Viewing angle, The bottom of a large theater screen is 3 ft above your eye level and the top of the screen is 10 ft above your eye level. Assume you walk away from the screen (perpendicular to the screen) at a rate of 3 ft/s while looking at the screen. What is the rate of change of the viewing angle 0 when you are 30 ft from the wall on which the screen hangs, assuming the floor is flat
You would do most parts in Steve's answer. However, the question uses the time derivative (in relations to t), not the x derivative. He forgot to take into account that dx/dt = -3ft/sec. The given rate (dx/dt) would be negative in the question's context.
Since Steve typed most of the steps already, lets skip to θ = B - A
dθ/dt = (dB/dt - dA/dt) dx/dt
We'll solve for (dB/dt - dA/dt) first
= -3/(x^2+9) + 10/(x^2+100)
= 7(x^2-30)/((x^2+9)(x^2+100))
At x = 30
= 7(30^2-30)/((30^2+9)(30^2+100))
= 203/30300
Now you have
dθ/dt = (203/30300)dx/dt
Lets substitute dx/dt in the equation and solve for dθ/dt
dθ/dt = (203/30300)(-3)
= -203/10100 or -0.02009 . . .
How can dx/dt be negative when you are walking away from the screen and adding more distance between you and the screen?? The answer to the problem is correct as stated -0.02 radians per second. The negative sign comes from the differential d(alpha)/dt = -dx/dt*(tan(alpha))/(x *(sec(alpha)*sec(alpha)) the other equation is similar but with the angle Beta. Arriving at d(theta)/dt = d(Beta)/dt - d(alpha)/dt
d(theta)/dt = -0.03 - (-0.01) giving us d(theta)/dt = -0.02
To find the rate of change of the viewing angle, we need to consider the geometry of the situation. Let's break it down step by step.
1. First, we need to find the distance between your eyes and the screen. This is equal to the distance from the floor to the bottom of the screen plus your eye level. In this case, the bottom of the screen is 3 ft above your eye level, so the distance from your eyes to the bottom of the screen is 3 ft + your eye level.
2. Next, we need to find the height of the screen. The top of the screen is 10 ft above your eye level, so the height of the screen is 10 ft.
3. Now, let's use the concept of similar triangles. The viewing angle is the angle between your line of sight and the horizontal plane (floor). This angle can be expressed as the ratio of the height of the screen to the distance between your eyes and the screen.
So, the viewing angle θ can be calculated as: θ = atan(height of the screen / distance from your eyes to the screen)
4. To find the rate of change of the viewing angle (dθ/dt), we can differentiate this expression with respect to time (t) using the chain rule:
dθ/dt = (d/dt) [atan(height of the screen / distance from your eyes to the screen)]
5. Since you are walking away from the screen perpendicular to it, the distance from your eyes to the screen is changing with time. Let's call this distance x. So, we have:
distance from your eyes to the screen = sqrt(x^2 + height of the screen^2)
6. We can now substitute this expression back into the equation for the viewing angle:
θ = atan(height of the screen / sqrt(x^2 + height of the screen^2))
7. Differentiating the viewing angle expression with respect to time (t) using the chain rule, we get:
dθ/dt = d/dt [atan(height of the screen / sqrt(x^2 + height of the screen^2))]
8. We know that you are walking away from the screen at a rate of 3 ft/s, so the rate of change of the distance from your eyes to the screen is dx/dt = 3 ft/s.
9. Now, we can substitute this information into our expression for the rate of change of the viewing angle:
dθ/dt = d/dt [atan(height of the screen / sqrt(x^2 + height of the screen^2))]
= d/dx [atan(height of the screen / sqrt(x^2 + height of the screen^2))] * dx/dt
= [1 / (1 + (height of the screen / (x^2 + height of the screen^2)))] * dx/dt
10. Finally, we can evaluate the rate of change of the viewing angle at the given distance of 30 ft (x = 30 ft) by substituting the known values:
dθ/dt = [1 / (1 + (10 ft / (30 ft^2 + 10 ft^2)))] * (3 ft/s)
Simplifying this expression will give us the rate of change of the viewing angle at a distance of 30 ft from the screen.
As usual, draw a diagram. You can easily see that if you are x away from the wall,
the angle of elevation of the bottom of the screen (A) is
cotA = x/3
A = arccot(x/3)
angle B to the top is
cotB = x/10
B = arccot(x/10)
So, since θ = B-A
dθ/dt = dB/dt - dA/dt
= -3/(x^2+9) + 10/(x^2+100)
= 7(x^2-30)/((x^2+9)(x^2+100))
so, at x=30
dθ/dt = 203/30300