Problem 30
Question
It is known that homing pigeons fly faster over land than over water. Assume that they fly 10 meters per second over land but only 8 meters per second over water. If a pigeon is located at the edge of a straight river 500 meters wide and must fly to its nest, located 1300 meters away on the opposite side of the river (Figure \(4.58\) ), what path would minimize its flying time?
Step-by-Step Solution
Verified Answer
The optimal path involves flying approximately 1015.4 meters over water.
1Step 1: Understand the Problem
To minimize the flying time, we need to figure out the path that involves flying the least over the slower medium (water) and as much as possible over the faster medium (land). We need to balance the distances flown over each medium to optimize the total time.
2Step 2: Define Variables and Equations
Let the distance flown over water be \( x \) meters. Since the river is 500 meters wide, the remaining distance to fly over land is \( \sqrt{1300^2 - x^2} \) meters due to the right triangle formed. We must compute the total flight time, \( T(x) = \frac{x}{8} + \frac{\sqrt{1300^2 - x^2}}{10} \).
3Step 3: Differentiate and Set Derivative to Zero
To find the minimum time, differentiate the function \( T(x) \) with respect to \( x \), and solve for \( x \) when the derivative equals zero: \( T'(x) = \frac{1}{8} - \frac{x}{10\sqrt{1300^2 - x^2}} = 0 \).
4Step 4: Solve the Derivative Equation
Solve \( \frac{1}{8} = \frac{x}{10\sqrt{1300^2 - x^2}} \) to find \( x \). Rearrange to \( 10\sqrt{1300^2 - x^2} = 8x \) and solve for \( x \), which gives \( \sqrt{1300^2 - x^2} = 0.8x \).
5Step 5: Simplify and Calculate \( x \)
Squaring both sides, we get \( 1300^2 - x^2 = 0.64x^2 \). Solve for \( x \), giving \( x = \frac{1300}{\sqrt{1.64}} \). Calculating gives \( x \approx 1015.4 \) meters.
6Step 6: Compute and Verify Total Time
Substitute \( x = 1015.4 \) back into the original total time formula to compute \( T(x) \). Make sure the computed time using this \( x \) results in the minimum possible time compared to other values.
Key Concepts
Optimization ProblemDifferentiationHoming PigeonsRight TriangleFlight Path
Optimization Problem
In the world of Calculus, an Optimization Problem is one where we seek to find the best solution among a set of possible choices. In our problem with a homing pigeon, we look for the shortest time path for a pigeon crossing a river. We need to balance two speeds: the pigeon's speed over land (10 meters per second) and over water (8 meters per second).
The goal is to find out which combination of land and water paths will minimize the time taken by the pigeon to reach its destination. Optimization problems often involve creating a model using variables and then applying calculus to determine the most efficient solution. Here, we define a path that considers both land and water traversal to achieve the desired minimum time.
The goal is to find out which combination of land and water paths will minimize the time taken by the pigeon to reach its destination. Optimization problems often involve creating a model using variables and then applying calculus to determine the most efficient solution. Here, we define a path that considers both land and water traversal to achieve the desired minimum time.
Differentiation
Differentiation is a crucial mathematical process in optimization. It helps us find out how a function changes with respect to its variables. In this problem, we have a function representing the total travel time of the pigeon. The time depends on how much distance the pigeon covers over water and land.
To solve, we differentiate the function of time, represented by \( T(x) = \frac{x}{8} + \frac{\sqrt{1300^2 - x^2}}{10} \), where \( x \) is the distance covered over water. The differentiated function, or the derivative, tells us how the time changes as \( x \) changes. By setting this derivative to zero, \( T'(x) = 0 \), we can find the value of \( x \) that minimizes the total time. Differentiation, so to say, zeros in on the optimal scenario.
To solve, we differentiate the function of time, represented by \( T(x) = \frac{x}{8} + \frac{\sqrt{1300^2 - x^2}}{10} \), where \( x \) is the distance covered over water. The differentiated function, or the derivative, tells us how the time changes as \( x \) changes. By setting this derivative to zero, \( T'(x) = 0 \), we can find the value of \( x \) that minimizes the total time. Differentiation, so to say, zeros in on the optimal scenario.
Homing Pigeons
Homing pigeons, known for their ability to find their way home over long distances, also play into our optimization problem. They are capable of flying at different speeds over different terrains. This problem assumes the pigeon flies faster over land than over water. Real-life scenarios like these help in understanding optimization problems by applying them to practical animal behavior.
The challenge is modeled by considering physical constraints, like the speed difference and distance, reflecting how pigeons might naturally choose a path minimizing energy or time. Such applications in biology and ecology make optimization problems engaging and relevant to understanding animal navigation.
The challenge is modeled by considering physical constraints, like the speed difference and distance, reflecting how pigeons might naturally choose a path minimizing energy or time. Such applications in biology and ecology make optimization problems engaging and relevant to understanding animal navigation.
Right Triangle
A right triangle is a geometric figure crucial to resolving the optimization problem. In our scenario, the path that includes both water and land travel forms a right triangle. With one side representing the river's width and another the total distance to the destination, the hypotenuse is the part the pigeon flies over land.
To determine the land travel distance, we use the Pythagorean theorem, which tells us: \( \sqrt{1300^2 - x^2} \), where \( x \) is the distance across the river. Recognizing the right triangle in such problems helps in building equations to describe relationships between distances effectively. This leads to easier resolution of the optimization problem.
To determine the land travel distance, we use the Pythagorean theorem, which tells us: \( \sqrt{1300^2 - x^2} \), where \( x \) is the distance across the river. Recognizing the right triangle in such problems helps in building equations to describe relationships between distances effectively. This leads to easier resolution of the optimization problem.
Flight Path
The flight path is the route the pigeon will take to minimize its travel time. It's directly affected by how fast it flies over different terrains.
By computing the optimal values using differentiation and the relationship of the right triangle, we establish the best flight path. The path balances the river crossing and land flying, finding the point where spending slightly more time in the slower medium (water) results in the shortest total travel time.
The calculated path ensures that the pigeon doesn't stray too far down the river, optimizing its route to reach the nest with minimal flight time, showcasing effective use of calculus in determining optimal navigation strategies.
By computing the optimal values using differentiation and the relationship of the right triangle, we establish the best flight path. The path balances the river crossing and land flying, finding the point where spending slightly more time in the slower medium (water) results in the shortest total travel time.
The calculated path ensures that the pigeon doesn't stray too far down the river, optimizing its route to reach the nest with minimal flight time, showcasing effective use of calculus in determining optimal navigation strategies.
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