Connect changing quantities

Connect rates using the chain rule

If $y=f(x)$ and both x and y vary with time t, $dx/dt$ is the rate of change of x with respect to time, and $dy/dt$ is the rate of change of y with respect to time. The source connects them by the chain rule:

$\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}.$

Differentiate before substituting the instant

Use this order:

1. Name the changing variables and their units.
2. Write an equation connecting them and eliminate extra variables if needed.
3. Differentiate with respect to time.
4. Substitute the values at the requested instant, then state the rate with its sign and units.

A given value such as $r=5$ is an instant’s radius, not a statement that r stays constant.

Keep units on the final rate.

Example 12 preserves both source methods: the Teacher’s completed chain-rule method and the implicit time-differentiation method already printed in both copies.

Choose lengths from the geometry

For triangle areas in Examples 10–11, use intercepts to find the base and height, rather than substituting a line’s slope directly into an area formula. Example 10’s base is the distance between two x-intercepts; Example 11’s base and height are the positive axis intercepts. Exact surds and radians are retained.

Connect the volume rate to the changing depth

A fixed conical tank has height 12 cm and top radius 6 cm. Water enters at $2\ \mathrm{cm}^3/\mathrm{s}$. Find the rate of increase of water depth when it is 4 cm. Predict how this rate changes at depth 8 cm, with the same inflow, and justify the comparison.

After your attempt: compare the reasoning

Similar triangles give $r=h/2$ for the water surface, so

$V=\frac13\pi r^2h=\frac{\pi h^3}{12}.$

Differentiate the relation before substituting the depth:

$\frac{dV}{dt}=\frac{\pi h^2}{4}\frac{dh}{dt}.$

At $h=4$, $dh/dt=1/(2\pi)\ \mathrm{cm}/\mathrm{s}$. At $h=8$, it is $1/(8\pi)\ \mathrm{cm}/\mathrm{s}$, one quarter as large. The surface area through which the level rises is proportional to $h^2$; the volume rate stays fixed while the depth rate changes.