1.Therefore, 300 / 1 + 1 = 301 lines.Yes, that's correct.Therefore, the total additional length is 301 * (π^2 / 800) meters.Total additional time is 301 * (π^2 / 800) * 5 minutes.So, to express the answers:1. Total additional length: (301/800) π² meters ≈ 3.703 meters.2. Total additional time: (301/160) π² minutes ≈ 18.515 minutes.But perhaps the problem expects the answer in terms of π², so we can write:1. (301/800) π² meters.2. (301/160) π² minutes.Alternatively, simplifying the fractions:301 and 800 have no common factors, so 301/800 is simplest.Similarly, 301/160 is simplest.Alternatively, we can write it as:1. (301 π²) / 800 meters.2. (301 π²) / 160 minutes.But perhaps the problem expects a numerical approximation.Given that, 3.703 meters and 18.515 minutes.But let me compute (301/800) π²:301/800 = 0.37625.0.37625 * π² ≈ 0.37625 * 9.8696 ≈ 3.703 meters.Similarly, 301/160 = 1.88125.1.88125 * π² ≈ 1.88125 * 9.8696 ≈ 18.515 minutes.Therefore, the answers are approximately 3.703 meters and 18.515 minutes.But since the problem might expect an exact answer, perhaps in terms of π², we can leave it as (301/800) π² meters and (301/160) π² minutes.Alternatively, if we consider that the number of lines is 300, then:1. (3/8) π² meters ≈ 3.7011 meters.2. (15/8) π² minutes ≈ 18.5055 minutes.But given the correct number of lines is 301, I think the answers should be based on 301.Therefore, the final answers are:1. Approximately 3.703 meters.2. Approximately 18.515 minutes.But to express them exactly, we can write:1. (301/800) π² meters.2. (301/160) π² minutes.Alternatively, if we want to write it as a single fraction:1. (301 π²)/800 meters.2. (301 π²)/160 minutes.But let me check the problem statement again.It says, "the entire field in straight parallel lines, each line spanning the width of the field (200 meters) and spaced 1 meter apart."So, the field is 300 meters long, and lines are spaced 1 meter apart along the length.Therefore, the number of lines is 300 / 1 + 1 = 301.Yes, that's correct.Therefore, the total additional length is 301 * (π² / 800) meters.Total additional time is 301 * (π² / 800) * 5 minutes.So, the answers are:1. (301 π²)/800 meters ≈ 3.703 meters.2. (301 π²)/160 minutes ≈ 18.515 minutes.But perhaps the problem expects the answer in terms of π², so we can write:1. (301/800) π² meters.2. (301/160) π² minutes.Alternatively, if we want to write it as a decimal, we can round it to two decimal places:1. Approximately 3.70 meters.2. Approximately 18.52 minutes.But since the problem might expect an exact answer, perhaps in terms of π², we can leave it as is.Alternatively, if we consider that the number of lines is 300, then:1. (3/8) π² meters ≈ 3.7011 meters.2. (15/8) π² minutes ≈ 18.5055 minutes.But given the correct number of lines is 301, I think the answers should be based on 301.Therefore, the final answers are:1. (301/800) π² meters ≈ 3.703 meters.2. (301/160) π² minutes ≈ 18.515 minutes.But to express them as exact values, we can write:1. boxed{dfrac{301 pi^2}{800} text{ meters}}2. boxed{dfrac{301 pi^2}{160} text{ minutes}}Alternatively, if the problem expects a numerical approximation, we can write:1. boxed{3.70 text{ meters}}2. boxed{18.52 text{ minutes}}But since the problem might expect an exact answer, I think the first option is better.Therefore, the answers are:1. boxed{dfrac{301 pi^2}{800} text{ meters}}2. boxed{dfrac{301 pi^2}{160} text{ minutes}}</think>

🧠:<think>Okay, so I've got this problem about a mortgage lender processing 100 loan applications. They've modeled the loans using Gaussian distributions for both the loan amounts and interest rates. I need to solve two parts here.Starting with part 1: Calculate the probability that a randomly selected loan has an amount between 250,000 and 350,000 and an interest rate between 4% and 6%. They mentioned that loan amounts and interest rates are independent, so I can handle each distribution separately and then multiply the probabilities.First, for the loan amounts. The mean is 300,000 and the standard deviation is 50,000. I need to find the probability that a loan is between 250,000 and 350,000. Since it's a normal distribution, I can standardize these values to Z-scores.For 250,000:Z = (250,000 - 300,000) / 50,000 = (-50,000) / 50,000 = -1For 350,000:Z = (350,000 - 300,000) / 50,000 = 50,000 / 50,000 = 1So, I need the probability that Z is between -1 and 1. From the standard normal distribution table, the area from -1 to 1 is approximately 0.6827. So, about 68.27% chance for the loan amount.Next, the interest rates. The mean is 5% and standard deviation is 1%. We need the probability that the rate is between 4% and 6%.Again, standardizing:For 4%:Z = (4 - 5) / 1 = -1For 6%:Z = (6 - 5) / 1 = 1Same as the loan amount, the probability is about 0.6827 or 68.27%.Since the two variables are independent, the joint probability is the product of the two individual probabilities. So, 