A varies directly as b\(^2\) when A = 4, b = 1. Find A when b = 2

When a quantity A varies directly as \( b^2 \), it means that as \( b^2 \) increases or decreases, A changes in exactly the same proportion. This type of relationship can be written mathematically as:

\[ A = k \cdot b^2 \] where k is a constant of proportionality.

We are given that when \( A = 4 \), \( b = 1 \). We can use these values to find the value of \( k \):

\[ 4 = k \cdot (1)^2 \\ 4 = k \cdot 1 \\ k = 4 \]

Now, we want to find the value of \( A \) when \( b = 2 \). Substitute \( k = 4 \) and \( b = 2 \) into the equation:

\[ A = 4 \cdot (2)^2 \\ A = 4 \cdot 4 \\ A = 16 \]

So, when \( b = 2 \), \( A = 16 \).

• This answer shows that as b doubles (from 1 to 2), b² becomes four times larger, and thus A is also four times larger than when \( b = 1 \), since \( A \) is directly proportional to \( b^2 \).