1.) A man 6 ft tall walks at a rate of 5 ft per sec. from a light that is 15 ft above the ground. At what rate is the top of his shadow changing?
2.) An airplane flies at an altitude of 5 miles toward a point directly over an observer. The speed of the plane is 600 mph. Find the rate at which the angle of elevation is changing when the angle is at 60 degrees.
Make a diagram showing the lamppost and the man's position any time.
Join the top of the lamppost to the man's top and extend it to the ground.
let the distance between the lamppost and the man along the ground be x ft, let the length of his shadow on the ground be y ft.
Since we have similar triangles, use the ratio
(x+y)/15 = y/6
6x + 6y = 15y
6x = 9y
2x = 3y
then 2dx/dt = 3dy/dt
2(5) = 3dy/dt
dy/dt = 10/3
So it appears that the position of the man is irrelevant and the shadow is changing at 10/3 ft/sec
Had it asked "how fast is his shadow moving ?" you would have added the 5ft/sec to the above answer.
2nd problem:
Again, make a diagram showing the plane moving in a line parallel to the ground.
Let the distance of the plane's path be x miles and the angle of elevation be Ø
I see that
tan Ø = 5/x
x tanØ = 5
x sec^2 Ø dØ/dt + tanØ dx/dt = 0
so when Ø = 60° , x = 5/√3
(5/√3)(1/3) dØ/dt + √3(600) = 0
dØ/dt = -1080 radians/hour
or 3/10 radians/sec , (I divided by 3600 to get to seconds)
(since derivatives in trig only work with radians, the answer obtained would be in radians)
1.) To find the rate at which the top of the man's shadow is changing, we need to use similar triangles. Let's denote the height of the man as 'h', the height of the light as 'L', and the length of the shadow as 'x'. We also need to find the rate at which the length of the shadow is changing, which we can denote as 'dx/dt'. We are given that the man is walking at a rate of 5 ft/sec, which we can denote as 'dy/dt'.
Using similar triangles, we have the following relationship:
(h+L)/x = h/(x+dx)
Cross-multiplying gives us:
x(h+L) = h(x+dx)
Expanding the equation gives us:
hx + xL = hx + hdx
Simplifying the equation gives us:
xL = hdx
Dividing both sides of the equation by 'dt' gives us:
(dh/dt)x + h(dx/dt) = 0
As we are looking for the rate at which the top of the shadow is changing, we are interested in finding dx/dt when h = 6 ft.
Given that L = 15 ft, h = 6 ft, and dh/dt = 5 ft/sec, we can substitute into the equation to solve for dx/dt:
(5 ft/sec)(6 ft) + (6 ft)(dx/dt) = 0
30 + 6(dx/dt) = 0
6(dx/dt) = -30
dx/dt = -5 ft/sec
Therefore, the top of the man's shadow is changing at a rate of -5 ft/sec.
2.) To find the rate at which the angle of elevation is changing, we need to use trigonometry. Let's denote the angle of elevation as 'θ' and the rate at which the angle is changing as 'dθ/dt'. We are given that the altitude of the plane is 5 miles, which we can convert to feet by multiplying by 5280 ft/mile, giving us 26,400 ft. We are also given that the speed of the plane is 600 mph, which we can convert to ft/sec by multiplying by 5280 ft/mile and dividing by 3600 sec/hour, giving us approximately 880 ft/sec. We are interested in finding dθ/dt when θ = 60 degrees.
Using trigonometry, we have the relationship:
tan(θ) = altitude/horizontal distance
Differentiating both sides of the equation with respect to time gives us:
sec^2(θ)(dθ/dt) = (d(altitude)/dt)/(horizontal distance)
Since we are looking for dθ/dt when θ = 60 degrees, we can substitute the given values:
sec^2(60 degrees)(dθ/dt) = (d(26,400 ft)/dt)/(horizontal distance)
The sec^2(60 degrees) is equal to 4, so the equation becomes:
4(dθ/dt) = (d(26,400 ft)/dt)/(horizontal distance)
The horizontal distance is equal to the distance the plane travels in the x-direction, which is equal to the speed (880 ft/sec) multiplied by the time it takes for the plane to fly directly over the observer.