0.6827 * 0.6827.Let me calculate that: 0.6827 * 0.6827. Hmm, 0.68 * 0.68 is 0.4624, and the extra 0.0027 on each would add a bit more. Let me do it more accurately:0.6827 * 0.6827. Let's compute 0.6827 squared.0.6827 * 0.6827:First, 0.6 * 0.6 = 0.360.6 * 0.0827 = 0.049620.0827 * 0.6 = 0.049620.0827 * 0.0827 ≈ 0.00684Adding them up: 0.36 + 0.04962 + 0.04962 + 0.00684 ≈ 0.4661Wait, that seems a bit rough. Alternatively, using a calculator approach:0.6827 * 0.6827:Multiply 6827 * 6827, then adjust the decimal.But maybe it's easier to use the fact that (0.68 + 0.0027)^2 ≈ 0.68^2 + 2*0.68*0.0027 + 0.0027^2 ≈ 0.4624 + 0.003672 + 0.00000729 ≈ 0.46607929So approximately 0.4661, which is about 46.61%.So, the probability is approximately 46.61%.Moving on to part 2: The lender wants to maintain a default rate of less than 5%. The probability of default is given by a logistic function: P(default) = 1 / (1 + e^{-(a + bX)}), where X is the interest rate, a = -4, b = 0.5.We need to find the maximum average interest rate such that the default rate is below 5%. So, we need to set P(default) = 0.05 and solve for X.So, 0.05 = 1 / (1 + e^{-( -4 + 0.5X )})Let me write that equation:0.05 = 1 / (1 + e^{-( -4 + 0.5X )})First, take reciprocals on both sides:1 / 0.05 = 1 + e^{-( -4 + 0.5X )}Which is 20 = 1 + e^{-( -4 + 0.5X )}Subtract 1:19 = e^{-( -4 + 0.5X )}Take natural logarithm on both sides:ln(19) = -( -4 + 0.5X )Simplify the right side:ln(19) = 4 - 0.5XNow, solve for X:0.5X = 4 - ln(19)Calculate ln(19): ln(19) ≈ 2.9444So,0.5X = 4 - 2.9444 = 1.0556Multiply both sides by 2:X = 2.1112So, approximately 2.1112%. But wait, that seems low. Let me double-check.Wait, the logistic function is P = 1 / (1 + e^{-(a + bX)}). So, when X increases, the exponent increases, so e^{-(a + bX)} decreases, so P increases. So, higher interest rates lead to higher default probabilities? That seems counterintuitive, but maybe because higher interest rates are riskier, so more likely to default.But in the problem, they want to maintain a default rate below 5%, so they need to set the interest rate such that the default probability is 5%. So, solving for X when P=0.05.Wait, but in my calculation, I got X ≈ 2.11%, which is lower than the mean interest rate of 5%. That seems odd because if the average interest rate is 5%, but to get a 5% default rate, you need to set the interest rate to 2.11%? That would mean lowering the interest rates to reduce defaults, which makes sense because lower interest rates are less risky, so less likely to default.But the question is asking for the maximum average interest rate to keep the default rate below 5%. So, if I set the average interest rate to 2.11%, then the default rate would be 5%. If I set it higher, the default rate would go above 5%. So, the maximum average interest rate is approximately 2.11%.But let me check my steps again.Starting equation:0.05 = 1 / (1 + e^{-( -4 + 0.5X )})Multiply both sides by denominator:0.05 * (1 + e^{-( -4 + 0.5X )}) = 1Divide both sides by 0.05:1 + e^{-( -4 + 0.5X )} = 20Subtract 1:e^{-( -4 + 0.5X )} = 19Take ln:-( -4 + 0.5X ) = ln(19)So,4 - 0.5X = ln(19)Then,-0.5X = ln(19) - 4Multiply both sides by -2:X = (4 - ln(19)) * 2Compute 4 - ln(19):ln(19) ≈ 2.94444 - 2.9444 ≈ 1.0556Multiply by 2:X ≈ 2.1112Yes, so approximately 2.11%. So, the maximum average interest rate is about 2.11% to keep the default rate below 5%.Wait, but the mean interest rate is 5%, which is higher than 2.11%. So, if the average is 5%, the default rate would be higher than 5%, which is not acceptable. Therefore, to keep the default rate below 5%, the average interest rate needs to be set lower, at 2.11%.So, that's the answer for part 2.But let me think again: is the logistic function correctly applied? The default probability increases as X increases because the exponent becomes less negative, so e^{-(a + bX)} decreases, so 1 / (1 + e^{-(a + bX)}) increases. So, higher X leads to higher default probability. Therefore, to have a lower default probability, we need a lower X. So, to have P=0.05, X must be lower than the mean of 5%.Yes, that makes sense. So, the maximum average interest rate is approximately 2.11%.But let me compute it more accurately.Compute ln(19):ln(19) ≈ 2.944438979So,4 - ln(19) ≈ 4 - 2.944438979 ≈ 1.055561021Multiply by 2:X ≈ 2.111122042%So, approximately 2.1111%, which we can round to 2.11%.Therefore, the maximum average interest rate is approximately 2.11%.But wait, the problem says "maximum average interest rate". So, if the average is 2.11%, then the default rate is exactly 5%. To keep it below 5%, the average interest rate should be slightly less than 2.11%. But since we're asked for the maximum, it's 2.11%.Alternatively, maybe they want the exact value, so 2.1111%, which is approximately 2.11%.So, summarizing:1. The probability is approximately 46.61%.2. The maximum average interest rate is approximately 2.11%.I think that's it.</think>