When the top of the ladder slides down the vertical wall at a rate of 0.15 m/s, and at the same moment, the bottom of the ladder is 3 m away from the wall, we can analyze the movement using related rates in calculus.
Let’s denote the length of the ladder as L, the height of the top of the ladder from the ground as y, and the distance from the bottom of the ladder to the wall as x. Given that at the moment the bottom of the ladder is 3 m away from the wall (x = 3 m), we can apply the Pythagorean theorem:
L2 = x2 + y2
As the top slides down, the height y is decreasing, which means dy/dt will be negative, specifically dy/dt = -0.15 m/s. We want to find out how fast the bottom of the ladder is moving away from the wall (dx/dt) at the instant when x = 3 m.
First, we differentiate the Pythagorean theorem equation with respect to time (t):
2L(dL/dt) = 2x(dx/dt) + 2y(dy/dt)
Since the ladder’s length L is constant, dL/dt = 0. Rearranging gives:
0 = 2x(dx/dt) + 2y(dy/dt)
Dividing through by 2 simplifies this to:
0 = x(dx/dt) + y(dy/dt)
Rearranging for dx/dt we get:
dx/dt = – (y/dx) * (dy/dt)
At the moment x = 3 m, we can find y using the Pythagorean theorem:
L2 = 32 + y2 => L = √(9 + y2)
Since we haven’t defined L, we can choose a length based on the height when the ladder is vertical. Assuming a specific ladder length allows us to calculate y. However, as a generalization, we need to assess how high the ladder could theoretically be depending on its slant. Let’s consider the case when it’s at a 45-degree angle, which gives us equal x and y. Assuming L= √(18) gives y = √(L2 – 9) under the context of our current situation.
Ultimately, by plugging values back into the derived equation, we can determine dx/dt, revealing the speed at which the bottom of the ladder moves away from the wall as the top slides down.
Thus, the scenario illustrates how the relationships between changing distances in geometry due to motion provide insight through the principles of calculus.