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The cube of $y$ varies directly as $z$. When $y=8, z=8$. Heather's work finding the value of $y$ when $z=64$ is shown:



$\begin{aligned}

y^3 & =k z \\

8 & =k \cdot 8 \\

1 & =k \\

y^3 & =1 \cdot 64 \\

y^3 & =64 \\

y & =4

\end{aligned}$



What is the first error, if any, in Heather's work?


AShe did not make any errors.
BShe cubed the wrong variable when making her equation.
CShe substituted incorrectly when calculating the constant of variation.
DShe used an equation that models inverse variation instead of direct variation.

Verified Answer :

  • The relationship is defined by \(y^3 = kz\), where \(y=8, z=8\) gives \(8^3 = k(8)\), so \(k=64\).
  • Heather incorrectly substituted \(y=8\) into the equation as \(8 = k \cdot 8\) instead of \(8^3 = k \cdot 8\).
  • The first error occurred during the substitution step when calculating the constant of variation.
  • The correct choice is: \(\boxed{\text{She substituted incorrectly when calculating the constant of variation.}}\)

Explanation

  1. Determining the Correct Constant of Variation

The problem states that the cube of \(y\) varies directly as \(z\). This is represented by the equation \(y^3 = k z\), where \(k\) is a constant of variation. We are given the initial conditions \(y = 8\) and \(z = 8\). To find the value of \(y\) when \(z = 64\), we must first correctly determine the constant \(k\). Substituting the given values into the equation, we have:
\[(8)^3 = k \cdot 8\]
\[512 = 8k\]
\[k = \frac{512}{8} = 64\]
Now, using \(k = 64\) and the second condition \(z = 64\), we find \(y\):
\[y^3 = 64 \cdot 64\]
\[y^3 = 4096\]
\[y = \sqrt[3]{4096} = 16\]

  1. Identifying the Error in Heather's Work

Heather's work began correctly with the expression \(y^3 = k z\). However, in the next step, she substituted \(y = 8\) directly into the equation instead of calculating \(y^3\). Her calculation was:
\[8 = k \cdot 8\]
This led to the incorrect constant \(k = 1\). This was an error in substitution when calculating the constant of variation. By substituting \(y\) instead of \(y^3\), she failed to properly account for the relationship between the cube of \(y\) and \(z\). The correct substitution should have been \(8^3 = k \cdot 8\).

Examples

In physics and engineering, direct and inverse variation are fundamental concepts. For example, the power dissipated by a resistor in a circuit follows the relationship \(P = I^2 R\), where power \(P\) varies directly as the square of the current \(I\). Understanding these relationships is crucial for correctly modeling electrical circuits, where incorrectly substituting values—like forgetting to square the current—would lead to significant errors in design calculations and could potentially lead to equipment failure or safety risks. Mastering these basic algebraic relationships ensures accurate modeling and safe application of scientific laws.

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