Question

One of several empirical formulas that relates the surface area \(S\) of a human body to the height \(h\) and weight \(w\) of the body is the Mosteller formula \(S(h, w)=\frac{1}{60} \sqrt{h w},\) where \(h\) is measured in \(\mathrm{cm}, w\) is measured in \(\mathrm{kg}\), and \(S\) is measured in square meters. Suppose \(h\) and \(w\) are functions of \(t\). a. Find \(S^{\prime}(t)\) b. Show that the condition under which the surface area remains constant as \(h\) and \(w\) change is \(w h^{\prime}(t)+h w^{\prime}(t)=0\) c. Show that part (b) implies that for constant surface area, \(h\) and \(w\) must be inversely related; that is, \(h=C / w,\) where \(C\) is a constant.

Short answer

Answer: For the surface area to remain constant, the condition \(w h^{\prime}(t)+h w^{\prime}(t)=0\) must be met. This condition implies an inverse relationship between height and weight, i.e. \(h = \frac{C}{w}\), where C is a constant.

Step-by-step solution

Find \(\frac{dS}{dt}\)

We start by finding the derivative of surface area, \(S\), with respect to \(t\). Since \(S\) is a function of \(h\) and \(w\), which are both functions of \(t\), we use the chain rule. \(\frac{dS}{dt} = \frac{\partial S}{\partial h} \frac{dh}{dt} + \frac{\partial S}{\partial w} \frac{dw}{dt}\). #Step 2: Partial derivatives#

Find \(\frac{\partial S}{\partial h}\) and \(\frac{\partial S}{\partial w}\)

Now, we find the partial derivatives of \(S\) with respect to \(h\) and \(w\). We have \(\frac{\partial S}{\partial h} = \frac{1}{60} \frac{w}{\sqrt{h w}} = \frac{w}{60 \sqrt{h w}}\) and \(\frac{\partial S}{\partial w} = \frac{1}{60} \frac{h}{\sqrt{h w}} = \frac{h}{60 \sqrt{h w}}\). #Step 3: Substitute partial derivatives and simplify#

Substitute and simplify \(\frac{dS}{dt}\)

Now we substitute the partial derivatives back into the expression for \(\frac{dS}{dt}\): $$\frac{dS}{dt} = \frac{w}{60 \sqrt{h w}} \frac{dh}{dt} + \frac{h}{60 \sqrt{h w}} \frac{dw}{dt} = \frac{1}{60 \sqrt{h w}} (w \frac{dh}{dt} + h \frac{dw}{dt})$$ #Step 4: Constant surface area condition#

Find the condition for constant surface area

For the surface area to remain constant, we require \(S'(t)=0\). This implies the following condition: $$\frac{1}{60 \sqrt{h w}} (w \frac{dh}{dt} + h \frac{dw}{dt}) = 0$$ Thus, we have: $$w\frac{dh}{dt} + h\frac{dw}{dt} = 0$$ #Step 5: Prove inverse relationship#

Prove \(h = \frac{C}{w}\) for constant surface area

If the surface area is constant, we know \(w\frac{dh}{dt} + h\frac{dw}{dt} = 0\). Divide both sides of the equation by \(h\) and \(w\) to get $$\frac{1}{h}\frac{dh}{dt} = - \frac{1}{w}\frac{dw}{dt}$$ We can now integrate both sides with respect to \(t\), $$\int \frac{1}{h} \frac{dh}{dt} dt = -\int \frac{1}{w} \frac{dw}{dt} dt$$ Now, change variables to integrate with respect to h and w respectively: $$\int \frac{1}{h} dh = - \int \frac{1}{w} dw$$ Integrating both sides yields, $$\ln(h) = -\ln(w) + C$$ where C is the constant of integration. Taking exponentials of both sides, we get: $$h = \frac{C}{w},$$ where \(C=e^C\) is another constant. This proves the inverse relationship between h and w for a constant surface area.

Key concepts

Chain Rule

When dealing with functions that depend on multiple variables, and those variables are functions of another variable, using the chain rule is crucial. In this exercise, the surface area \(S(h, w)\) of the human body depends on height \(h\) and weight \(w\), which are further functions of time \(t\). To track how \(S\) changes over time, we need the derivative \(\frac{dS}{dt}\).
According to the chain rule, we find the total derivative by calculating:

• \(\frac{\partial S}{\partial h} \frac{dh}{dt}\): The rate of change of \(S\) with respect to \(h\), multiplied by the rate of change of \(h\) with respect to \(t\), and
• \(\frac{\partial S}{\partial w} \frac{dw}{dt}\): The rate of change of \(S\) with respect to \(w\), multiplied by the rate of change of \(w\) with respect to \(t\).

Adding these together gives \(\frac{dS}{dt} = \frac{\partial S}{\partial h} \frac{dh}{dt} + \frac{\partial S}{\partial w} \frac{dw}{dt}\). This systematic approach lets us see how multiple interconnected changes affect a single outcome, which is why the chain rule is indispensable.

Partial Derivatives

Partial derivatives help in understanding the sensitivity of a multivariable function to changes in one specific variable, while keeping others constant. For the surface area \(S(h, w)\), we compute \(\frac{\partial S}{\partial h}\) and \(\frac{\partial S}{\partial w}\).
This means we examine how changes in height \(h\) and weight \(w\) individually affect \(S\).

• \(\frac{\partial S}{\partial h} = \frac{w}{60 \sqrt{h w}}\): This partial derivative shows how the surface area changes with respect to a small change in height while keeping weight constant.
• \(\frac{\partial S}{\partial w} = \frac{h}{60 \sqrt{h w}}\): This tells us the change in surface area if weight changes slightly, while height remains constant.

Both derivatives together prepare us for calculating \(\frac{dS}{dt}\) using the chain rule, emphasizing the independent effects of \(h\) and \(w\) on \(S\).

Inverse Relationship

An inverse relationship implies that as one variable increases, the other decreases in such a way that their product remains constant. In the context of this exercise, if the surface area \(S\) of a body remains constant while \(h\) and \(w\) vary, these variables must be inversely related.
From the condition \(w \frac{dh}{dt} + h \frac{dw}{dt} = 0\), we observe that any increase in height with respect to time needs to be matched by a proportional decrease in weight, and vice versa. This can be expressed as:

• Divide by \(hw\) to get \(\frac{1}{h}\frac{dh}{dt} = -\frac{1}{w}\frac{dw}{dt}\).
• Integrate both sides to get \(\ln(h) = -\ln(w) + C\), where \(C\) is a constant.
• Exponentiating both sides leads to \(h = \frac{C}{w}\).

This tells us that height is inversely proportional to weight, meaning the product \(hw\) remains constant, retaining the same surface area under these transformations.

Constant Surface Area

Keeping the surface area constant as \(h\) and \(w\) change requires the rate of change \(S'(t)\) to be zero. This results in the condition \(w\frac{dh}{dt} + h\frac{dw}{dt} = 0\).
It ensures that changes in one dimension are exactly offset by changes in another, preserving the overall surface area:

• If \(\frac{dh}{dt}\), the rate at which height changes, is positive, \(\frac{dw}{dt}\) must be negative enough to compensate, maintaining equilibrium.
• This scenario mirrors real-world situations, like resizing a balloon while keeping its outer material constant. Changing one dimension necessitates adjusting another.

For students, understanding this balance simplifies grasping the dynamics of variable changes under constant conditions, a crucial aspect of calculus and differential equations.