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Differentiation is a fundamental concept in calculus, particularly useful for finding the rate at which something changes. When dealing with motion, such as an object being tossed into the air, we often want to find how its position changes over time. This is where differentiation comes in.

If we have a position function, let's call it \(s(t)\), which describes the position of an object in terms of time \(t\), taking the derivative of this function gives us the velocity function \(v(t)\). In simple terms, this derivative tells us how fast and in which direction the position is changing.

• The derivative represents the slope of the tangent line to the curve at any point \(t\).
• To find the derivative, apply the power rule: for a function \(s(t) = 40t - 5t^2\), the derivative is \(v(t) = \frac{ds}{dt} = 40 - 10t\).

Differentiation helps us understand not just where things are, but how they move.

In the context of calculus, and particularly for this exercise, the velocity function \(v(t)\) is key to understanding motion. The velocity function is the derivative of the position function \(s(t)\). In our example, the position function \(s(t) = 40t - 5t^2\) differentiates to give us \(v(t) = 40 - 10t\).

Here's the breakdown of what the velocity function tells us:

• The value of \(v(t)\) tells us how fast the object is moving at time \(t\).
• If \(v(t) > 0\), the object is rising; if \(v(t) < 0\), it is falling.
• If \(v(t) = 0\), the object is at rest, meaning it has reached either the highest point (when rising) or the lowest point (when falling).

In our task, setting \(v(t) = 0\) helps determine when the object stops rising, which occurs at \(t = 4\) seconds.

To find the maximum height of an object launched upwards, we first need to determine when it stops rising. This is achieved by identifying when the velocity function equals zero. As calculated, this happens when \(t = 4\) seconds.

Once this critical time \(t\) is known, substitute it back into the position function to find the maximum height. With our position function \(s(t) = 40t - 5t^2\), plugging in \(t = 4\) gives:\[s(4) = 40(4) - 5(4)^2 = 160 - 80 = 80~\text{meters}\]This height represents the peak of the object's trajectory—the highest point reached before it begins to fall back down. Understanding how to calculate this maximum height is not only important for solving textbook problems, but also for understanding real-world motion scenarios.

The position function \(s(t)\) is a mathematical model representing the physical location of an object at any given time \(t\). In this problem, it is defined as \(s(t) = 40t - 5t^2\), where each term has a specific meaning:

• The term \(40t\) represents the linear part of the motion, essentially showing how the initial velocity propels the object in time.
• The term \(-5t^2\) is the quadratic part, indicating how gravity affects the object's ascent by slowing it down.

In general, a position function can model things like the trajectory of a projectile, where its shape is often a parabola.

The position function is central to determining both the velocity of an object and finding points of interest, like when it reaches a maximum height or where it eventually comes back to the starting level.