A 16-ft ladder is sliding down a wall as shown in the figures below. The variable h is the height of the ladder's top at time t, and x is the distance from the wall to the ladder's bottom. Suppose that the top is sliding down the wall at a rate 5 ft/s. Calculate dx/dt when h = 10. Give your answer correct to two decimal places.
dx/dt = ? ft/s
The ladder forms a rt. triangle with the wall and ground. The wall is the
height.When the height is 10ft,
X^2+10^2=16^2, X=12.49ft. When h goes to zero(ladder flat on ground),
X^2+0^2=16^2, X=16, t=10ft/5ft/s=2sec.
dx/dt=(16-12.49)ft/2s=1.75ft/s
To find dx/dt, we need to use similar triangles and apply the chain rule because the rate of change of x with respect to time is related to the rate of change of h with respect to time.
Let's denote the length of the ladder as L and the distance from the wall to the base of the ladder as x. We are given that L = 16 ft.
By using the similar triangles formed by the ladder, wall, and the ground, we can write the following equation:
h/x = L/(x + L)
Now, we differentiate both sides of the equation with respect to time (t):
(dh/dt)/x - h(dx/dt)/(x^2) = -L(dh/dt)/((x + L)^2)
Rearranging the equation, we can solve for dx/dt:
dx/dt = -(x^2)(dh/dt)/((x + L)^2) + (L(x^2)(dh/dt)/(x + L)^2) / h
Now, we can substitute the given values into the equation:
L = 16 ft
h = 10 ft
(dh/dt) = 5 ft/s
dx/dt = -((x^2)(5 ft/s))/((x + 16 ft)^2) + (16(x^2)(5 ft/s)/(x + 16 ft)^2) / 10 ft
Simplifying further, we get:
dx/dt = (-5x^2)/(x + 16)^2 + (8x^2)/(x + 16)^2
Now, we can plug in h = 10 ft into the equation to find dx/dt:
dx/dt = (-5(10)^2)/(10 + 16)^2 + (8(10)^2)/(10 + 16)^2
dx/dt = (-500)/(26)^2 + (800)/(26)^2
dx/dt = -500/676 + 800/676
dx/dt = 300/676 ft/s
Therefore, dx/dt is approximately equal to 0.44 ft/s (rounded to two decimal places).
To find dx/dt when h = 10 ft, we can use the given information that the top of the ladder is sliding down the wall at a rate of 5 ft/s.
First, we need to establish a relationship between the variables h and x. From the figure, we can see that the ladder, the wall, and the ground form a right triangle. By applying the Pythagorean theorem, we have:
x^2 + h^2 = 16^2
Differentiating both sides with respect to time t, we get:
2x(dx/dt) + 2h(dh/dt) = 0
Since we want to find dx/dt when h = 10 ft, we substitute h = 10 into the equation:
2x(dx/dt) + 2(10)(dh/dt) = 0
We also know that dh/dt = -5 ft/s because the top of the ladder is sliding down the wall at a rate of 5 ft/s (given in the problem). So, substituting dh/dt = -5 into the equation:
2x(dx/dt) + 2(10)(-5) = 0
Simplifying the equation:
2x(dx/dt) - 100 = 0
Now, solving for dx/dt:
2x(dx/dt) = 100
dx/dt = 100 / (2x)
To find dx/dt when h = 10 ft, we need to find the corresponding value of x using the Pythagorean theorem. Substitute h = 10 into the equation:
x^2 + 10^2 = 16^2
x^2 + 100 = 256
x^2 = 256 - 100
x^2 = 156
x ≈ 12.49 ft (rounded to two decimal places)
Now, substitute this value of x into the equation for dx/dt:
dx/dt = 100 / (2 * 12.49)
dx/dt ≈ 4.01 ft/s (rounded to two decimal places)
Therefore, dx/dt is approximately 4.01 ft/s.