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Chapter 5: Problem 1

Determine the percentage rate of change of the functions at the points indicated. $$ f(t)=t^{2} \text { at } t=10 \text { and } t=50 $$

Short Answer

Expert verified
At \( t=10 \), the percentage rate of change is 20%. At \( t=50 \), it is 4%.

Step by step solution

01

- Find the function values at specified points

Calculate the values of the function at the given points: 1. For \( t = 10 \): \[ f(10) = 10^2 = 100 \] 2. For \( t = 50 \): \[ f(50) = 50^2 = 2500 \]
02

- Calculate the derivative of the function

Determine the derivative of the function \( f(t) = t^2 \): \[ f'(t) = 2t \]
03

- Evaluate the derivative at the given points

Substitute the given points into the derivative to find the rate of change at those points: 1. For \( t = 10 \): \[ f'(10) = 2 \times 10 = 20 \] 2. For \( t = 50 \): \[ f'(50) = 2 \times 50 = 100 \]
04

- Calculate the percentage rate of change

Use the formula for percentage rate of change: \[ \text{Percentage Rate of Change} = \frac{f'(t)}{f(t)} \times 100\text{%} \]1. For \( t = 10 \): \[ \text{Percentage Rate of Change} = \frac{20}{100} \times 100\text{%} = 20\text{%} \] 2. For \( t = 50 \): \[ \text{Percentage Rate of Change} = \frac{100}{2500} \times 100\text{%} = 4\text{%} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Function Evaluation
Function evaluation involves finding the value of a function for specific inputs. In our exercise, we started by evaluating the function at given points, specifically at \( t=10 \) and \( t=50 \).

For a function \( f(t) = t^2 \), we substitute the given values of \( t \) to determine \( f(10) \) and \( f(50) \).
  • At \( t=10 \): \( f(10) = 10^2 = 100 \)
  • At \( t=50 \): \( f(50) = 50^2 = 2500 \)
Function evaluation helps us understand the behavior of the function at different points.
Derivative Calculation
Derivative calculation is fundamental in calculus, allowing us to find the rate of change of a function. In our function \( f(t) = t^2 \), we calculated the derivative as follows:

The derivative of \( f(t) = t^2 \) is found using the power rule, which states that the derivative of \( t^n \) is \( nt^{n-1} \).
Therefore, \( f'(t) = 2t \).

This derivative, \( f'(t) = 2t \), provides the slope of the function at any point \( t \). It's crucial for understanding how the function changes over time or at specific instances.
Rate of Change Formula
\( \text{Percentage Rate of Change} = \frac{f'(t)}{f(t)} \times 100 \)
  • For \( t=10 \): \( \frac{20}{100} \times 100 = 20\% \)
  • For \( t=50 \): \( \frac{100}{2500} \times 100 = 4\% \)
This formula gives a clear, normalized view of how much the function's output changes at specific points.
Calculus Applications
Calculus applications are vast and practical. Understanding the percentage rate of change through derivatives is essential in many fields like physics, economics, and biology. For instance, in our exercise, we determined how fast \( f(t)=t^2 \) changes at \( t=10 \) and \( t=50 \), resulting in changes of 20% and 4% respectively.

Such calculations are crucial in:
  • Predicting population growth rates
  • Analyzing the speed of accelerating objects
  • Assessing economic trends
  • Evaluating rates of drug absorption in the body
Mastering these concepts not only helps in academic settings but also lays the foundation for understanding real-world phenomena.

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Most popular questions from this chapter

The decay constant for the radioactive element cesium 137 is .023 when time is measured in years. Find its half-life.

A person is given an injection of 300 milligrams of penicillin at time \(t=0 .\) Let \(f(t)\) be the amount (in milligrams) of penicillin present in the person's bloodstream \(t\) hours after the injection. Then, the amount of penicillin decays exponentially, and a typical formula is \(f(t)=300 e^{-.6 t}\). (a) Give the differential equation satisfied by \(f(t)\). (b) How much will remain at time \(t=5\) hours? (c) What is the biological half-life of the penicillin (that is, the time required for half of a given amount to decompose) in this case?

Convince yourself that daily compounding is nearly the same as continuous compounding by graphing \(y=100[1+(.05 / 360)]^{360 x}\), together with \(y=100 e^{.05 x}\) in the window \([0,64]\) by \([250,2500] .\) The two graphs should appear the same on the screen. Approximately how far apart are they when \(x=32 ?\) When \(x=64\) ?

A model incorporating growth restrictions for the number of bacteria in a culture after \(t\) days is given by \(f(t)=5000\left(20+t e^{-.04 t}\right)\). (a) Graph \(f^{\prime}(t)\) and \(f^{\prime \prime}(t)\) in the window \([0,100]\) by \([-700,300]\) (b) How fast is the culture changing after 100 days? (c) Approximately when is the culture growing at the rate of \(76.6\) bacteria per day? (d) When is the size of the culture greatest? (e) When is the size of the culture decreasing the fastest?

Ten thousand dollars is invested at \(6.5 \%\) interest compounded continuously. When will the investment be worth $$\$ 41,787 ?$$
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