Question

The Effect of Flight Maneuvers on the Heart The amount of work done by the heart's main pumping chamber, the left ventricle, is given by the equation $$W=P V+\frac{V \delta v^{2}}{2 g}$$ where \(W\) is the work per unit time, \(P\) is the average blood pressure, \(V\) is the volume of blood pumped out during the unit of time, \(\delta("\) delta") is the density of the blood, \(v\) is the average velocity of the exiting blood, and \(g\) is the acceleration of gravity. When \(P, V, \delta,\) and \(v\) remain constant, \(W\) becomes a function of \(g,\) and the equation takes the simplified form $$W=a+\frac{b}{g}(a, b\( constant \))$$ As a member of NASA's medical team, you want to know how sensitive \(W\) is to apparent changes in \(g\) caused by flight maneuvers, and this depends on the initial value of \(g\) . As part of your investigation, you decide to compare the effect on \(W\) of a given change \(d g\) on the moon, where \(g=5.2 \mathrm{ft} / \mathrm{sec}^{2},\) with the effect the same change \(d g\) would have on Earth, where \(g=32\) \(\mathrm{ft} / \mathrm{sec}^{2} .\) Use the simplified equation above to find the ratio of \(d W_{\mathrm{moon}}\) to \(d W_{\mathrm{Earth}}\)

Short answer

The ratio of \(dW_{moon}\) to \(dW_{Earth}\) is \((32/5.2)^2\).

Step-by-step solution

Differentiate the equation

First step in finding \(dW_{moon}\) and \(dW_{Earth}\) is to differentiate the equation \(W = a + \frac{b}{g}\) with respect to \(g\). By doing so, one can obtain \(dW\), the infinitesimal change in \(W\) for a small change in \(g\). The derivative of a constant is 0 and the derivative of \(1/g\) is \(-1/g^2\). Therefore, the derivative of \(W\) with respect to \(g\) can be expressed as \(dW/dg = -b/g^2\).

Apply the definition of derivatives

By the definition of a derivative, \(dW = (dW/dg) * dg\). Thus, \(dW = \frac{-b}{g^2} * dg\).

Calculate \(dW_{moon}\) and \(dW_{Earth}\)

Now, we will substitute the values of \(g\) for moon and earth into the \(dW\) equation. For the moon, \(dW_{moon} = -b/(5.2)^2 * dg\). For Earth, \(dW_{Earth} = -b/(32)^2 * dg\). The \(dg\) terms in both equations will cancel each other in the ratio of \(dW_{moon}\) to \(dW_{Earth}\).

Compute the ratio

The ratio of \(dW_{moon}\) to \(dW_{Earth}\) is therefore \(dW_{moon}/dW_{Earth} = (-b/(5.2)^2 * dg)/(-b/(32)^2 * dg) = (32/5.2)^2.\)