Problem 70 - Applied Calculus

Question

A subject can perform a task at the rate of \(\sqrt{2 t+1}\) tasks per minute at time \(t\) minutes. Find the total number of tasks performed from time \(t=0\) to time \(t=12\).

Step-by-Step Solution

Answer

41.33 tasks are performed from time \(t=0\) to \(t=12\).

Step 1: Set up the Problem

We need to integrate the rate function to find the total number of tasks from time \(t=0\) to \(t=12\). Our rate function is \(\sqrt{2t+1}\).

Step 2: Determine the Integral

We find the integral of the rate function \(\int_0^{12} \sqrt{2t+1} \, dt\).

Step 3: Perform u-Substitution

Let \(u = 2t+1\), then \(\frac{du}{dt} = 2\) or \(dt = \frac{1}{2} du\). When \(t=0\), \(u=1\) and when \(t=12\), \(u=25\). The integral becomes \(\int_1^{25} \frac{1}{2} \sqrt{u} \, du\).

Step 4: Integrate the New Expression

The integral becomes \(\frac{1}{2}\int_1^{25} u^{1/2} \, du\). The antiderivative of \(u^{1/2}\) is \(\frac{2}{3}u^{3/2}\).

Step 5: Evaluate the Integral

Apply the limits of integration to the antiderivative: \[\frac{1}{2} \left[ \frac{2}{3}u^{3/2} \right]_1^{25} = \frac{1}{3} \left[(25)^{3/2} - (1)^{3/2}\right].\]

Step 6: Calculate Each Expression

Compute \((25)^{3/2}\), which is \(125\), and \((1)^{3/2}\), which is \(1\). So, \( \frac{1}{3} (125 - 1) = \frac{1}{3} \times 124\).

Step 7: Final Calculation and Result

The result is \( \frac{124}{3} = 41.333\). Therefore, the total number of tasks performed from time \(t=0\) to time \(t=12\) is approximately 41.33.

Key Concepts

Rate of Change

The rate of change in mathematics is like the heartbeat of a problem. It tells us how fast something is happening. In this exercise, our rate of change is given by the function \(\sqrt{2t+1}\). It reflects how quickly the tasks are being performed at any given minute \(t\).
To understand this better, think of it like a car's speedometer showing speed at each moment. Here, instead of speed, we have a rate of tasks being completed over time. Breaking it down:

• Rate function: \(\sqrt{2t+1}\)
• Variable: \(t\) (time in minutes)

Our job is to take this rate and figure out how many tasks are completed during a specific time interval, from \(t=0\) to \(t=12\). This requires integration, a powerful tool in calculus.

u-Substitution

u-Substitution is a handy technique in calculus that helps to simplify integration. It works like finding a simpler path through a tangled maze. This method transforms complicated expressions into something more manageable.
In this problem, our rate function under the integral is \(\sqrt{2t+1}\). To simplify it, we introduce a new variable \(u\):

• Let \(u = 2t + 1\)
• Then, \(dt = \frac{1}{2} du\)

This transformation helps in integrating complex functions, as it reduces the expression to a simpler form. After substitution, redefining the limits from \(t\) to \(u\), we can tackle the integral with less effort and potential for error.

Definite Integrals

Definite integrals are like adding up slices to find the whole pie, but with a twist—it involves area under a curve. In this exercise, we use them to determine the total tasks performed over a set time.
For our problem, the definite integral is \[ \int_0^{12} \sqrt{2t+1} \, dt. \]This integral is computed across the interval from \(t=0\) to \(t=12\), representing the period of task completion.

• Lower limit: \(t=0\) (beginning of task)
• Upper limit: \(t=12\) (end of task)

Once we've completed the integration, we evaluate it by applying these limits to determine the total number of tasks performed, which is approximately 41.33.

Antiderivatives

Antiderivatives, also known as indefinite integrals, are functions that reverse differentiation. In essence, finding an antiderivative is like asking what function could've given us a certain derivative.
When we integrated our modified expression with respect to \(u\), \[\int u^{1/2} \, du,\]we found its antiderivative, \(\frac{2}{3}u^{3/2}\). This function represents the accumulation of our rate over time.

• Expression: \(u^{1/2}\)
• Antiderivative: \(\frac{2}{3}u^{3/2}\)

Finally, when we plug in our limits from \(u=1\) to \(u=25\), we've effectively found the accumulated total, which tells us how many tasks were performed during our given interval.