At 60 degrees, the angle between the horizontal direction and the line of sight to the plane is 90 - 60 = 30 degrees. This means that the observer is looking up at an angle of 30 degrees from the horizontal.
Given that the speed of the plane is 880 ft/sec, we can calculate the time it takes for the plane to fly directly over the observer by using the equation:
horizontal distance = speed * time
horizontal distance = 880 ft/sec * time
horizontal distance = 880t ft
Since the tangent of 30 degrees is equal to sqrt(3)/3, we have:
4(dθ/dt) = (d(26,400 ft)/dt)/(880t) * sqrt(3)/3
d(26,400 ft)/dt is equal to the speed of the plane, which is 600 mph converted to ft/sec, giving us 880 ft/sec.
Therefore, we have:
4(dθ/dt) = (880 ft/sec) / (880t) * sqrt(3)/3
Simplifying the equation gives us:
4(dθ/dt) = sqrt(3)/3t
Dividing both sides of the equation by 4 gives us:
dθ/dt = (sqrt(3)/3t)/4
Substituting θ = 60 degrees into the equation gives us:
dθ/dt = (sqrt(3)/3(60))/4
Calculating the value gives us:
dθ/dt ≈ 0.00974 radians/sec
Therefore, the rate at which the angle of elevation is changing when the angle is 60 degrees is approximately 0.00974 radians/sec.
To solve both of these problems, we will use similar triangles and the concept of rates of change.
1.) Let's start by defining our variables:
- Let h be the height of the man.
- Let d be the distance of the man from the light.
- Let s be the length of the shadow.
We need to find the rate at which the top of the man's shadow is changing, which means we need to determine ds/dt, the rate of change of the shadow length with respect to time.
To find ds/dt, we need to relate the variables involved. Since we are dealing with similar triangles, we can use the property that corresponding sides of similar triangles are in proportion.
Considering the triangle formed between the man, the top of the shadow, and the top of the light, we have:
(man height + shadow length) / distance to the light = light height / distance to the light
(h + s) / d = 15 / d
By cross-multiplication, we get (h + s) = 15.
Differentiating both sides with respect to time (t), we have:
d(h + s)/dt = d(15)/dt
dh/dt + ds/dt = 0
Since the man's height is constant, dh/dt = 0. Thus, we can simplify the equation to:
ds/dt = -dh/dt = 0
Therefore, the rate at which the top of the man's shadow is changing is 0 ft/sec.
2.) Let's define our variables:
- Let h be the altitude of the airplane.
- Let d be the distance between the observer and the airplane.
- Let θ be the angle of elevation.
We need to find the rate at which the angle of elevation (θ) is changing when the angle is 60 degrees, which means we need to determine dθ/dt, the rate of change of the angle with respect to time.
To find dθ/dt, we need to relate the variables involved. Since we are dealing with similar triangles, we can again use the property that corresponding sides of similar triangles are in proportion.
Considering the triangle formed between the observer, the airplane, and the point directly overhead, we have:
h / d = 5 / (d + x)
Here, x represents the horizontal distance from the observer to the point directly overhead. Since we want to find the rate when θ = 60 degrees, we know that x = h * tan(θ).
Substituting x into the equation, we have:
h / d = 5 / (d + h * tan(θ))
Cross-multiplying, we get:
h * (d + h * tan(θ)) = 5d
Differentiating both sides with respect to time (t), we obtain:
dh/dt * (d + h * tan(θ)) + h * (ddt + dh/dt * tan(θ)) = 5ddt
Rearranging the equation and solving for dθ/dt, we have:
dθ/dt = (5ddt - dh/dt * (d + h * tan(θ))) / (h^2 + d * h / cos(θ))
Now, substitute the given values when θ = 60 degrees and solve for dθ/dt by plugging in values for dh/dt, dd/dt, h, θ, and d.