Introduction

Welcome to the guide of Solving Related Rates Problems in FiveHive's AP Calculus course. In this guide, we will learn how to solve related rate problems

Examples

In Topic 4.4, you already saw some examples of related rate problems. In this section, we will go over more questions relating to the concept of related rate.

Example 1: In physics, isothermal process refers to a thermodynamic process in which the temperature of gas is constant, meaning that $PV=C$, where $P$ is the pressure of the gas and $V$ is the volume of the gas. $C=100\mathrm{L}$. When $V=1.5\mathrm{L}$ and the volume is increasing by $1.5L/sec$, find the change in pressure at this instant.

Solution:

Rearrange the equation so that $P=\frac{100}{V}$.

Now, take derivative respect to $t$ on both side $\frac{dP}{dt}=-100\frac{1}{V^2}\frac{dV}{dt}$. 

Finally, plug $V=1.5\mathrm{L}$ and $\frac{dV}{dt}=1.5L/sec$, we obtain the numerical value for the change in pressure with respect to time.

$\frac{dP}{dt}=-100\frac{1}{1.5^2}\cdot 1.5=66.67pascal/sec$

Physically, this solution makes sense because as volume increases, pressure would decrease.

Example 2: The radius of a balloon is decreasing by $0.5cm/sec$. Find the rate of change of the volume of the balloon when $r=3\mathrm{cm}$. ($V_{\text{sphere}}=\frac{4}{3}\pi r^3$)

Solution: 

Directly take derivatives with respect to time $t$. $\frac{dV}{dt}=\frac{4}{3}\pi \cdot 3r^2\frac{dr}{dt}=4\pi r^2\frac{dr}{dt}$.

Substitute $r=3\mathrm{cm}$ and $\frac{dr}{dt}=-0.5cm/sec$ (because the radius is decreasing), we obtain $\frac{dV}{dt}=4\pi \cdot 3^2\cdot -0.5=-18\pi \mathrm{c}{\mathrm{m}}^3/sec$

Example 3: An upside down cone has an opening angle of $30\mathrm{deg}$. A tap supplies the cone with water at a rate of $1\mathrm{c}{\mathrm{m}}^3/sec$. When the water height is at $h=3\mathrm{cm}$, at what rate is the height of the water level in the cone increasing? The volume of a cone with radius $r$ and height $h$ is $V=\frac{1}{3}\pi r^{2}h$

Solution:

Geometrically, the height of the cone and the radius at that height is related through the tangent function, where $r=h\tan \alpha$, where $h$ is the height and $\alpha$ is the opening angle. Substitute this into the volume of the cone, we have

$V=\frac{1}{3}\pi h^2{\tan }^2\alpha h=\frac{1}{3}\pi h^3{\tan }^{2}30\mathrm{deg}=\frac{1}{9}\pi h^3$. 

Now, take derivatives on both side:

$\frac{dV}{dt}=\frac{1}{9}\pi \cdot 3h^2\frac{dh}{dt}$

Isolate $\frac{dh}{dt}$, one get $\frac{dh}{dt}=\frac{3}{\pi h^2}\frac{dV}{dt}$.

Substitute the values in, $\frac{dh}{dt}=\frac{3}{3^2\pi }\cdot 1=\frac{1}{3\pi }cm/sec$. 

It is important to do a sanity check. From the calculated expression that relates $\frac{dV}{dt}$ and $\frac{dh}{dt}$, we see that as $h$ gets larger, $\frac{dh}{dt}$ gets smaller. This is expected because as the height gets larger, the radius also gets larger, making the overall volume of the cone larger.

Practice Questions