Problem 11 - Calculus for Biology and Medicine

Question

In the following examples quantities \(x\) and \(y\) are given. Interpret the role of change dy/dx in words.11. A car moves along a straight road. Its location at time \(t\) is given by $$ s(t)=10 t^{2}, 0 \leq t \leq 2 $$ where \(t\) is measured in hours and \(s(t)\) is measured in kilometers.

Step-by-Step Solution

Answer

The derivative \(\frac{ds}{dt} = 20t\) shows the car's velocity, increasing linearly with time.

Step 1: Identify Given Information

We are provided with a function for the location of a car: \(s(t) = 10t^2\), where \(t\) is time in hours and \(s(t)\) is location in kilometers. We need to interpret \(\frac{dy}{dx}\), which in this context is the derivative \(\frac{ds}{dt}\).

Step 2: Find the Derivative

The rate of change of the location with respect to time is the derivative of \(s(t) = 10t^2\). Use the power rule to find \(\frac{ds}{dt}\):\[\frac{ds}{dt} = \frac{d}{dt}[10t^2] = 2 \cdot 10 \cdot t^{2-1} = 20t.\]

Step 3: Interpreting the Derivative

The derivative \(\frac{ds}{dt} = 20t\) represents the car's velocity at any time \(t\). This means the car's speed is increasing linearly with time and at any moment \(t\), it is moving at \(20t\) kilometers per hour.

Key Concepts

Derivative

To understand the concept of the derivative, think of it as a tool to find the rate at which one quantity changes with respect to another. In our example, we are given a function, \(s(t) = 10t^2\), that provides the location of a car at any time \(t\). The derivative of this function, \(\frac{ds}{dt}= 20t\), will tell us how the car's location changes as time progresses.

The process of taking a derivative involves applying rules like the power rule, chain rule, or product rule, depending on the function you are working with. For \(s(t) = 10t^2\), we use the power rule, which states when you have \(t^n\), the derivative is \(n \cdot t^{n-1}\). Hence, \(\frac{ds}{dt} = 20t\). After differentiating, this new function shows how the car's position change accelerates as time increases.

Rate of Change

The phrase 'rate of change' describes how one quantity changes relative to another over a particular interval. In the context of our example with the function \(s(t) = 10t^2\), the rate of change of location is represented by the derivative, \(\frac{ds}{dt}\). Essentially, this is the amount of distance the car covers per unit of time.

Mathematically, the rate of change helps to answer questions like: How fast is the car moving at a specific time? Is the speed increasing, decreasing, or constant? By evaluating the derivative at various points, you can compare these rates at different times. For instance, at \(t = 1\) hour, \(\frac{ds}{dt} = 20 \times 1 = 20\) km/h, indicating the car covers 20 kilometers in the next hour. Similarly, at \(t = 2\), \(\frac{ds}{dt} = 40\) km/h. This linear increase gives insight into the changing rate.

Velocity

Velocity is a specific application of the rate of change concept, particularly when discussing movement. Unlike speed, which only considers how fast an object is moving, velocity also accounts for direction, although in this particular exercise, we're considering movement along a straight line which simplifies direction to straightforward positive or negative movement.

In the example, the function \(s(t) = 10t^2\) provides how far the car has traveled at any time \(t\). The derivative, \(20t\), tells us the velocity of the car at any moment. This means at time \(t\), the car's velocity is \(20t\) kilometers per hour. This implies not only how fast the car is moving but also that the speed increases as time goes by. Thus, if you're analyzing the car's velocity at \(t = 2\), an increasing velocity of \(40\) km/h indicates the car is accelerating as it travels, making velocity a dynamic measure of motion in this scenario.