Question

for an average person, the rate of change of weight w (in pounds) with respect to height h (in inches) is given approximately by the following formula.

$\frac{dw}{dh}=0.0018h^{2}$

find w(h) if w(80) = 287.2 pounds. also, find the weight of a person who is 5 feet, 6 inches tall.

$w(h)=\square$

Explanation:

Step1: Integrate the derivative

We know that if \(\frac{dW}{dh}=0.0018h^{2}\), then by the power - rule of integration \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)), we integrate \(\frac{dW}{dh}\) with respect to \(h\).

\(W(h)=\int0.0018h^{2}dh\)

Since \(\int0.0018h^{2}dh=0.0018\int h^{2}dh\), and \(\int h^{2}dh=\frac{h^{3}}{3}+C\) (where \(C\) is the constant of integration), then \(W(h)=0.0018\times\frac{h^{3}}{3}+C\)

Simplify \(0.0018\times\frac{h^{3}}{3}\): \(0.0018\div3 = 0.0006\), so \(W(h)=0.0006h^{3}+C\)

Step2: Use the initial condition to find \(C\)

We are given that \(W(80)=287.2\). Substitute \(h = 80\) and \(W(80)=287.2\) into \(W(h)=0.0006h^{3}+C\)

\(287.2=0.0006\times(80)^{3}+C\)

First, calculate \(0.0006\times(80)^{3}\): \((80)^{3}=80\times80\times80 = 512000\), then \(0.0006\times512000 = 307.2\)

So, \(287.2=307.2 + C\)

Solve for \(C\): \(C=287.2-307.2=-20\)

Answer:

\(W(h)=0.0006h^{3}-20\)

Now, to find the weight of a person who is \(5\) feet \(6\) inches tall. Since \(1\) foot \( = 12\) inches, \(5\) feet \(6\) inches \(=5\times12 + 6=66\) inches.

Substitute \(h = 66\) into \(W(h)=0.0006h^{3}-20\)

\(W(66)=0.0006\times(66)^{3}-20\)

\((66)^{3}=66\times66\times66 = 287496\)

\(0.0006\times287496=172.4976\)

\(W(66)=172.4976-20=152.4976\approx152.5\) pounds.