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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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x and y are the sides of two squares such that $y=x-x^2$.Find the rate of change of the area of second square with respect to the area of first square.

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Toolbox:
  • If $ y = f(x)$, then $ \large\frac{dy}{dx}$ measures the rate of change of $y$ w.r.t $ x $.
  • $ \bigg(\large \frac{dy}{dx} \bigg)_{x=x_0}$ represents the rate of change of $y$ w.r.t $x$ at $x=x_0$
Step 1
Given : The sides of the two squares are $x$ and $y$ and
$ y = x-x^2$
Area of the I square is $ x^2$
Area of the II square is $ y^2$
$=(x-x^2)^2$
Let $A_1=x^2$
Differentiating w.r.t $x$ we get
$ \large\frac{dA_1}{dx}=2x$
$ A_2=(x-x^2)^2$
Differentiating w.r.t $x$ we get
$ \large\frac{dA_2}{dx}=2(x-x^2)(1-2x)$
$ =4x^3-6x^2+2x$
$=2x(2x^2-3x+1)$
$\large\frac{dA_1}{dx} (2x^2-3x+1)$
Hence rate of change in area w.r.t the area of the first square is
$ 2x^2-3x+1$
answered Aug 2, 2013 by thanvigandhi_1
 

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