A 20 foot ladder is sliding down a vertical wall at a constant rate of 2 feet per second.

a) How fast is the ladder moving away from the wall when the base of the ladder is 12 feet away from the wall?

b) Find the rate of change at which the angle between the ladder and the ground is changing when the base of the ladder is 16 feet from the wall.

Did you make your sketch?

label:
foot of ladder is x ft from wall
top of ladder is y ft above ground
length of ladder (hypotenuse) is 20

x^2 + y^2 = 400
2x dx/dt + 2y dy/dt = 0

when x = 12
144 + y^2= 400
y = 16

24 dx/dt + 32(-2) = 0
dx/dt = 64/24 or 8/3 ft/s

b) let the angle be Ø
cosØ = x/20
x = 20cosØ
dx/dt = -20sinØ dØ/dt

when x= 16, cosØ = 16/20 = 4/5
then sinØ = 3/5 , (recognize the 3-4-5 right-angled triangle)

using a) when x= 16, y = 12 , dy/dt = -2
32 dxdt + 24(-2) = 0
dx/dt = 48/32 = 3/2

so in dx/dt = -20sinØ dØ/dt
3/2 = -20(3/5) dØ/dt
dØ/dt = (3/2)(5/3)(-1/20) = -1/8

so at that moment, the angle is decreasing at a rate of 1/8 radians/second

Just noticed that Steve had already answered the same question above.

At least we both ended up with the same answer.

a) Well, this sounds like a classic case of ladder drama! Let's see what we can do. We know that the ladder is sliding down the wall at a constant rate of 2 feet per second. So, the speed at which the ladder is moving away from the wall can be found using the Pythagorean theorem.

Let's call the distance between the bottom of the ladder and the wall "x." According to the Pythagorean theorem, we have x^2 + 12^2 = 20^2. Solving for x, we get x = 16.

Now, to find how fast the ladder is moving away from the wall, we can take the derivative of both sides with respect to time. We get 2x(dx/dt) + 0 = 0.5*dx/dt. Plugging in x = 16, we find that the ladder is moving away from the wall at a rate of 0.5*(dx/dt) = 0.5*(16/20) = 0.4 feet per second. So, the ladder is moving away from the wall at a rate of 0.4 feet per second when the base is 12 feet away from the wall.

b) Ah, the rate at which the angle between the ladder and the ground is changing! Well, let me think of a good joke to make this calculation a bit more enjoyable.

Why did the angle between the ladder and the ground want to change so quickly? It felt like it was missing out on all the right angles!

Alright, let's tackle this. We need to find how fast the angle θ is changing with respect to time t. Using some trigonometry, we know that tan(θ) = x/12. Let's differentiate both sides with respect to t:

sec^2(θ)*(dθ/dt) = (1/12)*(dx/dt).

Now we just need to plug in the given information. When the base of the ladder is 16 feet from the wall, we found earlier that x = 16. Solving for (dθ/dt), we get (dθ/dt) = (1/12)*(dx/dt)*sec^2(θ). Since the ladder is sliding down the wall at a constant rate of 2 feet per second, we know that (dx/dt) = -2.

Now, all we need is to find the value of sec^2(θ). Using the Pythagorean theorem again, we know that sec^2(θ) = 20^2/(12^2 + x^2). Plugging in x = 16, we find sec^2(θ) = 400/(144 + 256) = 400/400 = 1.

Substituting the values we have, we get (dθ/dt) = (1/12)*(-2)*1 = -1/6 radians per second.

So, the angle between the ladder and the ground is changing at a rate of -1/6 radians per second when the base of the ladder is 16 feet from the wall.

Hope these answers brought a smile to your face!

To solve these problems, we can use related rates, which require us to find the rate of change of one variable with respect to another variable. In this case, the variables we are looking at are the distance between the ladder and the wall (base of the ladder) and the angle between the ladder and the ground.

a) To find how fast the ladder is moving away from the wall at a specific distance, we can create a right triangle with the ladder as the hypotenuse, the distance between the ladder and the wall as the base, and an unknown height. We can apply the Pythagorean theorem to find the height of the triangle.

Let's denote:
- The distance between the ladder and the wall as x (in feet)
- The height of the triangle as h (in feet)

According to the Pythagorean theorem, we have:
x^2 + h^2 = 20^2

Differentiating both sides with respect to time (t), we get:
2x(dx/dt) + 2h(dh/dt) = 0

Since the ladder is sliding down the wall at a constant rate, we know that dh/dt (the rate at which the height changes) equals -2 feet per second. The dx/dt (the rate at which the base changes) is what we need to find.

At x = 12 feet, we can substitute the values into our equation:
(12)^2 + h^2 = (20)^2

Simplifying the equation, we get:
144 + h^2 = 400
h^2 = 256
h = 16 feet

Substituting the values of x = 12 feet, h = 16 feet, and dh/dt = -2 feet per second into our equation, we can solve for dx/dt:
2(12)(dx/dt) + 2(16)(-2) = 0

Simplifying the equation, we get:
24(dx/dt) - 64 = 0
24(dx/dt) = 64
dx/dt = 64/24
dx/dt = 8/3

Therefore, when the base of the ladder is 12 feet away from the wall, the ladder is moving away from the wall at a speed of 8/3 feet per second.

b) To find the rate of change at which the angle between the ladder and the ground is changing, we can use trigonometric ratios. Let's denote the angle between the ladder and the ground as θ (in radians).

