Problem 217 - Calculus Volume 1

Question

[T] The velocity \(V\) (in centimeters per second) of blood in an artery at a distance \(x \mathrm{cm}\) from the center of the artery can be modeled by the function $$ V=f(x)=500\left(0.04-x^{2}\right) \text { for } 0 \leq x \leq 0.2 $$ a. Find \(x=f^{-1}(V)\) b. Interpret what the inverse function is used for. c. Find the distance from the center of an artery with a velocity of \(15 \mathrm{cm} / \mathrm{sec}, 10 \mathrm{cm} / \mathrm{sec}\) , and 5 \(\mathrm{cm} / \mathrm{sec}\) .

Step-by-Step Solution

Answer

The distance is approximately 0.18 cm for 15 cm/sec, 0.16 cm for 10 cm/sec, and 0.14 cm for 5 cm/sec.

Step 1: Define the Inverse Function

The inverse function is found by switching the dependent and independent variables of the given function. So, start by setting \( f(x) = V \) as given: \( V = 500(0.04 - x^2) \). We need to solve for \( x \): \( x = f^{-1}(V) \).

Step 2: Solve for x in terms of V

Rearrange \( V = 500(0.04 - x^2) \) to express \( x \) in terms of \( V \). Start by dividing both sides by 500: \( \frac{V}{500} = 0.04 - x^2 \). Next, solve for \( x^2 \): \( x^2 = 0.04 - \frac{V}{500} \). Finally, take the square root of both sides: \( x = \sqrt{0.04 - \frac{V}{500}} \).

Step 3: Find x for Given Velocities

Using the inverse function from Step 2, \( x = \sqrt{0.04 - \frac{V}{500}} \), calculate \( x \) for each given velocity. First, for \( V = 15 \): \( x = \sqrt{0.04 - \frac{15}{500}} \). Calculate: \( x \approx 0.18 \). Second, for \( V = 10 \): \( x = \sqrt{0.04 - \frac{10}{500}} \). Calculate: \( x \approx 0.16 \). Third, for \( V = 5 \): \( x = \sqrt{0.04 - \frac{5}{500}} \). Calculate: \( x \approx 0.14 \).

Step 4: Interpret the Inverse Function

The inverse function \( f^{-1}(V) \) gives the distance \( x \) from the center of the artery when you know the blood velocity \( V \). It is used to determine where in the artery a certain blood velocity occurs.

Key Concepts

Blood Velocity

Understanding blood velocity is crucial when studying how blood flows through arteries. In simple terms, it refers to how fast blood is moving past a certain point inside an artery. It is often measured in centimeters per second (cm/s).
When doctors or scientists measure blood velocity, they can assess whether blood flow is normal or if there are any potential issues, such as blockages or narrowing of the arteries. This information is vital in diagnosing and treating cardiovascular diseases.
In the mathematical context, the velocity of blood as a function is defined by how it changes with different variables—like the distance from the artery's center. In this exercise, the velocity function is given by:

• \( V = f(x) = 500(0.04 - x^2) \)

Distance from Center of Artery

The distance from the center of an artery, denoted as \( x \), plays a significant role in determining blood velocity. Typically, the closer the measurement is to the center of the artery, the higher the velocity of the blood. This occurs due to the dynamics of fluid flow within cylindrical tubes like arteries.
Understanding how the blood velocity varies with \( x \) is valuable for medical professionals. Calculating this distance helps in pinpointing different velocities within the artery, providing insights into the internal conditions of the vessel.
In our scenario, by knowing the blood velocity, one can calculate the distance \( x \) from the artery center using the inverse function. This is done through the equation:

• \( x = \sqrt{0.04 - \frac{V}{500}} \)

This formula helps find \( x \) when specific velocities such as 15 cm/s, 10 cm/s, and 5 cm/s are given.

Function Inversion

Function inversion is a mathematical process used to find the input of a function given its output. For instance, if you know a particular output or result, the inverse function lets you work backwards to find the original input value.
To find the inverse of a function, switch the roles of the dependent and independent variables and solve for the new independent variable. In this exercise, this means starting with blood velocity (\( V \)) and figuring out the distance from the center of the artery (\( x \)). The original function, expressed as \( V = 500(0.04 - x^2) \), becomes inverted when solved as:

• \( x = f^{-1}(V) = \sqrt{0.04 - \frac{V}{500}} \)

This tells us how to find \( x \) given a velocity. It's a crucial step, especially in fields like physics and engineering, where internal characteristics need to be deduced from external measurements.

Solve for x

Solving for \( x \) when given a function such as blood velocity involves algebraic manipulation to express \( x \) in terms of the known value(s).
This exercise focuses on using the inverse function, \( f^{-1}(V) = \sqrt{0.04 - \frac{V}{500}} \), to determine the distance \( x \) from the center of the artery based on specific blood velocities.
For example, calculations are done for certain velocities:

• For a velocity of 15 cm/s, \( x \approx 0.18 \)
• For a velocity of 10 cm/s, \( x \approx 0.16 \)
• For a velocity of 5 cm/s, \( x \approx 0.14 \)

By solving for \( x \), we answer questions about the location within the artery where particular velocities occur. This process crosses over into real-world applications where precise measurements and natural deduction are essential.