Problem 79 - Precalculus

Question

When a person has a cavity filled, the dentist typically administers a local anesthetic. After leaving the dentist's office, one's mouth often remains numb for several more hours. If a shot of anesthesia is injected into the bloodstream at the time of the procedure \((t=0)\), and the amount of anesthesia still in the bloodstream \(t\) hours after the initial injection is given by \(A=A_{0} e^{-0.5 t}\) in how many hours will only \(10 \%\) of the original anesthetic still be in the bloodstream?

Step-by-Step Solution

Answer

Approximately 4.604 hours after the injection.

Step 1: Understand the problem

The problem asks us to find the time at which only 10% of the original amount of anesthetic remains in the bloodstream. The equation given is \(A = A_0 e^{-0.5t}\). We need to solve for \(t\) when \(A = 0.1A_0\).

Step 2: Set up the equation

Since we're looking for the time \(t\) when the anesthetic left is 10% of the original amount, set \(A = 0.1A_0\). The equation becomes \(0.1A_0 = A_0 e^{-0.5t}\).

Step 3: Simplify the equation

Cancel \(A_0\) from both sides of the equation, assuming \(A_0 eq 0\). This results in: \[ 0.1 = e^{-0.5t} \]

Step 4: Solve for \(t\) using logarithms

To solve for \(t\), take the natural logarithm (ln) of both sides:\[ \ln(0.1) = \ln(e^{-0.5t}) \]This simplifies using the properties of logarithms to:\[ \ln(0.1) = -0.5t \]

Step 5: Isolate \(t\)

Solve for \(t\) by dividing both sides by \(-0.5\):\[ t = \frac{\ln(0.1)}{-0.5} \]

Step 6: Calculate \(t\)

Using a calculator, find \(\ln(0.1)\). The value is approximately \(-2.302\). Substitute into the equation:\[ t = \frac{-2.302}{-0.5} \]Calculate to find:\[ t \approx 4.604 \]

Step 7: Finalize the answer

The time when only 10% of the anesthetic remains is approximately 4.604 hours after injection.

Key Concepts

Logarithms

Logarithms are mathematical tools that help us solve equations where the variable is in the exponent. In simpler terms, they can "undo" exponentiation. For example, to solve the equation \(0.1 = e^{-0.5t}\), we use a logarithm. Here’s why logarithms are useful:

• They enable us to convert multiplicative relationships into additive ones.
• They simplify complex exponential equations into manageable linear forms.

To solve the exercise problem, we needed to take the natural logarithm of both sides of the equation \( \ln(0.1) = \ln(e^{-0.5t}) \). Applying the property \( \ln(e^x) = x \) simplifies it to \( \ln(0.1) = -0.5t \).Remember, the natural logarithm \(\ln\) specifically refers to the logarithm to the base \(e\), which is a common mathematical constant approximately equal to 2.718. Using logarithms, we isolated \(t\) and calculated the time when only 10% of the anesthetic remains.

Natural Exponent

The natural exponent is denoted by the number \(e\), which is approximately equal to 2.71828. It's a fundamental constant in mathematics, much like the number \(\pi\). The use of \(e\) in equations typically signifies a process that is continuous and based on natural growth or decay.In the context of exponential decay, where quantities decrease over time, the natural exponent comes into play. The formula given in the exercise, \(A = A_0 e^{-0.5t}\), illustrates how the amount of anesthesia in the bloodstream decays over time.

• The term \(-0.5t\) is the exponent indicating the decay rate. The negative sign shows decay (a decrease over time).
• \(A_0\) signifies the initial amount of anesthesia.

Letting the natural exponent \(e\) handle the decay, this kind of equation can model many real-world phenomena beyond just anesthesia, such as radioactive decay and even cooling of objects.

Anesthesia Dosage

Understanding anesthesia dosage and its decay in the bloodstream is crucial for safe medical procedures. When a local anesthetic is administered, its effectiveness diminishes over time. This diminishing or decaying process follows an exponential pattern.The rate at which the anesthesia "wears off" is represented by an equation like \(A = A_0 e^{-0.5t}\). Here's what the components mean:

• \(A_0\) is the original amount of anesthetic injected.
• \(A\) is the remaining amount after time \(t\).
• The \(e^{-0.5t}\) factor shows how quickly the anesthesia reduces.

For patients, it's important to understand that even when the feeling returns to normal, a small amount of anesthesia might still be present in the body. For safety, knowing the duration and effectiveness helps both patients and healthcare providers manage post-procedure care effectively.