A man is walking towards a wall at 3 ft/sec while looking at a picture on the wall, at a height of 5 feet from the floor. How is the angle of observation changing when the man is 8 feet away from the screen.
You have to have the height the man's eyes are frm the floor to do this.
angle of observation= arctan(eye-5)/d
where distance d is the horizontal distance. eye = height of eyes off floor.
you are given d'=3ft/sec
theta= arctan(eye-5/d)
4ye-5=d*tanTheta
take the deriviative of each side.
0=d' tanTheta+ d (sec^2 theta) dTheta/dt
so solve for dTheta/dt
dTheta/dt= -3tanTheta /(d sec^2 theta)
so at 8 feet, d=8, tanTheta=(eye-5)/8,
secTheta=8/sqrt(64+(eye-5)^2). Note if the eye height is very near 5 ft, then secTheta=appx 1.0
Bob is wrong
dx/dt = 3ft/s, dƟ/dt = ?
5ft/x = tan Ɵ
Differentiating each side with respect to the time: (-5ft)/x^2 dx/dt=〖sec〗^2 ϑ dϑ/dt
So dϑ/dt=(-5ft)/x^2 dx/dt.〖cos〗^2 θ
We want to evaluate the change at the particular instant when x = 8ft (and so the hypotenuse is √89 and the cosƟ = 8/ √89 )
dϑ/dt=(-5ft)/(64〖ft〗^2 ) 3 ft/sec.64/89=15/89sec
The angle is changing at 15/89 radians per second
To determine how the angle of observation is changing, we need to use trigonometry and calculate the rate of change of the angle. Let's break down the problem step by step.
1. We know that the man is walking towards the wall at a speed of 3 ft/sec. This means his distance from the wall is changing over time. Let's denote this distance as "x" (in feet).
2. We are interested in finding the rate at which the angle of observation is changing, so we need to determine how this angle is related to the distance "x."
3. The angle of observation will be the angle formed between the man's line of sight and the horizontal line passing through the picture on the wall. Let's call this angle "θ."
4. When the man is 8 feet away from the picture, we can consider a right triangle that consists of the horizontal line through the picture, the line connecting the man's eyes to the picture, and the line connecting the man's feet to the picture.
5. The height of the picture from the floor is given as 5 feet. Since we know that the man is 8 feet away and walking toward the wall at a speed of 3 ft/sec, we can determine that the distance between the man's eyes and the floor (the opposite side of the right triangle) is decreasing at a rate of 3 ft/sec.
6. Let's use the tangent function to relate the angle of observation "θ" to the distance "x" and the height of the picture. The tangent of "θ" is equal to the opposite side (height of the picture) divided by the adjacent side (distance between the man's eyes and the picture on the wall).
tan(θ) = height of the picture / distance between man's eyes and picture
tan(θ) = 5 / x
7. Now, let's differentiate both sides of the equation with respect to time (t) to find the rate of change of the angle of observation.
d/dt(tan(θ)) = d/dt(5 / x)
sec²(θ) * dθ/dt = -5 / x² * dx/dt
dθ/dt = (-5 / x² * dx/dt) / sec²(θ)
8. We are given that the man is walking towards the wall at a speed of 3 ft/sec, so dx/dt = -3. When the man is 8 feet away, x = 8. We need to find dθ/dt when x = 8.
dθ/dt = (-5 / (8)² * (-3)) / sec²(θ)
dθ/dt = 15 / 64 / sec²(θ)
9. To determine the value of sec²(θ), we can use the Pythagorean theorem because we know the opposite side (5) and the adjacent side (8).
sec²(θ) = hypotenuse² / adjacent side²
sec²(θ) = (5² + 8²) / (8²)
sec²(θ) = 64 / 64
sec (θ) = 1
10. Now that we have the value of sec²(θ), we can substitute it into the equation for dθ/dt.
dθ/dt = 15 / 64 / 1
dθ/dt = 15 / 64
Therefore, when the man is 8 feet away from the picture, the angle of observation is changing at a rate of 15/64 radians per second.