Problem 45 - Calculus for Biology and Medicine

Question

The volume \(V\) of a spherical cell of radius \(r\) is given by $$V(r)=\frac{4}{3} \pi r^{3}$$ If you can determine the radius to within an accuracy of \(3 \%\), how accurate is your calculation of the volume?

Step-by-Step Solution

Answer

The volume calculation is accurate to within 9\%.

Step 1: Define the problem

We need to determine how the uncertainty in the radius affects the uncertainty in the volume of the sphere. The given volume formula is \( V(r) = \frac{4}{3} \pi r^3 \). The radius \( r \) has an uncertainty of 3\%.

Step 2: Differentiate the volume function

Find the derivative of the volume with respect to the radius: \[ \frac{dV}{dr} = \frac{d}{dr} \left( \frac{4}{3} \pi r^3 \right) = 4 \pi r^2. \] This derivative tells us how the volume changes with a small change in the radius.

Step 3: Determine the change in volume

The small change in volume \( \Delta V \) due to a change \( \Delta r \) in the radius is given by: \[ \Delta V \approx \frac{dV}{dr} \cdot \Delta r = 4 \pi r^2 \cdot \Delta r. \] Since \( \Delta r \) is 3\% of \( r \), we have \( \Delta r = 0.03r \).

Step 4: Calculate the volume accuracy

Substitute \( \Delta r = 0.03r \) into the expression for \( \Delta V \): \[ \Delta V \approx 4 \pi r^2 \cdot 0.03r = 0.12 \pi r^3. \]

Step 5: Express the volume change as a percentage

The original volume \( V \) is \( \frac{4}{3} \pi r^3 \). Thus, the relative change in volume is: \[ \frac{\Delta V}{V} = \frac{0.12 \pi r^3}{\frac{4}{3} \pi r^3} = 0.09. \] This means the volume calculation has an error of 9\%.

Key Concepts

Spherical Cell Volume

Understanding the concept of spherical cell volume is vital in various fields such as biology and medicine. A spherical cell is often conceptualized as a perfectly round ball. When calculating its volume, the formula applied is based on the classical geometry for a sphere:

• The formula is given by: \[ V(r) = \frac{4}{3} \pi r^3 \]
• Here, \(V\) represents the volume, \(\pi\) is a constant approximately equal to 3.14159, and \(r\) is the radius of the cell.

This formula allows you to compute how much space the spherical cell occupies.
In biological contexts, knowing this volume is crucial. It helps in understanding cellular processes involving volume changes, which can indicate functions or dysfunctions within cells.
With this formula, whenever you know the radius of the cell, you can instantly find out its volume. All you need to do is plug in the value of \(r\) into the equation and perform the calculation.

Differential Calculus Applications

Differential calculus plays a significant role in understanding how changes in one quantity affect another. In the context of spherical cell volume, it helps assess how a small change in the radius of a cell affects its volume.

• The derivative of the volume with respect to the radius is calculated as:\[ \frac{dV}{dr} = 4\pi r^2 \]
• It shows that the rate of change of volume depends not only on the constant \(4\pi\) but also on the square of the radius \(r\).

By using differential calculus, biologists and researchers can approximate how a minor variation in the size of the cell affects its total volume.
Such an understanding is useful for biological experiments where cells might swell or shrink due to environmental influences or experimental treatments.
This method provides a powerful tool to predict these volume changes and helps in making informed decisions in laboratory settings.

Uncertainty in Measurements

Uncertainty in measurements is a critical factor to consider in scientific studies. In the scenario of a spherical cell, there is always a degree of uncertainty when measuring physical dimensions like the radius. A 3% uncertainty in the radius can cause a significant effect on the calculated volume.

• If the radius \(r\) has a 3% uncertainty, \(\Delta r\) can be expressed as \(0.03r\).
• The resulting change in volume \(\Delta V\) due to this uncertainty is:\[ \Delta V \approx 4\pi r^2 \cdot 0.03r = 0.12\pi r^3 \]

When expressed as a percentage of the original volume, this leads to a 9% uncertainty in the volume.
Understanding this helps in appreciating how measurement errors can propagate and affect overall data quality.
In practical terms, reducing uncertainties involves improving measurement techniques and using precise instruments, which is crucial for achieving accurate and reliable results in any scientific research.