Question

A 5 meter long ladder leans against a wall. The bottom of the ladder slides away from the wall at the constant rate of 1 3 m/s. How fast is the angle θ between the ground and the ladder changing when the bottom of the ladder is is 3 meters from the wall?

Answer 1

The rate at which the angle between the ground and the ladder is changing when the bottom of the ladder is 3 meters from the wall is zero.

Step-by-step explanation:

Given that the ladder is 5 meters long and slides away from the wall at a constant rate of 1.3 m/s, we can find the rate at which the angle θ between the ground and the ladder changes when the bottom of the ladder is 3 meters from the wall.

Let x represent the distance between the bottom of the ladder and the wall. Using the Pythagorean theorem, we have:

x^2 + (5^2) = (3^2)

x^2 = 9

x = 3

Differentiating both sides of the equation concerning time gives us:

2x(dx/dt) = 0

Since dx/dt = 1.3 m/s, we can substitute x = 3 and solve for dθ/dt.

2(3)(1.3) = 0

dθ/dt = 0

Answer 2

Answer:9.75 m/s

Step-by-step explanation:

Given

Length of ladder
`(L)=5 m`

Foot the ladder is moving away with speed of
`\frac{\mathrm{d} x}{\mathrm{d} t}=13 m/s`

From diagram

`x^2+y^2=L^2`------1

at
`x=3`

`y^2=25-9=16`

`y=4 m`

Now differentiating equation 1 w.r.t time

`2x\frac{\mathrm{d} x}{\mathrm{d} t}+2y\frac{\mathrm{d} y}{\mathrm{d} t}=0`

`x\frac{\mathrm{d} x}{\mathrm{d} t}=-y\frac{\mathrm{d} y}{\mathrm{d} t}`

`3* 13=-4* \frac{\mathrm{d} y}{\mathrm{d} t}`

`\frac{\mathrm{d} y}{\mathrm{d} t}=-(3* 13)/(4)=-9.75 m/s`

negative indicates distance is decreasing with time