We know that tan(θ) = h / x, where h is the height of the triangle and x is the distance between the ladder and the wall. Differentiating both sides with respect to time (t), we get:

sec^2(θ) (dθ/dt) = (dx/dt)(h/x)

We are given dx/dt = -2 feet per second and x = 16 feet. To solve for dθ/dt (the rate at which the angle changes), we need to find h.

Using the Pythagorean theorem:
x^2 + h^2 = 20^2
(16)^2 + h^2 = (20)^2
256 + h^2 = 400
h^2 = 144
h = 12 feet

Substituting the values of dx/dt = -2 feet per second, x = 16 feet, and h = 12 feet into our equation, we can solve for dθ/dt:
sec^2(θ) (dθ/dt) = (-2)(12/16)
sec^2(θ) (dθ/dt) = -3/2

We need to find the value of sec^2(θ) to solve for dθ/dt. Since we know that cos(θ) = x / hypotenuse, we can find cos(θ) using the given values:
cos(θ) = 16 / 20
cos(θ) = 4/5
sec(θ) = 1 / cos(θ)
sec(θ) = 5/4

Substituting sec(θ) = 5/4 into our equation, we get:
(5/4)^2 (dθ/dt) = -3/2

Simplifying the equation, we have:
25/16 (dθ/dt) = -3/2
(dθ/dt) = -3/2 * 16/25
(dθ/dt) = -48/50
(dθ/dt) = -24/25

Therefore, when the base of the ladder is 16 feet from the wall, the angle between the ladder and the ground is changing at a rate of -24/25 radians per second.

To solve these problems, we can use the concept of related rates. Related rates involve finding the rates of change of two or more related variables. In this case, we are given the rate of change of the ladder sliding down the wall and we need to find the rate of change of the distance between the ladder and the wall (a) and the rate of change of the angle between the ladder and the ground (b).

a) To find how fast the ladder is moving away from the wall, we can consider the ladder, the wall, and the ground as a right triangle. Let's call the distance between the ladder and the wall "a" and the distance along the ground from the base of the ladder to the wall "x".

The Pythagorean theorem states that the sum of the squares of the two shorter sides of a right triangle is equal to the square of the hypotenuse (the ladder in this case). So, we have the equation:

x^2 + a^2 = (20 ft)^2

To find how fast the ladder is moving away from the wall (da/dt), we need to find the derivative of both sides of the equation. Taking the derivative with respect to time (t), we get:

2x(dx/dt) + 2a(da/dt) = 0

We are given that dx/dt = 2 ft/s and we want to find da/dt, so we can substitute these values into the equation:

2(12 ft)(2 ft/s) + 2a(da/dt) = 0

Simplifying, we get:

24 ft^2/s + 2a(da/dt) = 0

Since we want to find da/dt, we can isolate it by dividing both sides of the equation by 2a:

da/dt = -24 ft^2/s / (2a)

Now, we need to find the value of "a" when the base of the ladder is 12 feet away from the wall. Since we have a right triangle, we know that when the base is 12 feet away, the height is given by:

a = √[(20 ft)^2 - (12 ft)^2]

Calculating this, we find:

a = 16 ft

Now we can substitute this value of "a" into our equation for da/dt:

da/dt = -24 ft^2/s / (2 * 16 ft)

Simplifying, we get:

da/dt = -3/4 ft/s

Therefore, the ladder is moving away from the wall at a rate of 3/4 feet per second when the base of the ladder is 12 feet away from the wall.

b) To find the rate of change of the angle between the ladder and the ground (db/dt), we can use trigonometry. Let's call the angle between the ladder and the ground "θ".

Using the same right triangle as before, we can write:

cos(θ) = a / (20 ft)

To find db/dt, we need to find dθ/dt. Taking the derivative of both sides of the equation with respect to time (t), we get:

-sin(θ)(dθ/dt) = (da/dt) / (20 ft)

We know da/dt from part a), which is -3/4 ft/s, and we want to find dθ/dt. So, we can substitute these values into the equation:

-sin(θ)(dθ/dt) = (-3/4 ft/s) / (20 ft)

Simplifying, we get:

-sin(θ)(dθ/dt) = -3/80 ft/s

To find the value of θ when the base of the ladder is 16 feet away from the wall, we can use the equation:

cos(θ) = a / (20 ft)

a = √[(20 ft)^2 - (16 ft)^2] = 12 ft

cos(θ) = 12 ft / (20 ft) = 3/5

Solving for θ, we find:

θ = arccos(3/5)

Now we can substitute this value for θ into our equation for dθ/dt:

-sin(arccos(3/5))(dθ/dt) = -3/80 ft/s

Using the identity sin(arccos(x)) = √(1 - x^2), we can simplify the equation:

-√(1 - (3/5)^2)(dθ/dt) = -3/80 ft/s

Now we can solve for dθ/dt:

√(1 - (3/5)^2)(dθ/dt) = 3/80 ft/s

dθ/dt = (3/80 ft/s) / √(1 - (3/5)^2)

Simplifying, we get:

dθ/dt ≈ 3/40 ft/s

Therefore, the angle between the ladder and the ground is changing at a rate of approximately 3/40 feet per second when the base of the ladder is 16 feet from the wall.