Question

y varies directly as the square of r. if y is 8 when r is 7, find y when r is 28.
a) write the variation equation.
b) find y when r is 28.

a) how are these two variables related?

a.
y = \frac{k}{r^2}

b.
y = \frac{k}{r}

c. y = kr^2

d. y = kr

Explanation:

Part (a)

When a variable \( y \) varies directly as the square of another variable \( R \), the direct variation equation is of the form \( y = kR^2 \), where \( k \) is the constant of variation. Option A is inverse square variation, Option B is inverse variation, and Option D is direct variation with \( R \) (not its square). So the correct relation is \( y = kR^2 \).

Step 1: Find the constant \( k \)

We know that when \( y = 8 \) and \( R = 7 \), using the equation \( y = kR^2 \), we substitute these values:
\( 8 = k(7)^2 \)
\( 8 = 49k \)
To solve for \( k \), divide both sides by 49:
\( k=\frac{8}{49} \)

Step 2: Find \( y \) when \( R = 28 \)

Now, use the equation \( y = kR^2 \) with \( k=\frac{8}{49} \) and \( R = 28 \):
\( y=\frac{8}{49}\times(28)^2 \)
First, calculate \( 28^2 = 784 \):
\( y=\frac{8}{49}\times784 \)
Simplify \( \frac{784}{49}=16 \) (since \( 49\times16 = 784 \)):
\( y = 8\times16 \)
\( y = 128 \)

Answer:

C. \( y = kR^2 \)

Part (b)