A 13 foot ladder is sliding down a vertical wall at 1 ft/sec. How fast is the base of this ladder moving on a horizontal ground after 5 seconds?
To find the rate at which the base of the ladder is moving on the horizontal ground, we can use the concept of similar triangles.
Let's consider the triangle formed by the ladder, the vertical wall, and the horizontal ground. The ladder is the hypotenuse, the vertical wall is the vertical side, and the base on the ground is the horizontal side.
Since the ladder is sliding down the wall at a rate of 1 ft/sec, we can say that the length of the vertical side of the triangle is changing at a rate of 1 ft/sec.
The length of the ladder is given as 13 ft. Let's denote the length of the base on the ground as x.
Using the Pythagorean theorem, we have:
x^2 + y^2 = 13^2
Differentiating both sides with respect to time (t), we get:
2x(dx/dt) + 2y(dy/dt) = 0
Since we're interested in finding dx/dt, the rate at which x is changing with respect to time, we can solve the equation for dx/dt:
2x(dx/dt) = -2y(dy/dt)
dx/dt = (-2y(dy/dt)) / (2x)
Now, let's substitute the known values into the equation:
dx/dt = (-2y(dy/dt)) / (2x)
dx/dt = (-2y(dy/dt)) / (2x)
dx/dt = (-2 * y * (-1 ft/sec)) / (2 * x)
dx/dt = (2 * y * 1 ft/sec) / (2 * x)
dx/dt = (y / x) ft/sec
To find the value of y / x, we can use the similar triangles formed by the initial triangle and the triangle formed after 5 seconds:
y / 13 = (y + 5) / x
Cross-multiplying and simplifying the equation, we get:
xy + 5x = 13y
Simplifying further, we get:
x = (13y) / (y + 5)
Substituting this value back into the equation for dx/dt, we get:
dx/dt = (y / ((13y) / (y + 5))) ft/sec
dx/dt = ((y + 5) / 13) ft/sec
Now we can substitute the value of y = 13 into the equation to find dx/dt after 5 seconds:
dx/dt = ((13 + 5) / 13) ft/sec
dx/dt = (18 / 13) ft/sec
Therefore, the base of the ladder is moving at a rate of 18/13 ft/sec on the horizontal ground after 5 seconds.