Question

If $x$ and $y$ are sides of two squares such that $y=x-x^{2}$, then the rate of change of area of the second square with respect to that of the first square is}

• A. $2x^{2}+3x+1$
• B. $3x^{2}+2x-1$
• C. $2x^{2}-3x+1$
• D. $3x^{2}+2x+1$

Hint:

$\frac{dy}{du} = \frac{dy/dx}{du/dx}$ is a useful shortcut for rates of change between two variables.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
Let $A_1$ and $A_2$ be the areas of the first and second squares respectively. We need to find $\frac{dA_2}{dA_1}$.

Step 2: Key Formula or Approach:
1. $A_1 = x^2$
2. $A_2 = y^2$
3. Using chain rule: $\frac{dA_2}{dA_1} = \frac{dA_2/dx}{dA_1/dx}$

Step 3: Detailed Explanation:
Given $y = x - x^2$.
$A_2 = y^2 = (x - x^2)^2$.
Differentiating $A_1$ with respect to $x$:
$\frac{dA_1}{dx} = 2x$
Differentiating $A_2$ with respect to $x$:
$\frac{dA_2}{dx} = 2(x - x^2) \cdot \frac{d}{dx}(x - x^2) = 2(x - x^2)(1 - 2x)$
$\frac{dA_2}{dx} = 2x(1 - x)(1 - 2x)$
Now, calculate the ratio:
$\frac{dA_2}{dA_1} = \frac{2x(1 - x)(1 - 2x)}{2x} = (1 - x)(1 - 2x)$
Multiply the factors:
$(1 - x)(1 - 2x) = 1 - 2x - x + 2x^2 = 2x^2 - 3x + 1$

Step 4: Final Answer:
The rate of change is $2x^2 - 3x + 1$.