\displaystyle \textbf{Question 1: }\ \ \text{The side of an equilateral triangle is increasing at the rate of }2\ \text{cm/sec.} \\ \text{At what rate is its area increasing when the side of the triangle is }20\ \text{cm?}\ \text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle  \text{At any time } t,\ \text{let } x\ \text{cm be the length of a side of an equilateral triangle and } A\ \text{be its area. Then,}
\displaystyle A=\frac{\sqrt{3}}{4}x^2
\displaystyle \Rightarrow\ \frac{dA}{dt}=\frac{\sqrt{3}}{4}\times2x\frac{dx}{dt}=\frac{\sqrt{3}}{2}x\frac{dx}{dt}
\displaystyle \therefore\ \left(\frac{dA}{dt}\right)_{x=20}=\frac{\sqrt{3}}{2}\times20\times2=20\sqrt{3}\ \text{cm}^2/\text{sec}\ \ \ \left[\because\ \frac{dx}{dt}=2\ \text{cm/sec (given)}\right]
\displaystyle \text{Hence, the area is increasing at the rate of }20\sqrt{3}\ \text{cm}^2/\text{sec.}

\displaystyle \textbf{Question 2: }\ \ \text{A ladder }5\ \text{m long is leaning against a wall. The bottom of the} \\ \text{ladder is pulled along the}\text{ground away from the wall at the rate of }2\ \text{m/sec.} \\ \text{How fast is height on the wall decreasing when the foot} \text{of the ladder is }4\ \text{m} \\ \text{away from the wall?}\ \text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Let } AB \text{ be the position of the ladder at any time } t \text{ such that } OA=x \text{ and } OB=y.
\displaystyle \text{Then,}
\displaystyle OA^2+OB^2=AB^2\ \Rightarrow\ x^2+y^2=5^2\ \ \ \ (i)
\displaystyle \text{It is given that the bottom of the ladder is pulled along the ground}
\displaystyle \text{away from the wall at the rate of }2\ \text{m/sec.}
\displaystyle \therefore\ \frac{dx}{dt}=2\ \text{m/sec.}
\displaystyle \text{Now,}\ \ x^2+y^2=5^2
\displaystyle \Rightarrow\ 2x\frac{dx}{dt}+2y\frac{dy}{dt}=0
\displaystyle \Rightarrow\ 2x(2)+2y\frac{dy}{dt}=0\ \ \ \left[\because\ \frac{dx}{dt}=2\right]
\displaystyle \Rightarrow\ \frac{dy}{dt}=-\frac{2x}{y}\ \ \ \ (ii)
\displaystyle \text{Putting } x=4 \text{ in (i), we get: }\ y=\sqrt{25-16}=3.
\displaystyle \text{Putting } x=4 \text{ and } y=3 \text{ in (ii), we get: }\ \frac{dy}{dt}=-\frac{8}{3}\ \text{m/sec.}
\displaystyle \text{Hence, the rate of decrease in the height of the ladder on the wall is }\frac{8}{3}\ \text{m/sec.}

\displaystyle \textbf{Question 3: }\ \ \text{Sand is pouring from a pipe at the rate of }12\ \text{cm}^3/\text{sec. The falling} \\ \text{sand forms a cone on the} \text{ground in such a way that the height of the cone is always} \\ \text{one-sixth of the radius of the base. How fast is the} \text{height of the sand-cone} \\ \text{increasing when the height is }4\ \text{cm?}\ \text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Let } r \text{ be the radius, } h \text{ be the height and } V \text{ be the volume of the sand-cone at any time } t.
\displaystyle \text{Then, } V=\frac{1}{3}\pi r^2h
\displaystyle \Rightarrow\ V=\frac{1}{3}\pi(36h^2)h=12\pi h^3\ \ \ \left[\because\ r=6h\right]
\displaystyle \Rightarrow\ \frac{dV}{dt}=36\pi h^2\frac{dh}{dt}\ \ \ \ (i)
\displaystyle \Rightarrow\ \frac{dh}{dt}=\frac{1}{3\pi h^2}\ \ \ \left[\because\ \frac{dV}{dt}=12\ \text{(Given)}\right]
\displaystyle \Rightarrow\ \left(\frac{dh}{dt}\right)_{h=4}=\frac{1}{3\pi(4)^2}=\frac{1}{48\pi}
\displaystyle \text{Thus, the height of the sand-cone is increasing at the rate of }\frac{1}{48\pi}\ \text{cm/sec.}

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