A 6 meter ladder...

Solution 1:

The picture shows your problem (note: not necessarily to scale). $x$ is the distance from the base of the ladder to the wall, and $y$ is the height of the ladder on the wall. Also, you know by the Pythagorean theorem that $x^2+y^2=36$. You may differentiate both sides with respect to $t$ (I presume you know how to do this, keeping in mind that $x$ and $y$ are functions of $t$). You will get an equation in terms of $x$, $y$, $\frac {dx}{dt}$, and $\frac {dy}{dt}$. You know three of these, so solving for the one you seek is trivial at this point.

Let me know if you have further questions!

Solution 2:

Let $x=x(t)$ be the distance of the foot of the ladder from the foot of the building, and let $y$ be the height of the top of the ladder. Then by the Pythagorean Theorem we have $x^2+y^2=36$.

Differentiation of $x^2+y^2=6^2$ gives $2x\frac{dx}{dt}+2y \frac{dy}{dt}=0$, and now we can answer any of the typical qeustions.

Solution 3:

Thanks so what i understand is

$x^2+y^2=36$

$x\frac{dx}{dt}+y\frac{dy}{dt}=0$

$x=3\sqrt{3}\hspace{0.03cm},y=3\hspace{0.03cm},\frac{dx}{dt}=2$

Therefore $\hspace{0.03cm}$$\frac{dy}{dt}=-2\sqrt{3}$