Playing with squares and circles

Calculus Level 5

direction of velocity of square is as shown

A circle of unit radius is placed at origin as shown. A square of unit side length moves with constant velocity of $2$ units in the direction as shown. When $OC=\dfrac{1}{4}$ , if the rate of change of shaded area can be represented as $\dfrac{\sqrt{A . B}}{C}$ where $\color{blueviolet}{A<B}$ , then find the rate of change of shaded area when $OC=\dfrac{1}{A}$ . If your answer is of the form $\dfrac{X\sqrt{Y}}{Z}$ where Y is square free and $X$ and $Z$ are co-prime to each other, then enter the value of

$X . Y . Z$ .

The answer is 24.

1 solution

Sparsh Sarode
Jun 12, 2016

Let $OC=x$ , then the enclosed area:

$\displaystyle \int_{x}^{1} ydx$

$\displaystyle \int_{x}^{1} \sqrt{1-x^2}dx=M$ $\text{ }(x^2+y^2=1\Rightarrow y=\sqrt{1-x^2})$

Then the rate of change of area is $\dfrac{dM}{dt}=\dfrac{dM}{dx} \dfrac{dx}{dt}$

But $\dfrac{dx}{dt}=\text{velocity}=v$

Rate of change of area= $\color{#3D99F6}{\sqrt{1-x^2}. v}$

At $OC=\dfrac{1}{4}$ , $\sqrt{1-\dfrac{1}{4^2}}. 2$

$=\dfrac{\sqrt{3. 5}}{2} \Rightarrow A=3$

At $OC= \dfrac{1}{3}$ , Substituting $x=\dfrac{1}{3}$ , we get:

$\dfrac{4\sqrt{2}}{3}$

$\therefore 4. 2. 3=24$

Playing with squares and circles

The answer is 24.

1 solution

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