A ladder 20 feet long leans up against a house.
The bottom of the ladder starts to slip away from the house at 0.17 feet per second. How fast is the tip of the ladder along the side of the house slipping when the ladder is 7.2 feet away from the house? (Round to 3 decimal places.)
Answer: _____ ft/ sec
Consider the angle the bottom of the ladder makes with the ground.
How fast is the angle changing (in radians) when the ladder is 7.2 feet away from the house?
Answer 2: ________
To solve this problem, we can use related rates. Let's start with the first question:
We are given the rate at which the bottom of the ladder is slipping away from the house, which is 0.17 feet per second. We need to find how fast the tip of the ladder along the side of the house is slipping when the ladder is 7.2 feet away from the house.
Let's define the variables:
- Let x represent the distance from the base of the ladder to the house (the variable we're interested in).
- Let y represent the distance from the top of the ladder to the ground (the height of the ladder).
- Let z represent the length of the ladder (20 feet).
Using the Pythagorean theorem, we have the equation: x^2 + y^2 = z^2.
Differentiating the equation implicitly with respect to time (t), we get:
2x(dx/dt) + 2y(dy/dt) = 2z(dz/dt).
Since we are interested in finding dx/dt (the rate at which the tip of the ladder is slipping along the side of the house), we can rearrange the equation as follows:
dx/dt = (1/x)(-y(dy/dt)).
Now, let's plug in the values we know:
x = 7.2 ft,
y = √(z^2 - x^2) = √(20^2 - 7.2^2) ≈ 18.542 ft,
dy/dt = 0.17 ft/s.
Plugging these values into the equation, we get:
dx/dt = (1/7.2)(-18.542 * 0.17) ≈ -0.374 ft/s (rounded to 3 decimal places).
Therefore, the answer to the first question is approximately -0.374 ft/s.
Moving on to the second question:
We need to find how fast the angle that the ladder makes with the ground is changing (in radians) when the ladder is 7.2 feet away from the house.
Let's define the angle as θ.
Using trigonometry, we have the relationship: tan(θ) = y/x.
Differentiating this equation implicitly with respect to time (t), we get:
sec^2(θ)(dθ/dt) = (dy/dt)/x.
Rearranging the equation, we get:
dθ/dt = (1/x)(dy/dt)sec^2(θ).
Plugging in the values we know:
x = 7.2 ft,
y = 18.542 ft (from the previous calculation),
dy/dt = 0.17 ft/s,
We also need to find sec^2(θ). Since tan(θ) = y/x, we have:
sec^2(θ) = 1 + tan^2(θ) = 1 + (y/x)^2.
Plugging these values into the equation, we get:
sec^2(θ) = 1 + (18.542/7.2)^2 ≈ 5.676.
Plugging these values back into the equation, we have:
dθ/dt = (1/7.2)(0.17)(5.676) ≈ 0.158 radians/s (rounded to 3 decimal places).
Therefore, the answer to the second question is approximately 0.158 radians/s.