The infamous Wile E. Coyote has been chasing the Roadrunner again. His latest idea for a trap involved launching an enormous boulder (about 16ft in diameter) up to a height of 750 ft, hoping that it would come down crashing on the unsuspecting Roadrunner. Unfortunately, as these things usually go, it is the hungry Coyote who is standing right under the falling rock, unable to move. As he looks up, the boulder doesnt seem that big, because its perceived size depends on the angle that the boulder subtends with respect to the eyes of the Coyote. But it is rapidly approaching and, therefore, rapidly appearing bigger. When the center of the boulder is 100ft above the Coyote's head, it is falling at a speed of 95 mi/h. (Remember, 1mi= 5280ft). How fast is the angle subtended by the boulder increasing at that moment?( Give your answer in radians per second, rounded to 3 decimal places.)
Please help me I am so confused!!! AHH
I am not sure how to start this!
as usual, draw a diagram. They really haven't given you many pieces of data, so don't look for a complicated formula.
It doesn't matter how high the rock flew. All we care about is where it is now, and how fast it's falling.
When the rock is x ft up,
x/8 = cotθ
1/8 dx/dt = -csc^2 θ dθ/dt
when x=100, csc^2 θ = 1+cot^2 θ = 1+(100/8)^2 = 157.25
we want dθ/dt, right?
95 mi/hr = 139.33 ft/s
dx/dt < 0 because it's falling - height is decreasing
1/8 (-139.33) = -157.25 dθ/dt
dθ/dt = 0.111 rad/sec
I know you got 8 from because it is the radius but can you explain why you are using cot?
Kim, you weren't allowed to seek help outside. I'm sorry, but your exam2 - take home portion is canceled.
Please come to speak with me in my office before the next class, you'll be able to make up probably in the next final exam.
Nano
Haha! This is hilarious! Good job Hernando Tellez for catching students here! You're the coolest instructor!
To find the rate at which the angle subtended by the boulder is increasing, we can use the concept of related rates from calculus. Let's break down the problem step by step:
1. First, we need to identify the given information and assign variables to them:
- Diameter of the boulder: 16ft
- Height at which the boulder is launched: 750ft
- Distance from the center of the boulder to the Coyote at a given time: 100ft
- Speed of the falling boulder at that moment: 95 mi/h
2. We want to find the rate at which the angle subtended by the boulder is increasing when the center of the boulder is 100ft above the Coyote's head. Let's call this rate dθ/dt, where θ represents the angle.
3. To relate the variables, we need to consider the right triangle formed by the Coyote, the center of the boulder, and the highest point of the boulder. The angle θ we are interested in is the angle at the Coyote's position.
4. We can use the tangent function to relate the variables in terms of θ:
tan(θ) = (diameter of the boulder) / (height of the boulder)
Substituting the values we know:
tan(θ) = 16 / 750
5. Now we need to differentiate both sides of the equation with respect to time (t), and solve for dθ/dt:
d/dt[tan(θ)] = d/dt[16/750]
To differentiate the left side, we can use the chain rule:
sec^2(θ) * dθ/dt = 0
Since dθ/dt is what we want to find, we solve for it:
dθ/dt = 0 / sec^2(θ)
dθ/dt = 0
Therefore, the rate at which the angle is increasing (dθ/dt) is 0 radians per second at the given moment.
From the given information, it seems that the rate of change for the angle subtended by the boulder is 0 radians per second, indicating that the angle is not changing at that specific time. This result might be unexpected, but it's based on the information provided in the problem.