Question

an electrician is working at the top of a ladder that is 18 ft long and leaning against the wall when its bottom starts sliding away from the wall at a rate of 1 ft/s. let x denote the distance from the wall to the base of the ladder and y denote the distance from the ground to the top of the ladder.

step 2 of 3: as the ladder slides away from the wall, both x and y change with respect to time t. find the equation that relates \\(\frac{dy}{dt}\\) to \\(\frac{dx}{dt}\\).

Explanation:

Relate the variables using the Pythagorean theorem

Using the Related Rates knowledge point

$$ x^2 + y^2 = 18^2 = 324 $$

Differentiate both sides with respect to time t

Using the Related Rates knowledge point

$$ \frac{d}{dt}(x^2 + y^2) = \frac{d}{dt}(324) $$

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

Solve for the relationship between the rates

Using the Related Rates knowledge point

$$ 2y\frac{dy}{dt} = -2x\frac{dx}{dt} $$

$$ \frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt} $$

Answer:

Step 2 of 3: As the ladder slides away from the wall, both \(x\) and \(y\) change with respect to time \(t\). Find the equation that relates \(\frac{dy}{dt}\) to \(\frac{dx}{dt}\).

\(\frac{dy}{dt} =\) <blank>\(-\frac{x}{y}\)</blank> \(\frac{dx}{dt}\)