6. Rate of Change

Today, we are going to discuss the concept of rate of change. Can anyone tell me what you think it means?

Is it about how one thing changes compared to another?

Exactly! It's essentially the speed at which one quantity is changing in relation to another quantity. Mathematically, we express this as the derivative. Can anyone give me an example of where we see rate of change in our lives?

Like the velocity of a car?

Right! Velocity is the rate of change of displacement over time. So if you see a speedometer showing 60 mph, that means for each hour, the car travels 60 miles. Remember, we often denote this as $v = \frac{ds}{dt}$.

Now let's connect this concept to economics. Can anyone think of what we might call the rate of change of total cost with an increase in quantity produced?

Is that marginal cost?

Exactly! Marginal cost represents the additional cost incurred for producing one more unit. It's expressed mathematically as $MC = \frac{dC}{dQ}$, where $C$ is the total cost and $Q$ is the quantity produced.

So if our total cost increases quickly with each unit produced, the marginal cost will be high?

Precisely! Understanding marginal costs helps businesses optimize production and set prices effectively.

Let's move on to a specific example. If the radius of a sphere increases at 2 cm/s, I want to determine how fast the volume is changing when the radius is 3 cm. What formula do we use for the volume of a sphere?

It's $V = \frac{4}{3} \pi r^3$.

Good job! Now, let's differentiate this with respect to time. What do we get?

We'll use the chain rule to differentiate. It becomes $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$.

Exactly! Now plug in the values of $r = 3$ cm and $\frac{dr}{dt} = 2$ cm/s. What do we find?

It looks like $\frac{dV}{dt} = 4\pi (3^2)(2) = 72\pi$ cm³/s.

Yes! So at that moment, the volume of the sphere is increasing at a rate of $72\pi$ cm³/s.