A 40-foot ladder is leaning against a house when its base starts to slide away. By the time the base is 32 feet from the house the base is moving at a rate of 4ft/sec.
a. How fast is the top of the ladder moving down the wall at that moment?
b. At what rate is the area of the triangle formed by the ladder, wall, and ground changing at the moment?
c. At what rate is the angle between the ladder and the ground changing at that moment?
I wasn't here the day the teacher taught us this...plz help me i have no idea where to start
This is just an exercise in implict differentiation.
At the moment in question, we have a 24-32-40 right triangle, with height=24.
Let θ be the angle between the ladder and the ground.
Let the base be x, and the height be y.
a)
x^2 + y^2 = 40^2
2x dx/dt + 2y dy/dt = 0
when x=32,
2(32)(4) + 2(24) dy/dt = 0
dy/dt = -16/3
so, the top is sliding down at 16/3 ft/s
b)
a = 1/2 xy
da/dt = 1/2 (y dx/dt + x dy/dt)
da/dt = 1/2 (24(4) + 32(-16/3))
= -112/3 ft^2/s
c)
y/x = tanθ, so
y = x tanθ
dy/dt = tanθ dx/dt + x sec^2θ dθ/dt
-112/3 = 3/4 (4) + 32 (5/4)^2 dθ/dt
dθ/dt = -121/150 rad/s
To solve this problem, we will use related rates, which involves differentiating the given information with respect to time.
Let's go through each question one by one:
a. How fast is the top of the ladder moving down the wall at that moment?
We are given that the base of the ladder is moving away from the house at a rate of 4 ft/sec. To find how fast the top of the ladder is moving down the wall, we need to find its vertical velocity.
Let's assume h represents the height of the ladder from the ground, and x represents the distance from the base of the ladder to the house.
At any given moment, we know that the ladder forms a right-angled triangle with the ground and the wall. According to the Pythagorean theorem, we have:
x^2 + h^2 = 40^2
Now let's differentiate both sides of this equation with respect to time t:
2x(dx/dt) + 2h(dh/dt) = 0
We are given dx/dt (the rate at which the base is moving away from the house) as 4 ft/sec, and we need to find dh/dt (the rate at which the height is changing).
Substituting the given values into the equation, we have:
2(32)(4) + 2h(dh/dt) = 0
Simplifying the equation:
64 + 2h(dh/dt) = 0
2h(dh/dt) = -64
Divide both sides by 2h:
dh/dt = -32/h
To find the value of dh/dt, we need to know h (the height of the ladder) at that moment in time. Unfortunately, this information is not given in the problem statement. To proceed, we would need additional information about h or make an assumption.
b. At what rate is the area of the triangle formed by the ladder, wall, and ground changing at that moment?
The area A of the triangle can be calculated using the formula:
A = (1/2) * base * height
At any given moment, the base of the triangle is x and the height is h. We can express the area as:
A = (1/2) * x * h
Let's differentiate both sides with respect to time t:
dA/dt = (1/2) * (dx/dt * h + x * dh/dt)
We know dx/dt as 4 ft/sec, and we need to find dh/dt.
Using the equation 2x(dx/dt) + 2h(dh/dt) = 0 from part a, we can substitute dx/dt as -2h(dh/dt):
dA/dt = (1/2) * (-2h(dh/dt) * h + x(dh/dt))
= -h^2(dh/dt) + x(dh/dt)
= (-h^2 + x)(dh/dt)
To find dA/dt, we need to know the values of h and x at that moment in time. Without this information, we cannot calculate the rate of change of the area.
c. At what rate is the angle between the ladder and the ground changing at that moment?
To determine the rate at which the angle θ between the ladder and the ground is changing, we need to differentiate the equation involving the right-angled triangle formed by the ladder, the wall, and the ground.
Using the equation x^2 + h^2 = 40^2, we can differentiate both sides with respect to time t:
2x(dx/dt) + 2h(dh/dt) = 0
We already know dx/dt as 4 ft/sec (given in the problem statement). To find dh/dt, we need additional information about h.
Once we have the values for h, we can calculate the rate at which the angle is changing using trigonometry.