Question

Find the derivative of each function at the given point and interpret the physical meaning of this quantity. Include units in your answer. Suppose the speed of a car approaching a stop sign is given by \(v(t)=(t-5)^{2},\) for \(0 \leq t \leq 5,\) where \(t\) is measured in seconds and \(v(t)\) is measured in meters per second. Find \(v^{\prime}(3)\)

Short answer

Answer: The rate at which the car is decelerating at t = 3 seconds is -4 m/s², and it represents the rate of change of the speed of the car with respect to time, indicating that the car is slowing down as it approaches the stop sign.

Step-by-step solution

Find the general derivative of the function using the power rule

The power rule states that if \(f(x) = x^n\), then \(f'(x) = nx^{n-1}\). We will apply this rule to find the derivative of the given function \(v(t) = (t-5)^2\). Using the chain rule, we have: $$ v^{\prime}(t) = 2(t-5)^{2-1}\cdot1 = 2(t-5) $$

Compute the derivative at the given point t=3

Now we will find the value of \(v^{\prime}(3)\): $$ v^{\prime}(3) = 2(3-5) = 2(-2) = -4 $$ So the value of the derivative at \(t=3\) is \(-4\).

Interpret the physical meaning and include units

The computed value of the derivative, \(v^{\prime}(3)=-4\) represents the rate of change of the speed of the car with respect to time at \(t=3\) seconds. Since the derivative has a negative value, it signifies that the speed of the car is decreasing. In this case, the car is decelerating or slowing down as it approaches the stop sign. The units for the velocity function \(v(t)\) are meters per second (m/s), and the units for time \(t\) are seconds (s). So, the units for the derivative \(v^{\prime}(t)\) are meters per second squared (m/s²). Thus, the rate at which the car is decelerating at \(t=3\) is \(-4\) m/s².

Key concepts

Power Rule

When calculating the derivative of a function, the power rule is one of the most fundamental and frequently used tools. It is an effective method for finding the derivative of polynomial functions where you have an expression raised to a power. According to the power rule, if you have a function in the form of \( f(x) = x^n \), where \( n \) is any real number, the derivative, \( f'(x) \), will be \( nx^{n-1} \).

This means you simply multiply the power by the coefficient (if there is one) and subtract one from the power to determine the derivative. For example, the derivative of \( x^4 \) using the power rule would be \( 4x^3 \). This simplifies the process of differentiation greatly, especially for polynomial functions, making it quick and efficient to apply.

Chain Rule

At times, functions are composed of multiple other functions, leaving us with a composite function. We use the chain rule to differentiate these composite functions. It's a method for finding the derivative of the composition of two or more functions – the 'outer' function and the 'inner' function(s).

To apply the chain rule, you take the derivative of the outer function, leaving the inner function unchanged, and then multiply it by the derivative of the inner function. Consider the composite function \( f(g(x)) \); the chain rule tells us that the derivative, \( f'(g(x)) \), is \( f'(g(x)) \times g'(x) \). The chain rule is essential when dealing with more complex expressions where functions are layered on top of each other, like nested brackets in mathematics.

Rate of Change

The concept of the rate of change is central to the understanding of calculus and represents how a quantity changes with respect to another. In everyday terms, it's like observing how fast a car's speed changes over time or how a savings account grows with interest over the years. In calculus, the rate of change is what derivatives are all about.

More mathematically, if you have a function \( y = f(x) \), the rate of change of \( y \) with respect to \( x \) at any point is given by its derivative \( f'(x) \). This can tell us if the function is increasing or decreasing, and by how much, at any given point. A positive derivative indicates an increasing function, while a negative derivative indicates a decreasing function. This is profoundly important for understanding motion, economics, physics, and many other fields where change is a central concern.

Instantaneous Velocity

Imagine you're looking at a car's speedometer that reads the speed at every moment, giving you the instantaneous velocity – the velocity of the car at that instant in time. In physics and calculus, the instantaneous velocity is the rate of change of position with respect to time at a specific moment. It's the derivative of the position function with respect to time.

The calculation of instantaneous velocity is not just about the 'how fast' but also the 'in what direction', making it a vector quantity. This means while speed gives you a magnitude, velocity tells you the magnitude with an associated direction. For instance, if a car is accelerating, its instantaneous velocity is positive, indicating increasing speed in its direction of motion. Conversely, a negative instantaneous velocity reveals that the object is decelerating, as seen in our exercise where the derivative turned out to be negative, indicating the car is slowing down.