Question

The top of a ladder slides down a vertical wall at a rate of 0.675 m/s. At the moment when the bottom of the ladder is 6 m from the wall, it slides away from the wall at a rate of 0.9 m/s. How long is the ladder?

Answer 1

Answer:

The length of the ladder is 10 m.

Step-by-step explanation:

Let x shows the distance of the top of ladder from the bottom of base of the wall, y shows the distance of the bottom of ladder from the base of the wall and l is the length of the ladder,

Given,

$\frac{dx}{dt}=-0.675\text{ m/s}$

$\frac{dy}{dt}=0.9\text{ m/s}$

y = 6 m,

Since, the wall is assumed perpendicular to the ground,

By the pythagoras theorem,

$l^2=x^2+y^2$

Differentiating with respect to t ( time ),

$0=2x\frac{dx}{dt}+2y\frac{dy}{dt}$     ( the length of wall would be constant )

By substituting the value,

$0=2x(-0.675)+2(6)(0.9)$

$0=-1.35x+10.8$

$\implies x=\frac{10.8}{1.35}=8$

Hence, the length of the ladder is,

$L=\sqrt{x^2+y^2}=\sqrt{8^2+6^2}=\sqrt{64+36}=\sqrt{100}=10\text{ m}$

Answer 2

Answer:

The length of ladder=8m.

Step-by-step explanation:

Given

The rate at which the top of a ladder slides down a vertical wall,$\frac{\mathrm{d}z}{\mathrm{d}t}$= 0.675m/s

The distance of bottom of ladder from the wall,x=6m

The rate at which it slides away from the wall ,$\frac{\mathrm{d}x}{\mathrm{d}t}$=0.9m/s

Let length of ladder =z

Length of wall=y

Distance between foot of ladder and wall=x

By using pythogorous theorem

$x^2+y^2=z^2$

Differentiate w.r.t time

$x\frac{\mathrm{d}x}{\mathrm{d}t}=z\frac{\mathrm{d}z}{\mathrm{d}t}$

y does not change hence, $\frac{\mathrm{d}y}{\mathrm{d}t}=0$

$6\times 0.9=z\times 0.675$

$z=\frac{5.4}{0.675}$

z=8 m

Hence, the length of ladder=8m.