Related rate problems typically present a scenario where the rate of one change is given, and you are asked to determine the rate of another change when a specific event occurs. These problems often involve geometric relationships. Let’s illustrate this with an example:

Example. Consider a water tank in the shape of an inverted circular cone with a base radius of 2 m and a height of 4 m. Water is being pumped into the tank at a rate of 2 $\text{m}^3/\text{min}$. Find the rate at which the water level is rising when the water depth is 3 m.

  • Given rate: $\frac{dV}{dt} = 2$ ($\text{m}^3/\text{min}$).
  • Required rate: $\frac{dh}{dt}\mid_{h=3}$.
  • The relationship between $V$ and $h$ is determined by the cone’s geometry.

Solution: The volume of a cone with height $h$ and base radius $r$ is given by:

\[V = \frac{1}{3}\pi r^2h.\]

Moreover, we know that $r:h = 2:4$ using similar triangles, which implies $V = \frac{1}{3}\pi\left(\frac{1}{2}h\right)^2 h$. Since $r$ also changes as water is pumped in, $r$ is a function of time $t$, and we cannot treat it as a constant. Now, applying implicit differentiation, we find:

\[\frac{dV}{dt} = \frac{1}{4}\pi h^2\frac{dh}{dt},\]

so $\frac{dh}{dt}\mid_{h=3} = 2\cdot 4\cdot \frac{1}{3^2\pi}$.

Linear Approximation

The linearization of a function $f(x)$ at $x=a$ corresponds to the equation of the tangent line of the curve $y=f(x)$ at the point $(a,f(a))$, which is given by:

\[y = f(a) + f'(a)(x-a).\]

This equation allows us to approximate the function $f(x)$ near $a$ using the expression $f(a) + f’(a)(x-a)$, resulting in:

\[f(x)\approx f(a) + f'(a)(x-a).\]

It’s worth noting that the linearization of $f(x)$ at $x=a$ is also referred to as the 1st-degree Taylor polynomial.