Derivative applied to a ladder

[sporus]

Homework Statement

A 10 - foot ladder is placed against a vertical wall. The bottom of the ladder slides away from the wall at a constant rate of 2.9 ft/s. How fast is the top of the ladder sliding down the wall when the top of the ladder is 7.6 feet from the ground?
Solution: Let the y-axis represent the wall and let the x-axis represent the ground.

Homework Equations

100 = y^2 + x^2 = 7.6^2 + x^2
=> x= 6.5

y = (100 - x^2)^.5

The Attempt at a Solution

y' = 1/2 * (100 - x^2)^-.5 * -2x
y'(6.5) = -0.855

 

[mighty2000]

ft/s

Thank you for your question. I can provide you with a more detailed explanation and clarification of your solution.

First, let's define the variables in this problem. The length of the ladder is given as 10 feet, so we can label that as L. The rate at which the bottom of the ladder slides away from the wall is given as 2.9 ft/s, so we can label that as dx/dt. We are asked to find the rate at which the top of the ladder is sliding down the wall, which we can label as dy/dt.

Next, let's set up our coordinate system. As you correctly stated, we can use the y-axis to represent the wall and the x-axis to represent the ground. We can also label the distance between the top of the ladder and the ground as y, and the distance between the bottom of the ladder and the wall as x.

Now, let's use the Pythagorean theorem to relate the variables in our problem. We know that the ladder forms a right triangle with the wall and the ground, so we can use the equation c^2 = a^2 + b^2 to represent this relationship, where c is the length of the ladder, a is the distance between the top of the ladder and the ground (y), and b is the distance between the bottom of the ladder and the wall (x). This can be written as L^2 = y^2 + x^2.

Next, we can differentiate this equation with respect to time to relate the rates of change of our variables. Using the chain rule, we get 2L(dL/dt) = 2y(dy/dt) + 2x(dx/dt). Since we know the values of L and dx/dt, we can rearrange this equation to solve for dy/dt, which is what we are looking for. This gives us dy/dt = (L/x) * dx/dt - (y/x) * dL/dt.

Plugging in the given values, we get dy/dt = (10/6.5) * 2.9 - (7.6/6.5) * 0 = 1.77 ft/s.

I hope this explanation helps you understand the solution better. If you have any further questions, please don't hesitate to ask.