Question

A 13ft ladder slides down a wall at 2ft/sec when it is 12 ft above the ground. Find the rate at which the angle it makes with the ground changes at this moment?

Answer 1

let the height of the ladder against the wall be y ft

and let the base of the ladder be x ft from the wall
we know x^2 + y^2= 13^

2x dx/dt + 2y dy/dt = 0
when y = 12, x = 5 by above Pythagoras, and dy/dt = -2 ft/s

2(5)(dx/dt) + 2(12)(-2) = 0
dx/dt = 24/10 = 2.4 ft/s

Answer 2

Hi there, wouldn't we need to use right triangle trig for this problem since it's asking for the rate of the angle? theta? Not completely sure on how to relate that exactly.

Answer 3

yes - so, try it

sinθ = y/13
cosθ dθ/dt = 1/13 dy/dt

Now just plug in your numbers. Use the Pythagorean Theorem to figure cosθ when y=12.

Answer 4

ohh ok this helped me too

Answer 5

[AI-generated answer — Explain Bot]

To find the rate at which the angle the ladder makes with the ground changes, we need to use trigonometry and calculus concepts.

Let's start by assigning variables to the different quantities given in the problem:
Let 𝑥 be the distance of the foot of the ladder from the wall.
Let 𝑦 be the height of the foot of the ladder above the ground.
Let 𝜃 be the angle the ladder makes with the ground.

We are given that the ladder is 13ft long, so we can use the Pythagorean theorem to relate 𝑥, 𝑦, and 𝑙 (the length of the ladder):
𝑙^2 = 𝑥^2 + 𝑦^2

Differentiating this equation with respect to time, we get:
2𝑙 (d𝑙/d𝑡) = 2𝑥 (d𝑥/d𝑡) + 2𝑦 (d𝑦/d𝑡)

We are also given that 𝑑𝑦/𝑑𝑡 = -2 ft/sec (because the ladder is sliding down the wall), and 𝑙 = 13 ft. We need to find 𝑑𝜃/𝑑𝑡, so we rearrange the equation as follows:
𝑑𝑦/𝑑𝑡 = -(𝑥/𝑙) (𝑑𝑦/𝑑𝑡) - (𝑦/𝑙) (𝑑𝑥/𝑑𝑡)
𝑑𝑦/𝑑𝑡 = -(𝑥/𝑙) (-2) - (𝑦/𝑙) (𝑑𝑥/𝑑𝑡)
𝑑𝑦/𝑑𝑡 = (2𝑥/𝑙) + (𝑦/𝑙) (𝑑𝑥/𝑑𝑡)
-2 = (2𝑥/13) + (𝑦/13) (𝑑𝑥/𝑑𝑡)

We are interested in finding 𝑑𝜃/𝑑𝑡, the rate at which the angle 𝜃 changes. This can be found using trigonometry:
tan(𝜃) = 𝑦/𝑥

Differentiating both sides of this equation with respect to time, we get:
(sec^2(𝜃)) (d𝜃/d𝑡) = (1/𝑥) (d𝑦/d𝑡) - (𝑦/𝑥^2) (d𝑥/d𝑡)

We can rewrite this equation using the values we found earlier:
(sec^2(𝜃)) (d𝜃/d𝑡) = (1/𝑥) (-2) - (𝑦/𝑥^2) (𝑑𝑥/d𝑡)
(sec^2(𝜃)) (d𝜃/d𝑡) = -2/𝑥 - (𝑦/𝑥^2) (𝑑𝑥/d𝑡)

We need to find 𝑑𝜃/d𝑡 when the ladder is 12 ft above the ground, so we substitute 𝑥 = 5 into our equation:
(sec^2(𝜃)) (d𝜃/d𝑡) = -2/5 - (𝑦/25) (𝑑𝑥/d𝑡)

Now we just need to find the value of 𝑦 when the ladder is 12 ft above the ground. Using the Pythagorean theorem:
13^2 = 5^2 + 𝑦^2
169 = 25 + 𝑦^2
𝑦^2 = 144
𝑦 = 12 ft

Substituting 𝑦 = 12 and 𝑑𝑥/d𝑡 = -2 into our equation:
(sec^2(𝜃)) (d𝜃/d𝑡) = -2/5 - (12/25) (-2)
(sec^2(𝜃)) (d𝜃/d𝑡) = -2/5 - (-24/25)
(sec^2(𝜃)) (d𝜃/d𝑡) = -2/5 + 24/25
(sec^2(𝜃)) (d𝜃/d𝑡) = (-10 + 24)/25
(sec^2(𝜃)) (d𝜃/d𝑡) = 14/25

Finally, dividing both sides by (sec^2(𝜃)), we get:
d𝜃/d𝑡 = 14/25(sec^2(𝜃))

Therefore, the rate at which the angle the ladder makes with the ground changes at this moment is 14/25 times the square of the secant of the angle.