Application of Derivatives

A particle moves along the curve $$y=x^{\dfrac{3}{2}}$$ in the first quadrant in such a way that its distance from the origin increases at the rate of 11 units per second.The value of $$dx/dt$$ when $$x=3$$ is
A straight line is said to be an asymptote to the curve y=f(x) if the $$\perp $$ distance of point P(x, y) on the curve from the line tends to zero when $$x\rightarrow \infty $$ or $$y\rightarrow \infty $$ or both x & y$$\rightarrow \infty $$
(a) Asymptotes parallel to y-axis:
The line $$x=k$$ is asymptote to the curve y=f(x) If
$$\lim_{x\rightarrow k^{+}}f\left ( x \right )=\infty $$ or $$-\infty $$ & $$\lim_{x\rightarrow k^{-}}f\left ( x \right )=+\infty $$ or $$-\infty $$
b) Asymptotes parallel to x-axis:
The line $$y=k$$ is asymptotes to the curve $$y=f(x)$$
if $$\lim_{x\rightarrow \infty }y=k$$ or $$\lim_{x\rightarrow -\infty }y=k$$
(c) Oblique asymptotes:
Let y=mx+c be an asymptote of the curve y=f(x), then from any point P(x, y) on the curve, the $$\perp $$ distance is $$\displaystyle \frac{\left | y-mx-c \right |}{\sqrt{1+m^{2}}}$$ which $$\rightarrow 0$$ as x tends to $$+\infty $$ or $$+\infty $$
Thus $$y-mx-c\rightarrow 0$$ as $$x\rightarrow +\infty $$ or $$-\infty $$
Again $$\displaystyle y-mx-c=x\left ( \frac{y}{x}-\frac{c}{x}-m \right )=0$$
As $$x\rightarrow \infty $$ and $$y-mx-c\rightarrow 0$$
So $$\displaystyle \lim_{x\rightarrow \pm \infty }\left ( \frac{y}{x}-\frac{c}{x}-m \right )=0$$   $$\Rightarrow $$   $$\displaystyle \lim_{x\rightarrow \pm \infty }\frac{y}{x}=m$$
and $$\lim_{x\rightarrow \pm \infty }y-mx-c=0$$   $$\Rightarrow $$   $$\lim_{x\rightarrow \pm \infty }y-mx=c$$
So we come to the fact that if y=mx+c is an asymptote to the curve y=f(x). Then as $$x\rightarrow +\infty $$ or $$-\infty $$
$$\lim_{x\rightarrow \infty }\frac{y}{x}=m=\lim_{x\rightarrow \infty }\frac{f\left ( x \right )}{x}$$ & $$\lim_{x\rightarrow \infty }=y-mx=\lim_{x\rightarrow \infty }\left \{ f\left ( x \right )-mx \right \}=c$$
On the basis of above information answer the following questions.
A window of fixed perimeter (including the base of the arch) is in the form of a rectangle surmounted by a semi-circle. The semi-circular portion is fitted with coloured glass, while the rectangular portion is fitted with clear glass. The clear glass transmits three times as much light per square metre as the coloured glass.
Suppose that $$y$$ is the length and $$x$$ is the breadth of the
rectangular portion and $$p$$ the perimeter.

Water is flowing out at the rate of $$\displaystyle 6\:\:m^{3}/min$$ from a reservoir shaped like a hemispherical bowl of radius$$ R = 13$$ m The volume of water in the hemispherical bowl is given by $$\displaystyle v=\frac{\pi }{3}.y^{2}\left ( 3R-y \right )$$ when the water is $$y$$ meter deep Find
Suppose $$f(x)$$ is a real valued polynomial function of degree $$6$$ satisfying the following conditions:
$$f$$ has minimum value at $$x = 0$$ and $$2$$
$$f$$ has maximum value at $$x = 1$$ 
for all $$x$$, $$\displaystyle \lim_{x\rightarrow 0}\frac{1}{x}\ln\begin{vmatrix}\dfrac{f\left ( x \right )}{x} &1  &0 \\0  &\dfrac1x  &1 \\1  &0  &\dfrac1x \end{vmatrix} = 2$$