Non-Motion Problems Non-Motion Problems

Introduction

The bread and butter example of derivatives and second-order derivatives is motion (position/velocity/acceleration). However, these concepts can be applied to any example of rates of change.

The Melting Ice Cube

An ice cube is melting. Its size after

t
seconds is equal to the following function:
V(t) = 1000(1-0.05t)^3
V(t)
is in cubic centimeters and
t
is in minutes.

Do the following:
1. Find the volume of the ice cube at 10 minutes.
2. Find how fast the ice cube is shrinking at 10 minutes.
3. Find how fast the top face of the ice cube is shrinking at 10 minutes.

Find the size of the ice cube at 10 minutes.

We are looking for the volume of the ice cube at 10 minutes. It is given by

V(t) = 1000(1-0.05t)^3
We want
V(10)
.
\begin{align*} V(10) &= 1000(1-0.05(10))^3 \\ &= 1000(1-0.5)^3 \\ &= 1000(0.5)^3 \\ &= 1000(0.125) \\ &= 125\text{cm}^3 \end{align*}
The ice cube has a volume of 125 cubic centimeters after 10 minutes.

How fast is the ice cube shrinking at 10 minutes?

1. Understand what the question is asking.
We are looking for the rate that the ice cube is shrinking at 10 minutes. That is the rate the volume is decreasing per second.
2. Convert the request into mathematical notation.

V(t) = 1000(1-0.05t)^3
V'(t) = \frac{d}{dt}V(t) = \frac{d}{dt}1000(1-0.05t)^3
3. Solve the differential equation(s).
\begin{align*} V'(t) &= 3(1000)(1-0.05t)^2(-0.05) \\ &= -150(1-0.05t)^2 \\ V(10)&= -150(1-0.05(10))^2 \\ &= -150(1-0.5)^2 \\ &= -150(0.5)^2 \\ &= -150(0.25) \\ &= -37.5\text{cm}^3/\text{min} \end{align*}
4. Interpret the solution.
10 minutes after starting, the ice cube is shrinking at a rate of 37.5 cubic centimeters per minute.

How fast is the top face of the ice cube shrinking at 10 minutes?

\htmlClass{text-dark_primary dark:text-primary}{S(t)=\text{Length of One Side}} \quad\quad \htmlClass{text-primary dark:text-dark_primary}{A(t)=\text{Area of One Face}} \quad\quad V(t)=\text{Volume of the Cube}
1. Understand what the question is asking.
We are looking for the rate that the top face of the ice cube is shrinking at 10 minutes. In the image above, that is the rate at which the filled in square is shrinking.

We need to find the area of one face of the cube using the volume of the cube. The volume is the length of one side cubed. The area is the length of one side squared.

Then, we need to find the rate that the area is shrinking with respect to time.
2. Convert the request into mathematical notation.
S(t)
is the length of one side.
A(t)
is the area of one side.
V(t) = 1000(1-0.05t)^3
V(t)=S(t)^3
A(t)=S(t)^2
S(t)=\sqrt[3]{V(t)}

3. Solve the differential equation(s).
\begin{align*} S(t)&=\sqrt[3]{1000(1-0.05t)^3} \\ &=10(1-0.05t) \\ &=10-0.5t \\ A(t)&=S(t)^2 \\ &=(10-0.5t)^2 \\ A'(t)&=\frac{d}{dt}A(t) \\ &=\frac{d}{dt}(10-0.5t)^2 \\ &=2(10-0.5t)(-0.5) \\ &=-1(10-0.5t) \\ &=-10+0.5t \\ A'(10)&=-10+0.5(10) \\ &=-10+5 \\ &=-5\text{cm}^2/\text{min} \end{align*}
4. Interpret the solution.
10 minutes after starting, the top face of the ice cube is shrinking at a rate of 5 square centimeters per minute.

Bonus Method of Finding the Derivatives

\begin{align*} S(t)&=10-0.5t \quad& V(t)=&S(t)^3 \quad& A(t)=&S(t)^2 \\ \text{} \\ S'(t)&=-0.5t \quad& V'(t)=&3S(t)^2S'(t) \quad& A'(t)=&2S(t)S'(t) \\ \text{} \\ \end{align*} \\ \begin{align*} S(10)&=10-0.5(10) \\ &=10-5 \\ &=5 \\ S'(10)&=-0.5 \\ V'(10)&=3(5)^2(-0.5) \\ &=-37.5\text{cm}^3/\text{min} \\ A'(10)&=2(5)(-0.5) \\ &=-5\text{cm}^2/\text{min} \end{align*}