/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 For each function: a. Find the... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

For each function: a. Find the relative rate of change. b. Evaluate the relative rate of change at the given value(s) of \(t\) $$ f(t)=t^{3}, t=1 \text { and } t=10 $$

Short Answer

Expert verified
The relative rates of change are 3 at \( t = 1 \) and 0.3 at \( t = 10 \).

Step by step solution

01

Understand the Relative Rate of Change

The relative rate of change of a function \( f(t) \) at a point \( t \) is given by the formula \( \frac{f'(t)}{f(t)} \). This measures the rate of change of the function relative to its current value.
02

Differentiate the Function

First, compute the derivative of the function \( f(t) = t^3 \). The derivative is found using the power rule: \( f'(t) = 3t^2 \).
03

Substitute into the Relative Rate of Change Formula

Now substitute \( f(t) \) and \( f'(t) \) into the relative rate of change formula: \[ \frac{f'(t)}{f(t)} = \frac{3t^2}{t^3} \]. Simplify this to get \( \frac{3}{t} \).
04

Evaluate at \( t = 1 \)

Substitute \( t = 1 \) into the expression for the relative rate of change: \[ \frac{3}{1} = 3 \]. The relative rate of change at \( t = 1 \) is 3.
05

Evaluate at \( t = 10 \)

Substitute \( t = 10 \) into the expression for the relative rate of change: \[ \frac{3}{10} = 0.3 \]. The relative rate of change at \( t = 10 \) is 0.3.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Derivative
In calculus, the derivative is a powerful tool that helps us understand how a function changes as its input changes. It measures the instantaneous rate of change of the function with respect to its variable. For a given function, the derivative tells us how much the function's output is expected to increase or decrease when the input is increased by a small amount.
In the exercise, the function is given as \( f(t) = t^3 \). To find its derivative, we apply differentiation techniques, which is the process of finding the derivative. The derivative of a function is often denoted by \( f'(t) \) or \( \frac{df}{dt} \). Understanding how to compute derivatives is crucial in analyzing and predicting the behavior of functions in mathematical models.
Power Rule
When differentiating a polynomial function like \( f(t) = t^3 \), we commonly use the power rule. The power rule is a simple yet essential formula in calculus that allows us to differentiate expressions of the form \( x^n \).
  • The power rule states: If \( f(x) = x^n \), then \( f'(x) = n \cdot x^{n-1} \).
  • In our exercise, \( f(t) = t^3 \), meaning \( n = 3 \).
  • Applying the power rule gives us \( f'(t) = 3 \cdot t^{3-1} = 3t^2 \).
Using the power rule greatly simplifies the process of finding derivatives for functions with terms involving powers of \( t \), enabling us to quickly determine the rate of change for such functions.
Evaluate
Once we've found the expression for the relative rate of change, our next task is to evaluate it at specific values of \( t \). Evaluating a mathematical expression involves substituting given values into the expression and calculating the result.
In the exercise, we first found that the relative rate of change of the function \( f(t) = t^3 \) is \( \frac{3}{t} \). To evaluate this at \( t = 1 \), we simply substitute 1 in place of \( t \), resulting in \( \frac{3}{1} = 3 \).
Similarly, for \( t = 10 \), substitute 10 into the expression to get \( \frac{3}{10} = 0.3 \).
Evaluating expressions is a crucial step in applying mathematical formulas to real situations, allowing us to compute exact values for specific scenarios.
Function
A function is a fundamental concept in mathematics used to describe a relationship between two sets of elements. Specifically, a function \( f(t) \) assigns exactly one output value to each valid input value \( t \).
  • In our exercise, \( f(t) = t^3 \) is a simple polynomial function of degree 3.
  • The output of the function depends on the cube of the input \( t \), displaying how the function behaves as \( t \) changes.
  • Understanding functions and their properties, like domain and range, is crucial because they model real-world relationships and can be used to make predictions.
By analyzing functions and their derivatives, we can gain deeper insights into rates of change and the dynamics governing various phenomena.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

73-76. Use your graphing calculator to graph each function on the indicated interval, and give the coordinates of all relative extreme points and inflection points (rounded to two decimal places). [Hint: Use NDERIV once or twice together with ZERO.] (Answers may vary depending on the graphing window chosen.) $$ f(x)=\frac{x}{e^{x}} \text { for }-1 \leq x \leq 5 $$

1-44. Find the derivative of each function. $$ f(x)=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} $$

Why can't we define logs of negative numbers, such as \(\ln (-2)\) ? [Hint: If \(\ln (-2)=x\), what is the equivalent exponential statement? What is the sign of \(e^{x}\) ?]

Recall that the concentration of a drug in the bloodstream after \(t\) hours is \(c e^{-k t},\) where \(k\) is called the "absorbtion constant." If one drug has a larger absorbtion constant than another, will it require more or less time between doses? (Assume that both drugs have the same value of \(c .)\)

97-98. ATHLETICS: World's Record 100-Meter Run In 1987 Carl Lewis set a new world's record of 9.93 seconds for the 100 -meter run. The distance that he ran in the first \(x\) seconds was $$ 11.274\left[x-1.06\left(1-e^{-x / 1.06}\right)\right] \text { meters } $$ for \(0 \leq x \leq 9.93 .\) Enter this function as \(y_{1},\) and define \(y_{2}\) as its derivative (using NDERIV), so that \(y_{2}\) gives his velocity after \(x\) seconds. Graph them on the window [0,9.93] by [0,100] Trace along the velocity curve to verify that Lewis's maximum speed was about 11.27 meters per second. Find how quickly he reached a speed of 10 meters per second, which is \(95 \%\) of his maximum speed.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.