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In Activities 1 through \(26,\) write the formula for the derivative of the function. $$ v(t)=-e^{0.05} $$

Short Answer

Expert verified
The derivative of \( v(t) = -e^{0.05} \) is 0.

Step by step solution

01

Identify the Function Type

The given function is a constant function, as it does not contain any variable terms. The function is given by \( v(t) = -e^{0.05} \), where \(-e^{0.05}\) is a constant value.
02

Recall Derivative of a Constant Function

The derivative of a constant function is always zero. If \( f(x) = c \), where \( c \) is a constant, then \( f'(x) = 0 \).
03

Apply the Rule to the Given Function

Since \( v(t) = -e^{0.05} \) is a constant function, the derivative \( v'(t) \) is 0, based on the rule that the derivative of a constant is zero.
04

Conclusion

The derivative of the function \( v(t) = -e^{0.05} \) is \( v'(t) = 0 \).

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Key Concepts

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

Constant Function
A constant function is one of the simplest types of functions in mathematics. It is a function that does not change, no matter what value is plugged into it. This mean if you have a function expressed as \( f(x) = c \), where \( c \) is a constant number, the output will always be \( c \), regardless of the input \( x \). For example, in the function \( v(t) = -e^{0.05} \), the output is always \(-e^{0.05}\), a fixed value.

Constant functions are crucial for understanding other, more complex functions, as they serve as a simple baseline. They represent horizontal lines on a graph that run parallel to the x-axis. Because they do not vary, they are quite easy to differentiate, which brings us to our next topic.
Derivative Rules
In calculus, derivatives are used to determine the rate at which a function is changing at any given point. For different types of functions, different derivative rules apply. The simplest rule, however, is the rule for constant functions. This states that the derivative of any constant function is zero.

Why? Because a constant function does not change; its rate of change is zero. Essentially, if you have a function \( f(x) = c \), where \( c \) is a constant, the derivative of this function, noted as \( f'(x) \), is equal to 0. Therefore, when calculating the derivative of a constant such as \(-e^{0.05}\), it simplifies the problem tremendously because the answer is straightforward: \( v'(t) = 0 \).
  • Rule: \( f'(x) = 0 \), where \( f(x) = c \)
  • Application: Any constant value will have a derivative of zero when differentiated.
With these derivative rules in mind, you can solve many calculus problems where constant functions appear.
Calculus Problem Solving
Solving calculus problems, particularly those involving derivatives, often requires one to identify the form of the function given. Once the type of function is recognized, applying the correct derivative rule is usually a straightforward process.

Let's take a closer look at how this is applied to our example problem. The function \( v(t) = -e^{0.05} \) is identified as a constant. Recognizing this, we apply the derivative rule for constants, which tells us that the derivative is zero. Therefore, solving this situation is about recognizing the pattern: constant functions result in a zero derivative.
  • Identify the function form (constant in this case)
  • Apply the relevant derivative rule (constant rule: \( f'(x) = 0 \))
  • Conclude with the calculated derivative (\( v'(t) = 0 \))
By mastering these steps, students can approach similar calculus problems with confidence, knowing they are applying the right rules to find solutions quickly and correctly.

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Most popular questions from this chapter

In Activities 1 through \(30,\) for each of the composite functions, identify an inside function and an outside function and write the derivative with respect to \(x\) of the composite function. $$ f(x)=\ln (35 x) $$

Personal Consumption The amount spent by a consumer on nondurable goods can be modeled as \(n(x)=-1.1+1.64 \ln x\) thousand dollars and the amount spent on motor vehicles can be modeled as $$ m(x)=1.62\left(1.26^{x}\right) \text { hundred dollars } $$ where \(x\) thousand dollars is the amount spent by that same consumer for all personal consumption. (Source: Based on data from the U.S. Bureau of Labor Statistics) a. Use output values corresponding to \(\$ 4,500\) and \(\$ 10,500\) personal consumption to determine an appropriate model for personal consumption as a function of the amount spent on motor vehicles. b. Write a model giving the amount spent on nondurable goods as a function of the amount spent on motor vehicles. c. How much is spent on nondurable goods by somebody who spends \(\$ 340\) on his or her motor vehicle? At what rate is this amount changing with respect to motor vehicle spending? Write a sentence of interpretation for the results.

Airline Load Capacity The capacity of commercial large aircraft to generate revenue is measured in ton-miles. The capacity taken up by paying passengers on U.S. carriers can be modeled as $$ g(x)=16.2\left(1.009^{2.18 x+3.41}\right) \text { trillion ton-miles } $$ when a total of \(x\) trillion passenger ton-miles are flown, $$ \text { data from } 20 \leq x \leq 85 . $$ (Source: Based on data from Bureau of Transportation Statistics; applies to large certified aircraft only) a. Write a model for the rate of change of the capacity taken up by paying passengers. b. What are the capacity, the rate of change of capacity, and the percentage rate of change of capacity when 80 trillion passenger ton-miles are flown?

In Activities 1 through \(30,\) for each of the composite functions, identify an inside function and an outside function and write the derivative with respect to \(x\) of the composite function. $$ f(x)=2^{\ln x} $$

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