Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • x and y are the sides of two squares such that y = x – x^2. Find the rate of change of the area of the second square with respect to the area of the first square.

x and y are the sides of two squares such that y = x - x^2. Find the rate of change of the area of the second square with respect to the area of the first square.

Answers (1)

Given: two squares of sides x and y, such that y = x - x^2

To find: The rate of change of area of both the squares with respect to each other

Explanation: Take A_1 and  A_2  as the area of first and 2nd square respectively

Thus, the area of the 1st square will be

A_1 = x^2

Differentiating the equation with respect to time, we get

\\ \frac{\mathrm{d} \mathrm{A}_{1}}{\mathrm{dt}}=\frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{dt}} \\ \Rightarrow \frac{\mathrm{d} \mathrm{A}_{1}}{\mathrm{dt}}=2 \mathrm{x} \frac{\mathrm{d}(\mathrm{x})}{\mathrm{dt}} \ldots \ldots \text { (i) }

And the area of the second square is

A\textsubscript{2} = y\textsuperscript{2}$ \ldots $ $ \ldots $ .(ii)

But given, y = x-x\textsuperscript{2}

Now substituting the known value in equation (ii), 

A\textsubscript{2} = (x-x\textsuperscript{2}) ^{2} $ \ldots $ $ \ldots $ .(iii)

After differentiating equation (iii) with respect to time it results into,

$$ \frac{\mathrm{d} \mathrm{A}_{2}}{\mathrm{dt}}=\frac{\mathrm{d}\left(\left(\mathrm{x}-\mathrm{x}^{2}\right)^{2}\right)}{\mathrm{dt}} $$
Apply the power rule of differentiation to get,
$$ \frac{\mathrm{d} \mathrm{A}_{2}}{\mathrm{dt}}=2\left(\mathrm{x}-\mathrm{x}^{2}\right) \times \frac{\mathrm{d}\left(\mathrm{x}-\mathrm{x}^{2}\right)}{\mathrm{dt}} $$
Applying the sum rule of differentiation, we get

\begin{aligned} &\frac{\mathrm{d} \mathrm{A}_{2}}{\mathrm{dt}}=2\left(\mathrm{x}-\mathrm{x}^{2}\right)\left(\frac{\mathrm{d}(\mathrm{x})}{\mathrm{dt}}-\frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{dt}}\right)\\ &\text { Applying the derivative, we get }\\ &\frac{\mathrm{d} \mathrm{A}_{2}}{\mathrm{dt}}=2\left(\mathrm{x}-\mathrm{x}^{2}\right)\left(\frac{\mathrm{d}(\mathrm{x})}{\mathrm{dt}}-2 \mathrm{x} \frac{\mathrm{d}(\mathrm{x})}{\mathrm{dt}}\right)\\ &\Rightarrow \frac{\mathrm{d} \mathrm{A}_{2}}{\mathrm{dt}}=2 \mathrm{x}(1-\mathrm{x})\left(\frac{\mathrm{d}(\mathrm{x})}{\mathrm{dt}}(1-2 \mathrm{x})\right)\\ &\Rightarrow \frac{\mathrm{d} \mathrm{A}_{2}}{\mathrm{dt}}=2 \mathrm{x}(1-\mathrm{x})(1-2 \mathrm{x}) \frac{\mathrm{d}(\mathrm{x})}{\mathrm{dt}} \ldots \ldots(\mathrm{iv}) \end{aligned}

Since we need to find the rate of change of area of both the squares with respect to each other, which is

$$ \frac{\mathrm{d} \mathrm{A}_{2}}{\mathrm{~d} \mathrm{~A}_{1}}=\frac{\frac{\mathrm{d} \mathrm{A}_{2}}{\mathrm{dt}}}{\frac{\mathrm{d} \mathrm{A}_{1}}{\mathrm{dt}}} $$
Substituting the known values from equation (i) and (iv), we get
\frac{\mathrm{d} \mathrm{A}_{2}}{\mathrm{~d} \mathrm{~A}_{1}}=\frac{2 \mathrm{x}(1-\mathrm{x})(1-2 \mathrm{x}) \frac{\mathrm{d}(\mathrm{x})}{\mathrm{dt}}}{2 \mathrm{x} \frac{\mathrm{d}(\mathrm{x})}{\mathrm{dt}}}$
By cancelling the like terms, we get

\begin{aligned} &\frac{\mathrm{d} \mathrm{A}_{2}}{\mathrm{~d} \mathrm{~A}_{1}}=(1-\mathrm{x})(1-2 \mathrm{x})\\ &\Rightarrow \frac{\mathrm{d} \mathrm{A}_{2}}{\mathrm{~d} \mathrm{~A}_{1}}=1(1-2 \mathrm{x})-\mathrm{x}(1-2 \mathrm{x})\\ &\Rightarrow \frac{\mathrm{d} \mathrm{A}_{2}}{\mathrm{~d} \mathrm{~A}_{1}}=1-2 \mathrm{x}-\mathrm{x}+2 \mathrm{x}^{2}\\ &\Rightarrow \frac{\mathrm{d} \mathrm{A}_{2}}{\mathrm{~d} \mathrm{~A}_{1}}=2 \mathrm{x}^{2}-3 \mathrm{x}+1\\ &\text { Therefore, the rate of change of area of second square with respect to area of first square is }\\ &2 x^{2}-3 x+1 \end{aligned}

Posted by

infoexpert22

View full answer

Similar Questions

Latest Question