An electrician is working at the top of a ladder that is 15ft
long and leaning against the wall when its bottom starts sliding at a rate of 1ft/s
. Fortunately, a fellow worker catches it when the ladder’s bottom is 5ft
from the wall. At what speed is the top of the ladder (along with the electrician) sliding down the wall at that instant? Round any intermediate calculations, if needed, to no less than six decimal places, and round your final answer to two decimal places.

To solve the problem, we use related rates in calculus. Given the ladder length and the rate at which the bottom of the ladder is sliding away from the wall, we need to determine the rate at which the top of the ladder is sliding down the wall when the bottom is 5 feet away from the wall.

First, we set up the relationship using the Pythagorean theorem. Let \( x \) be the distance from the wall to the bottom of the ladder, and \( y \) be the height of the ladder on the wall. The length of the ladder is 15 feet, so we have:

\[
x^2 + y^2 = 15^2
\]

Differentiate both sides with respect to time \( t \):

\[
2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0
\]

We know that \( \frac{dx}{dt} = 1 \) ft/s (the rate at which the bottom of the ladder is sliding away from the wall). We need to find \( \frac{dy}{dt} \) when \( x = 5 \) feet.

First, solve for \( y \) when \( x = 5 \):

\[
5^2 + y^2 = 15^2
\]
\[
25 + y^2 = 225
\]
\[
y^2 = 200
\]
\[
y = \sqrt{200} = 10\sqrt{2}
\]

Next, substitute \( x = 5 \), \( y = 10\sqrt{2} \), and \( \frac{dx}{dt} = 1 \) into the differentiated equation:

\[
2(5)(1) + 2(10\sqrt{2}) \frac{dy}{dt} = 0
\]
\[
10 + 20\sqrt{2} \frac{dy}{dt} = 0
\]
\[
20\sqrt{2} \frac{dy}{dt} = -10
\]
\[
\frac{dy}{dt} = \frac{-10}{20\sqrt{2}}
\]
\[
\frac{dy}{dt} = \frac{-1}{2\sqrt{2}}
\]
\[
\frac{dy}{dt} = \frac{-1}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{-\sqrt{2}}{4}
\]

Therefore, the speed at which the top of the ladder is sliding down the wall when the bottom is 5 feet from the wall is:

\[
\frac{dy}{dt} = -0.35 \text{ ft/s}
\]

Would you like further details or have any other questions?

Here are some additional questions you might consider asking:
1. Can you explain related rates with another example?
2. What happens if the rate of the ladder sliding changes?
3. How does the length of the ladder affect the rates of movement?
4. What if the ladder was on a frictionless surface, how would that change the rates?
5. Can you explain the significance of the negative sign in the answer? 

Tip: Always draw a diagram for related rates problems to better visualize the relationships between the variables.

Explore a calculus example involving related rates: determining the speed at which the top of a ladder slides down a wall when its bottom is sliding away. Detailed step-by-step solution with differentiation rules and the Pythagorean theorem. Suitable for undergraduate students and anyone interested in calculus applications.

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