Problem 54 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

One model of the cardiovascular system relates \(V(t)\), the stroke volume of blood in the aorta at a time \(t\) during systole (the contraction phase), to the pressure \(P(t)\) in the aorta during systole by the equation \(V(t)=\left[C_{1}+C_{2} P(t)\right]\left(\frac{3 t^{2}}{T^{2}}-\frac{2 t^{3}}{T^{3}}\right)\) where \(C_{1}\) and \(C_{2}\) are positive constants and \(T\) is the (fixed) time length of the systole phase. Find a relationship between the rates \(\frac{d V}{d t}\) and \(\frac{d P}{d t}\).

Step-by-Step Solution

Answer

The relationship is: \( \frac{dV}{dt} = C_{2}\frac{dP}{dt}\bigg(\frac{3 t^{2}}{T^{2}}-\frac{2 t^{3}}{T^{3}}\bigg) + \big[C_{1} + C_{2} P(t)\big]\bigg(\frac{6 t}{T^{2}}-\frac{6 t^{2}}{T^{3}}\bigg) \)

Step 1: Determine the Derivative of V(t)

Start by differentiating the given equation for stroke volume with respect to time, t:\[ V(t)=\big[C_{1}+C_{2} P(t)\big]\bigg(\frac{3 t^{2}}{T^{2}}-\frac{2 t^{3}}{T^{3}}\bigg) \]Apply the product rule for differentiation, where \(u = C_{1}+C_{2} P(t)\) and \(v = \frac{3 t^{2}}{T^{2}} - \frac{2 t^{3}}{T^{3}}\).

Step 2: Apply the Product Rule

By the product rule, for functions \(u\) and \(v\), the derivative of their product is given by:\[ \frac{d}{dt} \big[u(t)v(t)\big] = u'(t)v(t) + u(t)v'(t) \]Here, \(u(t) = C_{1} + C_{2} P(t)\) and \(v(t) = \frac{3 t^{2}}{T^{2}} - \frac{2 t^{3}}{T^{3}}\).

Step 3: Differentiate Each Component Separately

First, compute \(u'(t)\):\[ u'(t) = \frac{d}{dt}\big[C_{1} + C_{2} P(t)\big] = C_{2} \frac{d P(t)}{d t} \]Next, compute \(v'(t)\):\[ v'(t) = \frac{d}{dt}\bigg(\frac{3 t^{2}}{T^{2}} - \frac{2 t^{3}}{T^{3}}\bigg) = \frac{6 t}{T^{2}} - \frac{6 t^{2}}{T^{3}} \]

Step 4: Combine Using the Product Rule

Now, combine both parts using the product rule:\[ \frac{d V}{d t} = u'(t) v(t) + u(t) v'(t) \]Substitute \(u(t)\), \(u'(t)\), \(v(t)\), and \(v'(t)\) into this formula:\[ \frac{d V(t)}{d t} = \bigg[C_{2}\frac{dP(t)}{d t}\bigg] \bigg[\frac{3 t^{2}}{T^{2}}-\frac{2 t^{3}}{T^{3}}\bigg] + \big[C_{1} + C_{2} P(t)\big]\bigg[\frac{6 t}{T^{2}}-\frac{6 t^{2}}{T^{3}}\bigg] \]

Step 5: Simplify the Expression

The complete differentiated equation is:\[ \frac{d V(t)}{d t} = \frac{d P(t)}{d t}\bigg[C_{2} \bigg(\frac{3 t^{2}}{T^{2}}-\frac{2 t^{3}}{T^{3}}\bigg)\bigg] + \big[C_{1} + C_{2} P(t)\big]\bigg[\frac{6 t}{T^{2}}-\frac{6 t^{2}}{T^{3}}\bigg] \]This is the relationship between the rates \( \frac{d V}{d t} \) and \( \frac{d P}{d t} \) for the given model.

Key Concepts

stroke volume

Stroke volume is a key concept in understanding the cardiovascular system. It refers to the amount of blood ejected by the left ventricle of the heart during each contraction. Simply put, it's the volume of blood pushed out of the heart with each beat.
In the given model, stroke volume is represented as a function of time, denoted by \(V(t)\). This model shows how stroke volume changes during systole, which is the contraction phase of the heart cycle when the ventricles pump blood out. The equation for stroke volume in this model is:
\[ V(t) = \big[C_1 + C_2 P(t)\big]\bigg(\frac{3 t^2}{T^2} - \frac{2 t^3}{T^3}\bigg) \]
Here:
- \(P(t)\) is the pressure in the aorta
- \(C_1\) and \(C_2\) are constants
- \(T\) is the fixed duration of the systole phase
This equation helps us understand how stroke volume depends not only on the pressure in the aorta but also on the passage of time during systole.

product rule

The product rule is a fundamental principle in calculus used to differentiate products of two functions. According to the product rule, if we have two functions \(u(t)\) and \(v(t)\), the derivative of their product is given by:
\[ \frac{d}{dt} \big[u(t) v(t)\big] = u'(t) v(t) + u(t) v'(t) \]
In our specific problem, the functions \(u(t)\) and \(v(t)\) are:
\[ u(t) = C_1 + C_2 P(t) \]
\[ v(t) = \frac{3 t^2}{T^2} - \frac{2 t^3}{T^3} \]
This equation tells us how to break down the differentiation process into manageable parts. By applying the product rule, we can find the rate of change for the product of these two functions.

differentiation

Differentiation is the process of finding the derivative of a function, which represents the rate of change of the function with respect to a variable, typically time. In this exercise, we are asked to find the derivative of the stroke volume function \(V(t)\) with respect to time \(t\).
We start by differentiating each component separately:
First, let's differentiate \(u(t)\):
\[ u'(t) = \frac{d}{dt}\big[C_1 + C_2 P(t)\big] = C_2 \frac{d P(t)}{d t} \]
Then, differentiate \(v(t)\):
\[ v'(t) = \frac{d}{dt}\bigg(\frac{3 t^2}{T^2} - \frac{2 t^3}{T^3}\bigg) = \frac{6 t}{T^2} - \frac{6 t^2}{T^3} \]
Combining these results using the product rule, we get the differentiated expression for \(V(t)\):
\[ \frac{d V(t)}{d t} = \big[C_2 \frac{d P(t)}{d t}\big] \big[\frac{3 t^2}{T^2} - \frac{2 t^3}{T^3}\big] + \big[C_1 + C_2 P(t)\big] \big[\frac{6 t}{T^2} - \frac{6 t^2}{T^3}\big] \]

rate of change

The rate of change in calculus refers to how a quantity changes over time. In the context of this problem, we are interested in the rates of change of stroke volume \( V(t) \) and pressure \( P(t) \) over time.
To find the relationship between the rates of change, we use the differentiated version of the original equation:
\[ \frac{d V(t)}{d t} = \bigg[C_2 \frac{d P(t)}{d t}\bigg] \bigg[\frac{3 t^2}{T^2} - \frac{2 t^3}{T^3}\bigg] + \big[C_1 + C_2 P(t)\big]\bigg[\frac{6 t}{T^2} - \frac{6 t^2}{T^3}\bigg] \]
This equation shows us how the rate of change of stroke volume \( \frac{d V}{d t} \) is related to the rate of change of aortic pressure \( \frac{d P}{d t} \).
By understanding these rates, we gain insights into the dynamics of blood flow and pressure in the cardiovascular system. This can help in analyzing heart function and diagnosing cardiovascular conditions.