Results for this Submission The answer is NOT correct. A 15 ft ladder leans against a wall. The bottom of the ladder is 3 ft from the wall at time $$ t=0 $$ and slides away from the wall at a rate of $$ 3 \mathrm{ft} / \mathrm{sec} $$. Find the velocity of the top of the ladder at time $$ t=2 $$. The velocity of ladder at time $$ t=2 $$ is 1.286 ft/sec. Preview My Answers Submit Answers Your score was recorded. You have attempted this problem 2 times.

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Let $$ x(t) $$ be the distance from the wall to the base of the ladder and $$ y(t) $$ be the height of the ladder on the wall. The ladder’s length is constant at $$ 15 $$ ft, so by the Pythagorean theorem we have

$$
x(t)^2 + y(t)^2 = 15^2 = 225.
$$

Differentiate both sides with respect to $$ t $$:

$$
2x(t) \frac{dx}{dt} + 2y(t) \frac{dy}{dt} = 0.
$$

Dividing by 2 gives

$$
x(t) \frac{dx}{dt} + y(t) \frac{dy}{dt} = 0.
$$

We know that at all times, the base is sliding away at the rate

$$
\frac{dx}{dt} = 3 \text{ ft/sec}.
$$

Solving for $$ \frac{dy}{dt} $$ we obtain

$$
\frac{dy}{dt} = -\frac{x(t)}{y(t)} \frac{dx}{dt}.
$$

At $$ t = 2 $$ seconds, the base of the ladder has moved from $$ 3 $$ ft (initial position at $$ t=0 $$) to

$$
x(2) = 3 + 3(2) = 9 \text{ ft}.
$$

Now, solve for $$ y(2) $$ using the Pythagorean relationship:

$$
y(2) = \sqrt{225 - x(2)^2} = \sqrt{225 - 9^2} = \sqrt{225 - 81} = \sqrt{144} = 12 \text{ ft}.
$$

Substitute $$ x(2) = 9 $$ ft, $$ y(2) = 12 $$ ft, and $$ \frac{dx}{dt} = 3 $$ ft/sec into the expression for $$ \frac{dy}{dt} $$:

$$
\frac{dy}{dt} = -\frac{9}{12} \cdot 3 = -\frac{9 \times 3}{12} = -\frac{27}{12} = -\frac{9}{4} \text{ ft/sec}.
$$

Thus, the velocity of the top of the ladder at $$ t = 2 $$ seconds is

$$
\frac{dy}{dt} = -2.25 \text{ ft/sec}.
$$

The negative sign indicates that the top of the ladder is sliding downward.
At time $$ t=2 $$ seconds, the top of the ladder is moving downward at a velocity of 2.25 ft/sec.

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