Respuesta :
Answer:
Therefore the top of the ladder sliding up the wall the bottom is 2 m from the wall is 0.45 m/s.
Step-by-step explanation:
Given that a ladder is 7 m long.
Let the distance the wall and the bottom of the ladder be x.
and the distance of the ground to the top of the ladder where it touches the wall be y.
The length of the ladder be L which is constant.
The ladder is inclination against a wall. So it will from a right angled triangle whose base = x, height = y and hypotenuse = L.
By the Pythagorean Theorem,
x²+y²=L².....(1)
Differentiating with respect to t
[tex]\Rightarrow 2x\frac{dx}{dt}+2y\frac{dy}{dt}=0[/tex] ......(2) [Since L= 7 m is a constant]
The bottom of the ladder is pushed horizontally towards wall at 1.5 m/s.
It means [tex]\frac{dx}{dt}=-1.5[/tex] m/s. [ Since it is decrease]
When the bottom of the ladder is 2 m away from the wall, Then the tip of the ladder is y.
Putting x=2 and L=7 in equation (1) we get
[tex]2^2+y^2=7^2[/tex]
[tex]\Rightarrow y^2=49-4[/tex]
[tex]\Rightarrow y=\sqrt{45}[/tex]
Now putting the value of x=2, y and [tex]\frac{dx}{dt}[/tex] in equation (2)
[tex]\Rightarrow 2\times 2 \times(- 1.5 )+2\sqrt{45}\frac{dy}{dt}=0[/tex]
[tex]\Rightarrow 2\sqrt{45}\frac{dy}{dt}=6[/tex]
[tex]\Rightarrow \frac{dy}{dt}=\frac{6}{2\sqrt{45}}[/tex]
[tex]\Rightarrow \frac{dy}{dt}=\frac{6}{6\sqrt{5}}[/tex]
[tex]\Rightarrow \frac{dy}{dt}=\frac{1}{\sqrt{5}}[/tex]
[tex]\Rightarrow \frac{dy}{dt}=0.45[/tex] (approx)
Therefore the top of the ladder sliding up the wall the bottom is 2 m from the wall is 0.45 m/s.
