A 25m ladder is leaning against a
vertical wall. The floor is s
lightly slippery and the foot
of the ladder slips away from the wall at a rate of 0.3m/
s. How fast is the top of the l
adder sliding down the wall when the top is 20m above the floor?
if the base of the ladder is x from the wall, and the height up the wall is y, then we have
x^2+y^2 = 25^2
so, when y=20, x=15.
Taking the derivative wrt t, we have
x dx/dt + y dy/dt = 0
Now just find dy/dt, knowing that dx/dt = 0.3
Note that dy/dt will be negative, since the top is sliding down.
x^2+y^2=25^2
2x+2y.dy/dx=0
dy/dx=-x/y
dy/dt=dx/dt.dy/dx
=(0.3).(-x/y)
=(0.3).(-15/20)
=-0.225m/s
To find how fast the top of the ladder is sliding down the wall, we can use related rates.
Let's denote the distance from the foot of the ladder to the wall as x (which is changing) and the distance from the floor to the top of the ladder as y (which we want to find the rate of change for).
We need to find dy/dt (the rate of change of y with respect to time) when y = 20m.
According to the Pythagorean theorem, we have:
x^2 + y^2 = 25^2
Differentiating both sides of the equation with respect to time, we get:
2x(dx/dt) + 2y(dy/dt) = 0
We are given that dx/dt = -0.3m/s (since the foot of the ladder is sliding away from the wall), and we want to find dy/dt when y = 20m.
Substituting the given values into the equation, we get:
2x(-0.3) + 2(20)(dy/dt) = 0
Simplifying the equation, we have:
-0.6x + 40(dy/dt) = 0
Now, substitute x = sqrt(25^2 - y^2) = sqrt(625 - 400) = sqrt(225) = 15.
Plugging in x = 15, we get:
-0.6(15) + 40(dy/dt) = 0
Simplifying further:
-9 + 40(dy/dt) = 0
40(dy/dt) = 9
dy/dt = 9/40
Therefore, when y = 20m, the top of the ladder is sliding down the wall at a rate of 9/40 m/s.
To find out how fast the top of the ladder is sliding down the wall, we will need to use the concept of related rates. Let's assign some variables to the information given in the problem:
Let x be the distance between the foot of the ladder and the wall.
Let y be the distance between the top of the ladder and the floor.
Let z be the length of the ladder.
We are given the following information:
dx/dt = 0.3 m/s (rate at which the foot of the ladder is moving away from the wall)
z = 25 m (length of the ladder)
We need to find dy/dt (rate at which the top of the ladder is sliding down the wall) when y = 20 m.
Using the Pythagorean theorem, we have:
x^2 + y^2 = z^2
Differentiating both sides of the equation with respect to time (t), we get:
2x(dx/dt) + 2y(dy/dt) = 2z(dz/dt)
Since dx/dt and dz/dt are given, we can substitute them into the equation:
2x(0.3) + 2y(dy/dt) = 2(25)(0)
Simplifying the equation, we have:
0.6x + 2y(dy/dt) = 0
Since we are interested in finding dy/dt when y = 20, we can substitute y = 20 into the equation:
0.6x + 2(20)(dy/dt) = 0
Now we can solve for dy/dt:
0.6x + 40(dy/dt) = 0
40(dy/dt) = -0.6x
(dy/dt) = -0.6x/40
We know that x is changing as the foot of the ladder moves away from the wall, but we need to find x when y = 20.
Using the Pythagorean theorem, when y = 20:
x^2 + 20^2 = 25^2
x^2 + 400 = 625
x^2 = 225
x = 15
Substituting x = 15 into the equation for dy/dt, we get:
(dy/dt) = -0.6(15)/40
(dy/dt) = -0.225 m/s
Therefore, the top of the ladder is sliding down the wall at a rate of 0.225 m/s.