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Magazine Chapter 5: Problem 70
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 \(t=12\).
Short Answer
Expert verified
41.33 tasks are performed from time \(t=0\) to \(t=12\).
Step by step solution
01
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 \(\sqrt{2t+1}\).
02
Determine the Integral
We find the integral of the rate function \(\int_0^{12} \sqrt{2t+1} \, dt\).
03
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 \(\int_1^{25} \frac{1}{2} \sqrt{u} \, du\).
04
Integrate the New Expression
The integral becomes \(\frac{1}{2}\int_1^{25} u^{1/2} \, du\). The antiderivative of \(u^{1/2}\) is \(\frac{2}{3}u^{3/2}\).
05
Evaluate the Integral
Apply the limits of integration to the antiderivative: \[\frac{1}{2} \left[ \frac{2}{3}u^{3/2} \right]_1^{25} = \frac{1}{3} \left[(25)^{3/2} - (1)^{3/2}\right].\]
06
Calculate Each Expression
Compute \((25)^{3/2}\), which is \(125\), and \((1)^{3/2}\), which is \(1\). So, \( \frac{1}{3} (125 - 1) = \frac{1}{3} \times 124\).
07
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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)
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 \(u\):
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, \(dt = \frac{1}{2} du\)
Definite Integrals
Definite integrals are like adding up slices to find the whole pie, but with a twist鈥攊t 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.
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)
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.
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: \(\frac{2}{3}u^{3/2}\)
