The Ladder!

$\Large \displaystyle \text{Problem Source: NCERT Mathematics Part - I}$

Let AB be the ladder and C is the junction of wall and ground, AB = 500 cm .

Let CA = x cm, CB = y cm

According to the equation, x increases, y decreases and $\frac{dx}{dt}$ = $2 cm/s$

By Pythagoras Theorem $AC^2 + BC^2 = AB^2$

⇒ $x^2 + y^2 = 250000$

Differentiate w.r.t 't'

⇒ $2x$ $\frac{dx}{dt}$ + $2y$ $\frac{dy}{dt}$ = 0

⇒ $4x$ + $2y$ $\frac{dy}{dt}$ = 0

⇒ $\frac{dy}{dt}$ = $\frac{-2x}{y}$

When $x$ = 400, $y$ = 300 $\boxed{By Pythagoras Theorem}$

∴ $\frac{dy}{dt}$ = $\frac{-2 * 400}{300}$

∴ $\frac{dy}{dt}$ = $\frac{-8}{3}$ ~ $\boxed{-2.667}$

Since, Decreased in mentioned in the question the answer is $\boxed{2.667}$

By pythagorean theorem, we have

$5^2=y^2+x^2$

Differentiating both sides with respect to time (t), we have

$0=2y\dfrac{dy}{dt}+2x\dfrac{dx}{dt}$ $\implies$ $2y\dfrac{dy}{dt}=-2x\dfrac{dx}{dt}$ $\implies$ $\dfrac{dy}{dt}=\dfrac{-x\dfrac{dx}{dt}}{y}$

When $x=4$ ,

$5^2=y^2+4^2$ $\implies$ $y=3$

Substitute,

$\dfrac{dy}{dt}=\dfrac{-4(2)}{3}=-2.667$

Since the direction is downward (negative direction), the desired answer is $-(-2.667)=\boxed{2.667}$