Question

please solve this for me right now

Answer 1

Given : The length of the ladder is l=5ml=5m

dxdt=dxdt=2cm/s2cm/s

The distance of the foot of the ladder from the foot x=4mx=4m

The height of the wall is y=52−42−−−−−−√=3y=52−42=3

(By pythagoras theorem)

When x=5,y=3x=5,y=3

Also x2+y2=52x2+y2=52

Step 2:

Differentiating w.r.t tt on both sides we get,

2x×dxdt2x×dxdt+2y.dydt+2y.dydt=0=0

⇒x.dxdt⇒x.dxdt+y.dydt+y.dydt=0=0

Step 3:

Now substituting the values for x,yx,yand dxdtdxdt we get,

4×2+3×dydt4×2+3×dydt=0=0

⇒dydt⇒dydt=−83=−83cm/scm/s

Hence the height of the ladder on the wall decrease at the rate of 8383cm/s

Answer 2

Let the foot of the ladder be at a distance of x metres from the wall.

Let y be the height of the wall at any time t.

By Pythagoras theorem, we get

⇒ x^2 + y^2 = (5)^2

⇒ x^2 + y^2 = 25.

⇒ y^2 = 25 - x^2

⇒ y = √25 - x^2

On differentiating, we get

$=>\frac{dy}{dx}=\frac{-x}{\sqrt{25 - x^2}}*\frac{dx}{dt}$

Given dx/dt = 2 cm/s .

$=>\frac{dy}{dt}=\frac{-2x}{\sqrt{25 - x^2}}$

Given, foot of ladder is 4m away from the wall. So, x = 4

$=>\frac{dy}{dt}=\frac{-2*4}{\sqrt{25 - 4^2}}$

$=>\frac{dy}{dt}=\frac{-8}{3}$

Therefore, height of the wall is decreasing at the rate of (8/3) cm/s.

Hope it helps!