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Chapter 4: Problem 9

For each function: a. Find the relative rate of change. b. Evaluate the relative rate of change at the given value(s) of \(t\) $$ f(t)=25 \sqrt{t}-1, t=6 $$

Short Answer

Expert verified
The relative rate of change at \( t = 6 \) is approximately 0.209.

Step by step solution

01

Find the derivative of the function

To find the relative rate of change, we first need the derivative of the function, which represents the rate of change of \( f(t) \) with respect to \( t \). The given function is \( f(t) = 25 \sqrt{t} - 1 \). The derivative, \( f'(t) \), is found by differentiating each term: \( f'(t) = \frac{d}{dt}[25 \sqrt{t}] - \frac{d}{dt}[1] = \frac{25}{2\sqrt{t}} \).
02

Calculate the relative rate of change

The relative rate of change of a function \( f(t) \) is given by the expression \( \frac{f'(t)}{f(t)} \). Substituting, we get \[ \frac{f'(t)}{f(t)} = \frac{\frac{25}{2\sqrt{t}}}{25\sqrt{t} - 1}. \] Simplify this to obtain the relative rate of change function.
03

Substitute \( t = 6 \) into the relative rate of change

Now that we have the formula for the relative rate of change, we substitute \( t = 6 \). This gives \[ \frac{\frac{25}{2\sqrt{6}}}{25\sqrt{6} - 1}. \] Evaluate this to find the relative rate of change at \( t = 6 \).
04

Simplify the expression

Calculate the necessary square roots and perform arithmetic to simplify: \[ f'(6) = \frac{25}{2\sqrt{6}}, \quad f(6) = 25\sqrt{6} - 1. \] Thus, \[ \frac{f'(6)}{f(6)} = \frac{\frac{25}{2\sqrt{6}}}{25\sqrt{6} - 1} \approx \frac{12.5}{60.7948 - 1}. \] Simplifying further yields the relative rate of change.

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Key Concepts

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

Derivative
In calculus, a derivative is a fundamental concept that describes how a function changes as its input changes. It represents the function's rate of change or the slope of the function at any given point. To find the derivative of a function like the one we have here, \(f(t) = 25\sqrt{t} - 1\), we use differentiation.
  • Differentiate each term separately. The constant term (-1) will disappear because its derivative is zero.
  • For the term \(25\sqrt{t}\), use the rule that the derivative of \(\sqrt{t}\) is \(\frac{1}{2\sqrt{t}}\). Multiply this by 25 to get \(\frac{25}{2\sqrt{t}}\).

Understanding derivatives is crucial for finding how a function behaves and changes over time, especially in applied contexts where such rates of change are relevant.
Simplification
Simplification is the process of reducing expressions into a simpler or more manageable form. After deriving the expression for the relative rate of change, it is essential to simplify it to make calculations easier.
For the relative rate of change of our function, the expression is \(\frac{\frac{25}{2\sqrt{t}}}{25\sqrt{t} - 1}\).
  • Combine and cancel out terms wherever possible.
  • Break down complex fractions into smaller parts, making them more understandable and easier to work with.

This process is very helpful for solving homework exercises more efficiently and is a useful skill in dealing with calculus problems.
Function Evaluation
Function evaluation involves substituting a specific value into a function to find the output at that point. Once we have simplified the expression for the relative rate of change, it's time to use function evaluation to determine the exact value for \(t = 6\).
This means placing \(t = 6\) into the derived expression \(\frac{\frac{25}{2\sqrt{6}}}{25\sqrt{6} - 1}\) to calculate the numeric result.
  • First, compute the square roots needed for the derivatives.
  • Next, complete the arithmetic to determine the relative rate at this specific point.

Using function evaluation helps move from generalized form to specific details about the behavior of the function for particular values of \(t\).
Applied Calculus
Applied calculus refers to the implementation of calculus concepts and techniques to solve real-world problems. Here, calculating the relative rate of change is an application of calculus that provides insights into how a function's value changes relative to its previous state at any given \(t\).
In real-world contexts, these calculations could represent anything from how fast a vehicle is speeding up at a given moment to how quickly a meteorological factor changes over time.
  • Understanding the derivative provides a powerful tool for modeling rates of change.
  • Simplification and function evaluation lead to precise calculations, equipping you to make informed predictions or decisions.

Overall, applied calculus transforms theoretical concepts into practical problem-solving techniques, expanding our ability to understand and change the world around us.

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Most popular questions from this chapter

\(61-62 .\) By calculating the first few derivatives, find a formula for the \(n\) th derivative of each function \((k\) is a constant \()\). $$ f(x)=e^{k x} $$

If an amount is invested at interest rate \(r\) compounded continuously, the doubling time (the time in which it will double in value) is found by solving the equation \(P e^{r t}=2 P\). The solution (by the usual method of canceling the \(P\) and taking logs) is \(t=\frac{\ln 2}{r} \approx \frac{0.69}{r} .\) For annual compounding, the doubling time should be somewhat longer, and may be estimated by replacing 69 by 72 . For example, to estimate the doubling time for an investment at \(8 \%\) compounded annually we would divide 72 by \(8,\) giving \(\frac{72}{8}=9\) years. The \(72,\) however, is only a rough "upward adjustment" of \(69,\) and the rule is most accurate for interest rates around \(9 \% .\) For each interest rate: a. Use the rule of 72 to estimate the doubling time for annual compounding. b. Use the compound interest formula \(P(1+r)^{t}\) to find the actual doubling time for annual compounding. $$ 1 \% $$

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.

\(63-68 .\) Find the differential of each function and evaluate it at the given values of \(x\) and \(d x\). $$ y=e^{-2 x} \text { at } x=3 \text { and } d x=0.5 $$

\(51-54 .\) Find the equation for the tangent line to the curve \(y=f(x)\) at the given \(x\) -value. $$ f(x)=x^{2} e^{x+1} \text { at } x=-1 $$
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