Problem 63
Question
In the study of the response to acetylcholine by a frog's heart, the formula $$ R(x)=\frac{x}{c+d x} $$ arises, where \(x\) denotes the concentration of the drug and \(c\) and \(d\) are positive constants. a. Find \(R(0)\) and \(R\) (2). What is the physical significance of \(R(0)\) ? b. Find a formula that expresses the concentration \(x\) in terms of \(R(x)\).
Step-by-Step Solution
Verified Answer
a. \(R(0) = 0\), \(R(2) = \frac{2}{c+2d}\). \(R(0)\) signifies no response at zero concentration. b. \(x = \frac{R(x)c}{1 - R(x)d}\).
1Step 1: Evaluate \(R(0)\)
To find \(R(0)\), we substitute \(x = 0\) into the given function: \(R(x) = \frac{x}{c + d x}\). Thus, \(R(0) = \frac{0}{c + d \, (0)} = \frac{0}{c} = 0\).
2Step 2: Evaluate \(R(2)\)
To find \(R(2)\), substitute \(x = 2\) into the function: \(R(2) = \frac{2}{c + d \, (2)} = \frac{2}{c + 2d}\).
3Step 3: Physical Significance of \(R(0)\)
\(R(0)\) represents the response of the system when the concentration of the drug is zero. Physically, it indicates that there is no response from the frog's heart without the presence of acetylcholine.
4Step 4: Express \(x\) in terms of \(R(x)\)
To solve for \(x\), start with \(R(x) = \frac{x}{c + d x}\). Rearrange it to: \(x = R(x)(c + d x)\). Expanding gives: \(x = R(x)c + R(x)dx\). Isolate terms involving \(x\): \(x(1 - R(x)d) = R(x)c\). Therefore, \(x = \frac{R(x)c}{1 - R(x)d}\).
Key Concepts
Rational FunctionsProblem-SolvingAlgebraic Manipulation
Rational Functions
Rational functions are a key concept in calculus and algebra. They are functions represented as the ratio of two polynomials. In the formula given by the exercise, \( R(x) = \frac{x}{c + dx} \), the numerator is \( x \) and the denominator is \( c + dx \). The behavior of rational functions can vary depending on the values and expressions in the numerator and denominator.
These functions can model relationships in which one quantity depends on another and where this dependency is not linear. For instance, the function in this exercise models the response of a frog's heart to a concentration of a drug. It is crucial to understand that rational functions can have undefined points, which occur where the denominator equals zero. To master rational functions, it is essential to:
These functions can model relationships in which one quantity depends on another and where this dependency is not linear. For instance, the function in this exercise models the response of a frog's heart to a concentration of a drug. It is crucial to understand that rational functions can have undefined points, which occur where the denominator equals zero. To master rational functions, it is essential to:
- Identify the numerator and the denominator.
- Simplify expressions when possible.
- Find the domain, taking note of where the function is undefined.
Problem-Solving
Problem-solving with rational functions involves understanding the function's behavior and interpreting it correctly. Let's illustrate this with the original exercise. Initially, the problem requires finding \( R(0) \) and \( R(2) \). Calculating \( R(0) \) involves substituting \( x = 0 \) into the function, resulting in \( \frac{0}{c} = 0 \). This indicates that there is no response without the drug concentration.
Next, for \( R(2) \), substituting \( x = 2 \) gives \( \frac{2}{c + 2d} \). This fraction describes the specific response at a concentration of 2 units. By addressing these individual values, we can interpret specific points in the function.
Next, for \( R(2) \), substituting \( x = 2 \) gives \( \frac{2}{c + 2d} \). This fraction describes the specific response at a concentration of 2 units. By addressing these individual values, we can interpret specific points in the function.
- Find specific values by substitution.
- Interpret these values in the context of the problem.
- Use these calculated points to gain insights into the overall behavior of the function.
Algebraic Manipulation
Algebraic manipulation is a fundamental skill in working with rational functions. It involves rearranging equations to solve for variables of interest. In the original exercise, the goal was to rearrange the function \( R(x) = \frac{x}{c + dx} \) to express \( x \) in terms of \( R(x) \).
This was achieved by first multiplying both sides by \( c + dx \) to eliminate the fraction: \( x = R(x)(c + dx) \). Expanding this, we get \( x = R(x)c + R(x)dx \). The next step involves isolating \( x \) by rearranging terms: \( x(1 - R(x)d) = R(x)c \). Finally, solving for \( x \) gives \( x = \frac{R(x)c}{1 - R(x)d} \).
This was achieved by first multiplying both sides by \( c + dx \) to eliminate the fraction: \( x = R(x)(c + dx) \). Expanding this, we get \( x = R(x)c + R(x)dx \). The next step involves isolating \( x \) by rearranging terms: \( x(1 - R(x)d) = R(x)c \). Finally, solving for \( x \) gives \( x = \frac{R(x)c}{1 - R(x)d} \).
- Eliminate fractions by multiplying both sides by a common denominator.
- Expand expressions to simplify the equation.
- Rearrange terms to isolate the desired variable.
Other exercises in this chapter
Problem 63
Solve the inequality. $$ |x+3| \geq 3 $$
View solution Problem 63
a. Describe the region consisting of all \((x, y)\) for which $$ (x, y)=(x,-y) . $$ b. Describe the region consisting of all \((x, y)\) for which $$ (x, y)=(-x,
View solution Problem 63
The revenue function \(R\) for a certain product is given by $$ R(x)=5 x^{2}-\frac{x^{4}}{10} $$ The cost function \(C\) is given by $$ C(x)=4 x^{2}-24 x+38 $$
View solution Problem 64
Solve the inequality. $$ |x-0.3|>1.5 $$
View solution