Related Rates

Related rates are a fascinating application of differentiation in calculus that helps us understand how different quantities change in relation to one another over time. In real life, many situations involve quantities that are related to one another and change together. For example, how the volume of air inside a balloon changes as its radius increases or how the shadow of a moving person changes its length over time. Related rate problems allow us to calculate how the rate of change of one quantity affects another.

Understanding the concept

Before we dive into the examples, let's consider the basic idea. In related rate problems, you usually have a mathematical relationship between two or more quantities that can change over time. You are given the rate at which one of these quantities changes and asked to find the rate of change of the other quantity.

Let us consider two fundamental steps when solving related rate problems:

1. Find the relationship: Identify a formula or equation that connects two or more changing quantities. This is usually a geometric or algebraic relationship.
2. Differentiate with respect to time: Once you understand this relationship, differentiate it with respect to time. This allows you to relate different rates of change of the quantities involved.

Example 1: Expanding circle

Visual example

Imagine a circle with a radius (r) that is increasing over time. We are often interested in how the area of a circle changes as the radius increases. We know that the relationship between the area (A) of a circle and its radius (r) is given by the equation:

A = pi r^2

Suppose you are given that the radius (r) is increasing at a constant rate of 5 cm/sec. We want to find out how fast the area of the circle is increasing when the radius is 10 cm.

Solution to the problem

1. Differentiate the relation with respect to time (t). Remember that (A) and (r) are both functions of time (t).

(frac{dA}{dt} = frac{d}{dt}(pi r^2) = 2pi r frac{dr}{dt})

2. You are given that (frac{dr}{dt} = 5) cm/sec and you have to find (frac{dA}{dt}) when (r = 10) cm.

(frac{dA}{dt} = 2pi times 10 times 5 = 100pi)

Therefore, when the radius is 10 cm, the area of the circle is increasing at the rate of (100pi) square cm per second.

Example 2: Ladder against a wall

Text example

Consider a classic problem where a ladder rests against a vertical wall. The length of the ladder is 15 feet. The foot of the ladder is being pulled away from the wall at a rate of 2 feet per second. We are tasked with finding how fast the top of the ladder is sliding down the wall when the base of the ladder is 9 feet from the wall.

Solution to the problem

1. Establish the relationship between the distance of the base from the wall (call it (x)), the height of the ladder on the wall (call it (y)), and the length of the ladder (which is constant and equal to 15 feet). Use the Pythagorean Theorem for this right triangle:

x^2 + y^2 = 15^2

2. Differentiate both sides of the equation with respect to time (t):

(frac{d}{dt}(x^2 + y^2) = frac{d}{dt}(15^2))

2xfrac{dx}{dt} + 2yfrac{dy}{dt} = 0

We know (x = 9), (frac{dx}{dt} = 2) ft/s and we need to find (frac{dy}{dt}). First, solve for (y) using (x = 9):

9^2 + y^2 = 15^2 \ y^2 = 144 \ y = 12 quad (text{choosing the positive value since height is positive})

Now substitute these values into the differentiated equation:

2(9)(2) + 2(12)frac{dy}{dt} = 0 \ 36 + 24frac{dy}{dt} = 0 \ 24frac{dy}{dt} = -36 \ frac{dy}{dt} = -frac{36}{24} = -1.5

Thus, the top of the ladder is sliding down the wall at a rate of 1.5 feet per second, while the base is 9 feet away from the wall.

Example 3: Removing water from a cone

Text example

Imagine a conical tank with an upside-down attitude from which water is flowing out. The radius of the top surface of the tank is 5 m and the height is 10 m. Water is flowing out of the tank at the rate of 3 cubic metres per minute. At a certain time, the water level is 6 m high. You have to find the rate of fall of the water level at that time.

Grooming the relationship

Initially, the volume (V) of a cone with base radius (r) and height (h) is given as:

V = frac{1}{3}pi r^2 h

In this setup, (r) and (h) change as the water is drained, but they maintain a constant ratio. This ratio is derived from the similarity of the triangles formed by the cone as it is drained:

frac{r}{5} = frac{h}{10} \ r = frac{h}{2}

Substitute this into the volume equation:

V = frac{1}{3}pileft(frac{h}{2}right)^2h = frac{1}{12}pi h^3

3. Differentiate the volume equation with respect to time:

frac{dV}{dt} = frac{1}{12}pi times 3h^2 frac{dh}{dt} \ frac{dV}{dt} = frac{pi}{4}h^2 frac{dh}{dt}

We are given (frac{dV}{dt} = -3) m³/min (negative because volume is decreasing) and (h = 6). Substitute these into the differentiated equation:

-3 = frac{pi}{4}(6)^2frac{dh}{dt} \ -3 = 9pi frac{dh}{dt} \ frac{dh}{dt} = -frac{3}{9pi} = -frac{1}{3pi}

This result indicates that when the height of water is 6 m, the water level in the tank is falling at the rate of about (-frac{1}{3pi}) m per minute.

Conclusion

Relative rate problems give us a powerful technique for understanding how the rate of change of one quantity can affect another. It is not only a fascinating mathematical exercise but also a vital tool for solving real-life problems in fields such as physics, engineering, and even biology. By establishing relationships between quantities and differentiating them with respect to time, we gain insight into the dynamic processes around us.

In these examples, you have seen how to transform a physical situation into a mathematical model, how to use calculus to extract information about rates of change, and how to interpret those results meaningfully in the context of the problem. Practicing with different scenarios increases your understanding of this concept, making it easier to apply it to more complex situations you encounter later.

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