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Chapter 3: Problem 17

Differentiate the functions in Problems 1-28. Assume that \(A\), \(B\), and \(C\) are constants. \(P(t)=3000(1.02)^{t}\)

Short Answer

Expert verified
The derivative is \( \frac{dP}{dt} = 3000 \times (1.02)^{t} \times 0.0198 \).

Step by step solution

01

Identify the Function Type

The function given is \( P(t) = 3000(1.02)^{t} \). This is an exponential function of the form \( a imes b^{t} \), where \(a = 3000\) and \(b = 1.02\).
02

Apply the Exponential Differentiation Rule

To differentiate \( a imes b^{t} \) with respect to \(t\), use the rule: \[ \frac{d}{dt} \left( a imes b^{t} \right) = a imes b^{t} imes \ln(b) \]. Here, \(a = 3000\) and \(b = 1.02\).
03

Differentiate the Function

Apply the differentiation rule to the function: \[ \frac{dP}{dt} = 3000 \times (1.02)^{t} \times \ln(1.02) \].
04

Simplify the Derivative

Calculate \(\ln(1.02)\). Using a calculator, \(\ln(1.02) \approx 0.0198\). Then, the derivative becomes: \[ \frac{dP}{dt} = 3000 \times (1.02)^{t} \times 0.0198 \].

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Key Concepts

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

Exponential Function
An exponential function is a special mathematical expression where a constant base is raised to a variable exponent. In our given function, \( P(t) = 3000(1.02)^{t} \), we have a typical form of an exponential function: \( a \times b^{t} \). Here:
  • \(a\) represents the initial amount, in this case, 3000.
  • \(b\) represents the base of the exponent, which is 1.02.
  • \(t\) is the exponent variable, representing time in our function.
  • \((1.02)^{t}\) shows how the quantity grows or decays exponentially. Since 1.02 is slightly more than 1, it indicates a growth process over time.
Understanding this form helps in identifying how exponential functions describe real-world phenomena such as population growth, radioactive decay, and financial investments. Each scenario relies on the base value and the rate of growth or decay to predict how the situation evolves over time.
Derivative Calculation
Differentiating an exponential function like \( P(t) = 3000(1.02)^{t} \) involves using specific rules that simplify the process. To calculate its derivative, identify the constants and variable parts first:
  • The constant multiplier is 3000, which stays unchanged throughout the differentiation.
  • The expression \((1.02)^{t}\) is where differentiation occurs since \(t\) is the variable.
We use the formula: \[ \frac{d}{dt} \left( a \times b^{t} \right) = a \times b^{t} \times \ln(b) \]Substituting the values, we find:
  • \(a = 3000\)
  • \(b = 1.02\)
  • \(\ln(1.02)\) is calculated using a calculator, which provides approximately 0.0198.
The result demonstrates how exponential growth or decay dynamically affects a quantity by introducing the natural log factor into the differentiation process.
Differentiation Rules
Differentiation rules are essential tools in calculus that guide us through finding derivatives efficiently. For exponential functions, the differentiation rules are straightforward yet powerful:
  • The function \( a \times b^{t} \) differentiates into itself multiplied by \( \ln(b) \). This uses the natural logarithm, a critical aspect in calculating derivatives for exponential growth or decay.
  • Constant factors such as \(3000\) in our example do not change when differentiating the variable part. This is due to the product rule, which implies that constants remain unaltered when multiplied by the derivative of the function.
Applying these rules helps streamline derivative calculations, making it more manageable even for complex functions. It also ensures accuracy in understanding how base changes affect the growth rate described by the exponential function.

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Most popular questions from this chapter

Carbon- 14 is a radioactive isotope used to date objects. If \(A_{0}\) represents the initial amount of carbon- 14 in the object, then the quantity remaining at time \(t\), in years, is $$ A(t)=A_{0} e^{-0.000121 t} $$ (a) A tree, originally containing 185 micrograms of carbon- 14 , is now 500 years old. At what rate is the carbon- 14 decaying? (b) In 1988 , scientists found that the Shroud of Turin, which was reputed to be the burial cloth of Jesus, contained \(91 \%\) of the amount of carbon- 14 in freshly made cloth of the same material. \({ }^{6}\) According to this data, how old was the Shroud of Turin in 1988 ?

In 2009 , the population, \(P\), of India was \(1.166\) billion and growing at \(1.5 \%\) annually. (a) Give a formula for \(P\) in terms of time, \(t\), measured in years since 2009 . (b) Find \(\frac{d P}{d t},\left.\frac{d P}{d t}\right|_{t=0}\), and \(\left.\frac{d P}{d t}\right|_{t=25}\). What do each of these represent in practical terms?

Find the derivative. Assume that \(a, b, c\), and \(k\) are constants. $$ f(t)=t e^{-2 t} $$

Find the derivative. Assume that \(a, b, c\), and \(k\) are constants. $$ y=\frac{e^{x}}{1+e^{x}} $$

If \(\frac{d}{d t}(t f(t))=1+f(t)\), what is \(f^{\prime}(t) ?\)
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