I am stuck on these problems. I have tried it multiple times and gotten it wrong.
1. A plane flies horizontally at an altitude of 3 km and passes directly over a tracking telescope on the ground. When the angle of elevation is π/6, this angle is decreasing at a rate of π/4 rad/min. How fast is the plane traveling at that time?
2. Two carts, A and B, are connected by a rope 39 ft long that passes over a pulley P (see the figure). The point Q is on the floor h = 12 ft directly beneath P and between the carts. Cart A is being pulled away from Q at a speed of 1 ft/s. How fast is cart B moving toward Q at the instant when cart A is 5 ft from Q? (Round your answer to two decimal places.)
3. Gravel is being dumped from a conveyor belt at a rate of 25 ft3/min, and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 6 ft high? (Round your answer to two decimal places.)
Too bad you didn't show your work.
#1. Let x be the distance of the plane from the point directly overhead. Then
3/x = tanθ
so, x = 3cotθ
dx/dt = -3csc^2θ dθ/dt
when θ = π/6, x = 3√3 and cscθ = 2
So, the plane's speed,
dx/dt = -3(2)(-π/4)
#2. At the moment in question, AP = 13, so BP = 26.
Let x = BQ, y=AQ. So we know that AP+BP=39, and
√(y^2+12^2) + √(x^2+12^2) = 39
y/√(y^2+144) dy/dt + x/√(x^2+144) dx/dt = 0
Plugging our numbers,
5/13 * 1 + √133/26 dx/dt = 0
#3. If the base of the cone has radius r, then the height h=2r, so the volume
v = 1/3 πr^2 h = 1/3 π (h/2)^2 h = 1/12 πh^3
dv/dt = 1/4 π h^2 dh/dt
Plugging in the numbers,
25 = π/12 * 36 * dh/dt
1. Well, it sounds like the plane is having a real "altitude-itude" problem! But don't worry, I'll help you land this question safely.
First, let's break it down. We have the angle of elevation decreasing at a rate of π/4 rad/min and we want to find the speed of the plane.
Now, let's use a little trigonometry. We know that tan(θ) = opposite/adjacent, where θ is the angle of elevation.
At the given angle of elevation of π/6, the opposite side is the altitude of the plane, which is 3 km. We want to find the rate of change of the opposite side with respect to time, which is the speed of the plane.
Now, taking the derivative of both sides, we get sec^2(θ) * dθ/dt = d/dt(3 km).
Substituting the given values, we have sec^2(π/6) * (π/4 rad/min) = d/dt(3 km).
Now, you can solve this equation to find the speed of the plane. Happy flying!
2. Ah, the good ol' cart problem. It's time to go for a little ride!
We have cart A being pulled away from point Q at a speed of 1 ft/s. We want to find how fast cart B is moving toward Q when cart A is 5 ft from Q.
Now, let's use a bit of trigonometry and some good old Pythagoras.
We know that the total length of the rope is 39 ft, and the height of point Q is 12 ft. So, the horizontal distance between point Q and cart A is sqrt(39^2 - 12^2) ft.
When cart A is 5 ft from Q, the horizontal distance between them is sqrt(39^2 - 12^2 - 5^2) ft.
Now, we can differentiate this expression with respect to time to find the rate at which cart B is moving toward Q.
Solve this equation and you'll be able to keep up with cart B's speed like a pro!
3. Ah, the joys of gravel dumping and pile formation. It's like watching a modern art masterpiece in progress!
We know that the gravel is being dumped at a rate of 25 ft^3/min and it forms a pile in the shape of a cone.
Now, we want to find how fast the height of the pile is increasing when it reaches 6 ft.
This calls for some voluminous knowledge! The volume of a cone is given by V = (1/3) * π * r^2 * h, where r is the radius of the base and h is the height.
Since the base diameter and height are always equal, we have r = h/2.
Now, we can find the volume V as a function of height h. Remember, V is also changing with time!
Then, differentiate V with respect to time to find dV/dt. But hey, we're interested in the rate of change of height, so we'll differentiate h with respect to time.
Using the chain rule, we'll have dV/dt = dV/dh * dh/dt.
Substituting the given values, we'll obtain dV/dh = (1/3) * π * (h^2/4).
Now that we have dV/dh, we can find dh/dt by dividing dV/dt with dV/dh.
Once you solve this lovely equation, the height's rate of change will be no longer a towering riddle! Keep piling up those answers!
1. To solve the first problem, you can use trigonometry and related rates.
Let's denote the distance between the telescope and the plane by x, and let's denote the angle of elevation by θ.
