The equation \(p=13+0.625d\) approximates the pressure \(p\), in pounds per square inch, exerted on a fish at a depth of \(d\) feet (ft) below the surface of the water. What is the increase
in depth that is necessary to increase the pressure by \(1\) pound per square inch?
It is easier to interpret these sorts of questions if we compare it to something that we're accustomed to.
Our model
$$p=13+0.625d$$
$$ \text{slope}=\frac{p}{d} $$
$$ \frac{\text{pressure}}{\text{depth}}= \frac{0.625\text{ pound per square inch} }{1\text{ feet} } $$
Standard linear model
$$y=mx+b$$
$$ \text{slope}=\frac{y}{x} $$
$$ \frac{y}{x}=\frac{m}{1} $$
We need to find the change in depth if pressure is \(1\). We can use a proportion to find this value.
$$ \frac{0.625\text{ pound per square inch} }{1\text{ feet} } = \frac{1 \text{ pound per square inch}}{ x \text{ feet}}$$
$$ 0.625x=1 $$
$$ x= \boxed{1.6 \text{ feet}} $$
We can test an arbitrary situation. For example, what is the change in \(d\) when we change \(p=0\) to \(p=1\).
\(p=0\)
$$p=13+0.625d$$
$$0=13+0.625d$$
$$-13=0.625d $$
$$ d=-20.8 $$
\(p=1\)
$$p=13+0.625d$$
$$1=13+0.625d$$
$$-12=0.625d $$
$$ d=-19.2$$
The difference is \(-19.2-(-20.8)=\boxed{1.6 \text{ feet}}\)