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Derivatives help us understand how a function changes as its variable changes. In this exercise, the position function of a car, \( s(t) = 5t^2 + 2 \), gives us the distance traveled over time. The derivative of this function, represented as \( v(t) = \frac{ds}{dt} \), tells us how the distance changes with time – or simply, the velocity of the car.
By differentiating the position function, we can see how quickly the position is changing at any given moment.

• Velocity as a Derivative: Velocity is simply the rate at which the position changes. When we derive the position function, we determine the velocity.
• Notation: While the position function is \( s(t) \), its derivative, or velocity, is \( v(t) \).

Understanding derivatives is crucial for solving problems involving rates of change, like calculating velocity.

The power rule is a quick way to find derivatives of polynomial functions. It's expressed as follows: if you have a term \( at^n \), its derivative is \( n \times at^{n-1} \).
This rule significantly simplifies computation, allowing us to swiftly find derivatives without complex calculations.
For example, in the given function \( s(t) = 5t^2 + 2 \), apply the power rule:

• \( \frac{d}{dt}(5t^2) = 10t \) (using the power rule on \( 5t^2 \))
• \( \frac{d}{dt}(2) = 0 \) (a constant term has a derivative of zero)

Therefore, the derivative of the position function is \( v(t) = 10t \). This use of the power rule allows us to easily transition from a position function to a velocity function.

A position function \( s(t) \) describes how far an object travels over time. It gives us a detailed description of an object's movement. In this exercise, the position function \( s(t) = 5t^2 + 2 \) represents how many feet the car travels in \( t \) seconds.
The position function provides invaluable insights into how an object moves. We can determine exactly where an object is at a given time and understand its journey.
Position functions do more than chart distance; they lay the groundwork for calculating velocity and acceleration by allowing us to use derivatives to find rates of change.

The rate of change answers the question: How fast is something changing? For motion problems like ours, the rate of change of position gives us the velocity.
If you think of a car's journey, the rate of change helps us understand speed at any point. Essentially, the velocity found from the derivative shows instantaneous speed.
With the position function \( s(t) = 5t^2 + 2 \), finding the rate of change means deriving this to get \( v(t)=10t \). This derivative gives us the formula to calculate how fast the car is moving at any chosen time \( t \).
By understanding the rate of change, you decode the object's behavior over time, predicting future positions, and effectively analyzing movement.