Browse Questions

# A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall ?

$\begin{array}{1 1}(A) \;\frac{8}{3}cm/s \\ (B)\frac{-8}{3}cm/s \\ (C)\;\frac{8}{3}cm^2/s \\ (D)\;\frac{8}{3}cm^2/s \end{array}$

Toolbox:
• If $y=f(x)$,then $\large\frac{dy}{dx}$ measures the rate of change of $y$ w.r.t $x$.
• $\big(\large\frac{dy}{dx}\big)_{x=x_0}$ represents the rate of change of $y$ w.r.t $x$ at $x=x_0$
Step 1:
Given : The length of the ladder is $l=5m$
$\large\frac{dx}{dt}=$$2cm/s The distance of the foot of the ladder from the foot x=4m The height of the wall is y=\sqrt{5^2-4^2}=3 (By pythagoras theorem) When x=5,y=3 Also x^2+y^2=5^2 Step 2: Differentiating w.r.t t on both sides we get, 2x\times \large\frac{dx}{dt}$$+2y.\large\frac{dy}{dt}$$=0 \Rightarrow x.\large\frac{dx}{dt}$$+y.\large\frac{dy}{dt}$$=0 Step 3: Now substituting the values for x,y and \large\frac{dx}{dt} we get, 4\times 2+3\times \large\frac{dy}{dt}$$=0$
$\Rightarrow \large\frac{dy}{dt}$$=-\large\frac{8}{3}$$cm/s$
Hence the height of the ladder on the wall decrease at the rate of $\large\frac{8}{3}$$cm/s$