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IB DP Maths AI SL Study Notes

5.4.3 Real-world Applications

Motion

Integration plays a pivotal role in analysing motion, particularly in computing displacement and distance travelled by an object. Given a velocity function, v(t), the displacement from time a to b is given by the integral of v(t) from a to b. The total distance travelled, on the other hand, is obtained by integrating the absolute value of v(t) over the interval [a, b].

Example 1: Displacement and Distance

Consider an object with a velocity function v(t) = t2 - 4t. Find the displacement and total distance travelled from t = 0 to t = 4.

  • Displacement: Integral from 0 to 4 of (t2 - 4t) dt
  • Distance: Integral from 0 to 4 of |t2 - 4t| dt

Solution:

  • Displacement = [t3/3 - 2t2] from 0 to 4 = 16/3 - 16 = -16/3 units
  • For distance, find when v(t) = 0: t(t-4) = 0, so t = 0 or 4. Integrate the absolute value in pieces: Integral from 0 to 4 of (4t - t2) dt = [2t2 - t3/3] from 0 to 4 = 32/3 units

Accumulation

Accumulation refers to the total amount accumulated over a certain interval. In calculus, the integral of a rate of change function, such as rate of flow or production, gives the total amount accumulated over a time period.

Example 2: Accumulation of Water

If a tank fills at a rate of r(t) = 3t2 litres per hour, find the total amount of water that enters the tank from t = 1 to t = 3 hours.

Solution:

  • Total water = Integral from 1 to 3 of 3t2 dt = [t3] from 1 to 3 = 26 litres

Average Value

The average value of a function f(x) from a to b is given by (1/(b-a)) multiplied by the integral from a to b of f(x) dx. This provides the mean value of the function over the interval [a, b].

Example 3: Average Temperature

Given a temperature function T(t) = 20 + 5sin(t) degrees Celsius, find the average temperature from t = 0 to t = pi.

Solution: To find the average temperature over a given interval [a, b] for a continuous temperature function T(t), we can use the following formula: Tavg = (1/(b-a)) * integral from a to b of T(t) dt

In this case:

  • T(t) = 20 + 5sin(t) (temperature function)
  • a = 0 (start time)
  • b = pi (end time)

So, we need to set up and evaluate the integral: Tavg = (1/(pi-0)) * integral from 0 to pi of (20 + 5sin(t)) dt

Evaluating this integral, we find that: Tavg = (10 + 20pi)/pi degrees Celsius.

FAQ

Units are crucial in real-world integration problems because they give meaning to the quantities we're calculating. When integrating a velocity function, for example, the resulting displacement will have units of distance, such as metres or kilometres. Keeping track of units ensures that our calculations make physical sense and allows us to interpret our results correctly. Moreover, it helps in preventing errors; if units don't align properly in an equation or calculation, it's a clear sign that something might be amiss.

Absolutely! While area and volume are common applications, integrals can represent a wide variety of accumulated quantities. For instance, in economics, integrals can be used to find total revenue over time given a rate of income. In physics, they can determine quantities like electric charge or magnetic flux. The key is to understand that integrals represent the total accumulated value of a function over an interval, and this concept can be applied in countless real-world scenarios beyond just area and volume.

The average value of a function over an interval provides a single value that summarises the behaviour of the function over that interval. In real-world applications, this can be useful for summarising data or making comparisons. For example, the average temperature over a day can give a general idea of the day's weather, or the average speed of a car over a journey can provide insight into the overall pace of the trip. In many cases, the average value can be more informative than individual data points, especially when making decisions or assessments based on the data.

Accumulation is essentially about adding up quantities over a certain interval, and this is precisely what integrals do. When we integrate a function, we're finding the accumulated total of that function's values over a specified range. In real-world terms, this could mean finding the total amount of water that flows through a pipe over a certain time, the total distance travelled by a moving object, or any other scenario where quantities build up over time or space. Integrals provide the mathematical tool to compute these accumulated totals.

Integrals are fundamental in modelling real-world motion problems because they allow us to compute quantities accumulated over time. For instance, if we have a function representing the velocity of an object, the integral of this function over a time interval gives us the object's displacement during that interval. Similarly, if we have an acceleration function, integrating it gives us the velocity function. This makes integrals invaluable for problems where we need to determine distances travelled, changes in velocities, or other accumulated quantities in motion-related scenarios.

Practice Questions

A tank is being filled with water at a rate of r(t) = 2t^2 + 3t litres per minute, where t is the time in minutes. Calculate the total amount of water that enters the tank from t = 2 to t = 5 minutes.

To find the total amount of water that enters the tank from t = 2 to t = 5 minutes, we need to integrate the rate function r(t) over this interval. The integral of a rate of change function gives the total amount accumulated over a specified time period. Mathematically, if r(t) represents the rate of change function, the total accumulation A from time a to b is given by: A = integral from a to b of r(t) dt Substituting the given rate function and limits of integration, we get: A = integral from 2 to 5 of (2t2 + 3t) dt = [(2/3)t3 + (3/2)t2] evaluated from 2 to 5 = [(2/3)(5)3 + (3/2)(5)2] - [(2/3)(2)3 + (3/2)(2)2] = [(250/3 + 75/2) - (16/3 + 6)] = [725/6 - 34/3] = [219/2] = 109.5 So, the total amount of water that enters the tank from t = 2 to t = 5 minutes is 109.2 litres.

The temperature in a room is modelled by the function T(t) = 22 + 3sin(t) degrees Celsius, where t is the time in hours. Calculate the average temperature in the room from t = 0 to t = 2π hours.

To find the average temperature from t = 0 to t = 2π, we use the formula for the average value of a function over an interval [a, b]: Average Value = (1/(b-a)) * integral from a to b of f(x) dx Substituting the given temperature function and limits of integration, we get: Average Temperature = (1/(2π - 0)) * integral from 0 to 2π of (22 + 3sin(t)) dt = (1/2π) * [22t - 3cos(t)] evaluated from 0 to 2π = (1/2π) * [22(2π) - 3cos(2π) - (22(0) - 3cos(0))] = (1/2π) * [44π - 3 + 3] = (44π/2π) = 22 Thus, the average temperature in the room from t = 0 to t = 2π hours is 22 degrees Celsius.

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