The area, A, in square metres, of a circular oil spill after t min is given by A = pt2.
The instantaneous rate of change of the area of the oil spill after 2 min is approximately

The formula for the area of the oil spill is given as:

\[
A = p t^2
\]

To find the **instantaneous rate of change** of the area after 2 minutes, we need to compute the derivative of \( A \) with respect to \( t \), and then evaluate it at \( t = 2 \).

1. First, take the derivative of \( A \) with respect to \( t \):

\[
\frac{dA}{dt} = \frac{d}{dt} \left( p t^2 \right) = 2p t
\]

2. Now, substitute \( t = 2 \) into the derivative to find the rate of change at that specific time:

\[
\frac{dA}{dt} \Bigg|_{t=2} = 2p(2) = 4p
\]

Thus, the instantaneous rate of change of the area of the oil spill after 2 minutes is approximately \( 4p \) square meters per minute.

Would you like further details or have any questions?

Here are 5 related questions:
1. How does the formula for area change if the spill spreads at a variable rate?
2. What is the interpretation of the derivative in this context?
3. How would the rate of change differ if \( t = 3 \) minutes?
4. Can we interpret the constant \( p \) in terms of physical properties of the oil?
5. How would the analysis change if the relationship between area and time was not quadratic?

**Tip**: The derivative gives the rate of change at any instant, which is crucial for understanding how a system evolves over time.

The area, A, in square metres, of a circular oil spill after t min is given by A = p * t^2. The instantaneous rate of change of the area of the oil spill after 2 min is approximately:

Learn how to calculate the instantaneous rate of change of the area of a circular oil spill given by A = p * t^2 after 2 minutes. This problem covers key calculus concepts such as derivatives and the power rule. Suitable for high school students in Grades 10-12.

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