Problem 51 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Gnown of A CaNcenous Tumon The volume of a spherical cancerous tumor is given by the function $$ V(r)=\frac{4}{3} \pi r^{3} $$ where \(r\) is the radius of the tumor in centimeters. Find the rate of change in the volume of the tumor when a. \(r=\frac{2}{3} \mathrm{~cm}\) b. \(r=\frac{5}{4} \mathrm{~cm}\)

Step-by-Step Solution

Answer

The rate of change in the volume of the tumor is given by the derivative with respect to the radius, which is $\dfrac{dV}{dr} = 4\pi r^{2}$. When $r=\frac{2}{3} \mathrm{~cm}$, the rate of change is $\dfrac{dV}{dr} = \frac{16}{9}\pi\ \text{ cm}^3/\text{cm}$, and when $r=\frac{5}{4} \mathrm{~cm}$, the rate of change is $\dfrac{dV}{dr} = \frac{25}{4}\pi\ \text{ cm}^3/\text{cm}$.

Step 1: Derive the volume function V(r) with respect to r

To find the rate of change in volume with respect to radius, we need to find the derivative of the volume function with respect to the radius (r). We are given the function: \[ V(r) = \frac{4}{3} \pi r^{3} \] Now, derive V(r) with respect to r to obtain dV/dr: \[ \dfrac{dV}{dr} = \dfrac{d}{dr}\left(\frac{4}{3} \pi r^{3}\right) \]

Step 2: Apply the power rule to differentiate V(r)

Apply the power rule (n*x^(n-1)) to find the derivative of the volume function: \[ \dfrac{dV}{dr} = \frac{4}{3} \pi \cdot 3r^{3-1} = 4\pi r^{2} \] Now, we have the rate of change of the volume as a function of the radius, given by: \[ \dfrac{dV}{dr} = 4\pi r^{2} \]

Step 3: Evaluate the rate of change for r = 2/3 cm

We need to find the rate of change in volume when the radius is 2/3 cm. Substitute r = 2/3 cm in the derivative: \[ \dfrac{dV}{dr} = 4\pi \left(\frac{2}{3}\right)^{2} = 4\pi \cdot\frac{4}{9} \] After simplifying, we get: \[ \dfrac{dV}{dr} = \frac{16}{9}\pi\ \text{ cm}^3/\text{cm} \] So, the rate of change in volume when r = 2/3 cm is 16/9π cm³/cm.

Step 4: Evaluate the rate of change for r = 5/4 cm

We need to find the rate of change in volume when the radius is 5/4 cm. Substitute r = 5/4 cm in the derivative: \[ \dfrac{dV}{dr} = 4\pi \left(\frac{5}{4}\right)^{2} = 4\pi \cdot\frac{25}{16} \] After simplifying, we get: \[ \dfrac{dV}{dr} = \frac{25}{4}\pi\ \text{ cm}^3/\text{cm} \] So, the rate of change in volume when r = 5/4 cm is 25/4π cm³/cm.

Key Concepts

Rate of Change

Rate of change is an important concept in mathematics and physics, describing how a quantity varies with respect to another variable. In the context of a spherical tumor, we look at how the volume changes as its radius changes. This is crucial in medical scenarios where understanding how quickly a tumor can grow provides insights for treatment planning.
- To find the rate of change of the volume of the tumor with respect to its radius, we calculate the derivative of the volume function. - As the radius increases or decreases, the rate of change tells us how rapidly the volume is growing or shrinking.
In this exercise, we're specifically looking at how the tumor's volume changes, which is essential for predicting the growth rate and developing appropriate medical responses.

Derivative

A derivative is a fundamental concept in calculus that measures how a function changes as its input changes. It represents the slope or the rate of change of a function at any point. In simpler terms, it tells us how steep a curve is at a particular point.
To calculate the derivative of the volume function of the spherical tumor, we focus on how the volume function, given by \( V(r) = \frac{4}{3} \pi r^{3} \), reacts to changes in the radius \( r \).
- Taking the derivative involves applying rules of differentiation, which in this case involves using the power rule, a specific technique for differentiating polynomials.- The result of the derivative gives us a new function, \( \frac{dV}{dr} = 4\pi r^{2} \), which illustrates how quickly the volume changes for a given change in the radius of the tumor.

Power Rule

The power rule is a basic but powerful shortcut in calculus used to differentiate functions of the form \( f(x) = x^n \). It states that the derivative of \( x^n \) is \( nx^{n-1} \). This rule simplifies the differentiation process considerably.
In the case of our spherical tumor, we employed the power rule to differentiate \( V(r) = \frac{4}{3} \pi r^{3} \). Here's how it is applied:

• The exponent, 3, is brought down as a multiplier.
• We subtract one from the original exponent, resulting in 2.
• This leads to the derivative \( \frac{dV}{dr} = 4\pi r^{2} \).

By using the power rule, we simplified the computation and obtained the rate of change of the tumor's volume with respect to its radius efficiently.

Spherical Geometry

Spherical geometry is the study of shapes on the surface of a sphere, which is part of a branch of mathematics known as geometry. In our exercise, we are looking at a three-dimensional spherical shape, specifically a tumor, and analyzing its properties and changes in volume.
Understanding spherical geometry helps us in grasping how volumes are computed in shapes like spheres.

• Volumes of spheres can be calculated using the formula \( V = \frac{4}{3} \pi r^{3} \).
• The formula considers the radius \( r \), which is the distance from the center of the sphere to its surface.
• This provides a basis for understanding how volumes change when the radius varies.

In this context, recognizing the relationship between the radius and volume of a sphere aids in comprehending the spatial characteristics of a spherical tumor, which is crucial in fields such as biology and medical physics.