We have the following information:
- Altitude of the plane: 3 km = 3000 m
- Angle of elevation: θ = π/6
- Rate at which the angle of elevation is changing: dθ/dt = π/4 rad/min
First, we can use the tangent function to relate the angle of elevation θ and the distance x:
tan(θ) = altitude / distance
tan(π/6) = 3000 / x
Next, we can differentiate both sides of this equation with respect to time (t):
d(tan(θ))/dt = d(3000/x)/dt
Using the chain rule and the derivative of tan(θ), we have:
sec^2(θ) * dθ/dt = (-3000/x^2) * dx/dt
Now, we can substitute the values we know into the equation:
sec^2(π/6) * (π/4) = (-3000/x^2) * dx/dt
Simplifying the equation, we find:
(4/√3) * π/4 = (-3000/x^2) * dx/dt
Canceling out the π/4 terms and rearranging the equation, we have:
1/√3 = (-3000/x^2) * dx/dt
Solving for dx/dt, we find:
dx/dt = (-√3 * x^2) / 3000
Now, we can substitute the given angle of elevation and altitude into the equation and solve for dx/dt when θ = π/6:
dx/dt = (-√3 * (3000^2)) / 3000
Simplifying further, we find:
dx/dt = -√3 * 3000
Therefore, the plane is traveling at a speed of -√3 * 3000 m/min at that time.
2. To solve the second problem, let's first define some variables:
- Distance between Cart A and point Q: x
- Distance between Cart B and point Q: y
We have the following information:
- Length of the rope: 39 ft
- Speed of Cart A: dx/dt = 1 ft/s
- Height of point Q: h = 12 ft
We can create a right triangle with the rope as the hypotenuse, Cart A as the adjacent side, and Cart B as the opposite side. This means we can relate x, y, and the length of the rope using the Pythagorean theorem:
x^2 + y^2 = (39)^2
Taking the derivative of both sides of the equation with respect to time, we get:
2x(dx/dt) + 2y(dy/dt) = 0
Since we want to find dy/dt when x = 5 ft, we can substitute this value into the equation:
2(5)(1) + 2y(dy/dt) = 0
Simplifying the equation, we find:
10 + 2y(dy/dt) = 0
Now, we can solve for dy/dt:
2y(dy/dt) = -10
dy/dt = -10 / (2y)
Using the Pythagorean theorem equation x^2 + y^2 = (39)^2, we can substitute the value of x = 5 ft:
(5)^2 + y^2 = (39)^2
25 + y^2 = 1521
y^2 = 1496
Taking the positive square root of both sides, we find:
y ≈ 38.66 ft
Finally, we can substitute this value of y into the equation for dy/dt:
dy/dt = -10 / (2 * 38.66)
Thus, at the instant when Cart A is 5 ft from Q, Cart B is moving toward Q at a speed of approximately -0.26 ft/s.
3. To solve the third problem, we can use related rates and the formula for the volume of a cone.
Let's denote the height of the pile as h at a particular time, and let's denote the rate at which the height is changing as dh/dt.
We have the following information:
- Rate at which gravel is being dumped from the conveyor belt: dV/dt = 25 ft^3/min
- Pile of gravel in the shape of a cone with base diameter and height always equal
- Height of the pile when the question asks for the rate: h = 6 ft
The volume of a cone is given by the formula:
V = (1/3) * π * r^2 * h
Since the base diameter and height are always equal, the radius (r) of the pile can be expressed as half of the height:
r = h/2
Applying the chain rule, we have:
dV/dt = (∂V/∂r) * (dr/dt) + (∂V/∂h) * (dh/dt)
Differentiating the volume formula with respect to r and h, we get:
dV/dt = (1/3) * π * (2r * dr/dt * h + r^2 * dh/dt)
Substituting the expressions for r and dh/dt, we have:
dV/dt = (1/3) * π * (2(h/2) * dr/dt * h + (h/2)^2 * dh/dt)
Simplifying the equation, we find:
dV/dt = (1/3) * π * (h^2 * dr/dt + (h^2)/4 * dh/dt)
Since the problem gives us dV/dt, we can substitute the known values into the equation to solve for dh/dt:
25 = (1/3) * π * (6^2 * dr/dt + (6^2)/4 * dh/dt)
Simplifying further, we find:
25 = (1/3) * π * (36 * dr/dt + 9 * dh/dt)
Now, we need to find the value of dr/dt. Since the base diameter and height are always equal, the radius (r) is equal to the height (h). Therefore, dr/dt can be expressed as dh/dt.
Substituting this in the equation, we have:
25 = (1/3) * π * (36 * dh/dt + 9 * dh/dt)
Simplifying even more, we find:
25 = (1/3) * π * (45 * dh/dt)
Now, we can solve for dh/dt:
dh/dt = 25 / [(1/3) * π * 45]
Calculating the value, we have:
dh/dt ≈ 0.047 ft/min
Therefore, the height of the pile is increasing at a rate of approximately 0.047 ft/min when the pile is 6 ft high.