X is directly proportional to the square of Y. When X is 12, then Y is 2. Find the value of X when Y is 3.
27
The question states that X is directly proportional to the square of Y. This means that as the square of Y increases, X increases at a constant rate, and vice versa. We can express this relationship mathematically.
When X is directly proportional to the square of Y, we can write the relationship as:
\(X \propto Y^2\)
To turn this proportionality into an equation, we introduce a constant of proportionality, let's call it \(k\). The equation becomes:
\(X = kY^2\)
Here, \(k\) is a constant value that relates X and the square of Y.
We are given that when X is 12, Y is 2. We can use these values to find the value of \(k\). Substitute X = 12 and Y = 2 into the equation \(X = kY^2\):
\(12 = k \times (2)^2\)
\(12 = k \times 4\)
Now, we can solve for \(k\):
\(k = \frac{12}{4}\)
\(k = 3\)
So, the constant of proportionality is 3. The specific relationship between X and Y is \(X = 3Y^2\).
Now that we know the relationship \(X = 3Y^2\), we can find the value of X when Y is 3. Substitute Y = 3 into the equation:
\(X = 3 \times (3)^2\)
\(X = 3 \times 9\)
\(X = 27\)
Therefore, when Y is 3, the value of X is 27.
The value of X when Y is 3 is 27.
| Concept | Description | Equation Form |
|---|---|---|
| Direct Proportionality | As one quantity increases, the other increases at a constant rate. Ratio is constant. | \(y \propto x\) or \(y = kx\) |
| Direct Proportionality to a Power | As one quantity increases, the other increases at a rate proportional to a power of the first. | \(y \propto x^n\) or \(y = kx^n\) |
| Constant of Proportionality (k) | The constant value relating the two proportional quantities. Found using one pair of values. | \(k = y/x\) or \(k = y/x^n\) |
Understanding proportionality is crucial in many mathematical and scientific contexts. Besides direct proportionality, two other common types are inverse proportionality and joint proportionality.
Being able to identify the type of proportionality and set up the correct equation with the constant \(k\) is the key to solving such problems.
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