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}
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}
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$\frac{dy}{du} = \frac{dy/dx}{du/dx}$ is a useful shortcut for rates of change between two variables.
Updated On: Jun 19, 2026
- $2x^{2}+3x+1$
- $3x^{2}+2x-1$
- $2x^{2}-3x+1$
- $3x^{2}+2x+1$
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The Correct Option is C
Solution and Explanation
Step 1: Formula
Area of first square $A_1 = x^2$. Area of second square $A_2 = y^2 = (x-x^2)^2$.
Step 2: Meaning
We need to find $\frac{dA_2}{dA_1}$. Using chain rule: $\frac{dA_2}{dA_1} = \frac{dA_2/dx}{dA_1/dx}$.
Step 3: Analysis
- $\frac{dA_1}{dx} = 2x$ - $\frac{dA_2}{dx} = 2(x-x^2)(1-2x) = 2(x - 2x^2 - x^2 + 2x^3) = 2(x - 3x^2 + 2x^3)$ - Ratio: $\frac{2(x - 3x^2 + 2x^3)}{2x} = \frac{x(1 - 3x + 2x^2)}{x} = 2x^2 - 3x + 1$.
Step 4: Conclusion
Hence, the rate of change is $2x^2 - 3x + 1$. Final Answer: (C)
Area of first square $A_1 = x^2$. Area of second square $A_2 = y^2 = (x-x^2)^2$.
Step 2: Meaning
We need to find $\frac{dA_2}{dA_1}$. Using chain rule: $\frac{dA_2}{dA_1} = \frac{dA_2/dx}{dA_1/dx}$.
Step 3: Analysis
- $\frac{dA_1}{dx} = 2x$ - $\frac{dA_2}{dx} = 2(x-x^2)(1-2x) = 2(x - 2x^2 - x^2 + 2x^3) = 2(x - 3x^2 + 2x^3)$ - Ratio: $\frac{2(x - 3x^2 + 2x^3)}{2x} = \frac{x(1 - 3x + 2x^2)}{x} = 2x^2 - 3x + 1$.
Step 4: Conclusion
Hence, the rate of change is $2x^2 - 3x + 1$. Final Answer: (C)
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