Problem 53 - Calculus and its applications

Question

Results of practice. A keyboarder's speed over a 5 -min interval is given by $$ W(t)=-6 t^{2}+12 t+90, \quad t \text { in }[0,5] $$ where \(W(t)\) is the speed, in words per minute, at time \(t\). a) Find the speed at the beginning of the interval. b) Find the maximum speed and when it occurs. c) Find the average speed over the 5 -min interval.

Step-by-Step Solution

Answer

a) The speed at the beginning is 90 words/min. b) Maximum speed of 96 words/min occurs at \( t = 1 \). c) Average speed over the interval is 70 words/min.

Step 1: Determine Speed at the Beginning of the Interval

To find the speed at the beginning of the interval, evaluate the function at \( t = 0 \):\[ W(0) = -6(0)^2 + 12(0) + 90 = 90. \]Hence, the speed at \( t = 0 \) is 90 words per minute.

Step 2: Determine the Maximum Speed

First, find the critical points by setting the derivative of \( W(t) \) to zero. Calculate \( W'(t) \):\[ W'(t) = -12t + 12. \]Set \( W'(t) = 0 \) and solve for \( t \):\[ -12t + 12 = 0 \implies t = 1. \]Evaluate \( W(t) \) at \( t = 1 \):\[ W(1) = -6(1)^2 + 12(1) + 90 = 96. \]Check endpoints \( t = 0 \) and \( t = 5 \):\( W(0) = 90 \) and\[ W(5) = -6(5)^2 + 12(5) + 90 = 60. \]Thus, the maximum speed, 96 words per minute, occurs at \( t = 1 \).

Step 3: Calculate the Average Speed Over the Interval

To find the average speed over the interval \([0, 5]\), use the formula for the average value of a function:\[\bar{W} = \frac{1}{5-0}\int_{0}^{5} W(t) \, dt.\]First, compute \( \int_{0}^{5} W(t) \, dt \):\[\int (-6t^2 + 12t + 90) \, dt = -2t^3 + 6t^2 + 90t + C.\]Evaluate:\[\left[ -2t^3 + 6t^2 + 90t \right]_{0}^{5} = \left( -2(5)^3 + 6(5)^2 + 90(5) \right) - \left( 0 \right) = -250 + 150 + 450 = 350.\]The average speed is:\[\bar{W} = \frac{350}{5} = 70 \text{ words per minute}.\]

Step 4: Final Step: Compile the Results

The speed at the beginning of the interval is 90 words per minute. The maximum speed is 96 words per minute at \( t = 1 \). The average speed over the 5-minute interval is 70 words per minute.

Key Concepts

Derivative

In calculus, the concept of a derivative provides us with the rate of change of a function with respect to a variable. Essentially, it tells us how fast a quantity is changing at any given point. In the context of the exercise, we examined the keyboarder's speed function over time, which is denoted as \(W(t) = -6t^2 + 12t + 90\). To find a critical attribute of this function, like when the speed is at its maximum, we first need to determine its derivative. The derivative of \(W(t)\) with respect to \(t\), giving us \(W'(t) = -12t + 12\), tells us how the speed changes over time. This new function can help us find moments when the speed's rate of increase or decrease reaches zero by solving for when \(W'(t) = 0\). This is crucial for pinpointing maximum or minimum points of a function.

Critical Points

Critical points of a function occur where the derivative is zero or undefined. These points provide valuable information about the function's behavior, such as where it might reach a local maximum or minimum. In the provided exercise, we sought to identify where the keyboarder's speed was at its peak. Since we have the derivative \(W'(t) = -12t + 12\), setting this equal to zero gives \(-12t + 12 = 0\), leading to a critical point at \(t = 1\). Determining whether this critical point is indeed a maximum involves examining the values of the original function \(W(t)\) at this critical point and at the endpoints of the interval of interest, which is from \(t = 0\) to \(t = 5\). This helps us ensure that \(t = 1\) is not only a local maximum but the maximum within the given interval.

Average Value of a Function

The average value of a function on a closed interval provides insight into the function's average behavior over that interval. For a function \(f(x)\) defined on \([a, b]\), the formula for calculating this average is:\[ \bar{f} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \].This essentially averages out the function's values across the interval. In the case of our exercise, the average speed of the keyboarder over a 5-minute interval was determined using this method. By computing the definite integral of the speed function \(W(t) = -6t^2 + 12t + 90\), we find the total speed over the interval \([0, 5]\), which calculates to be 350. Dividing by the length of the interval (5) provides the average speed \(\bar{W} = 70\) words per minute. This approach gives a big-picture view of how the speed behaves on average throughout the duration.