A man walks along a straight path at a speed of 3 ft/s. A searchlight is located on the ground 4 ft from the path and is kept focused on the man. At what rate is the searchlight rotating when the man is 3 ft from the point on the path closest to the searchlight

0.48 rad/sec

Step-by-step explanation:

dx/dt = 3ft

to find d∅/dt when x = 3

Using the right angled triangle

ΔABC

Now perpendicular (BC) = X FT

Base ( AC) 4 ft

so Hypotenuse(AB) = √ (BC² + AC²) = √ ( x² + 4²) = √ ( x² + 16)

and x/4 = tan∅ ; x = 4tan∅

now we differentiate each side

dx/dt = d/dt(4tan∅) = d/d∅(4tan∅) d∅/dt

⇒ dx/dt = (4sec²∅) d∅/dt  = ( 1/4 cos²∅) dx/dt

⇒ d∅/dt = cos²∅/4 × 3 = 3/4 cos²∅

now when x = 3, the length of the beam is

{from our initial equation (AB) = √ (BC² + AC²)}

AB = √(3² + 4²) = √(9 + 16) = 5ft

therefore cos∅ = 4/5

so we substitute in  (d∅/dt = ( 1/4 cos²∅) dx/dt)

d∅/dt = 1/4 ( 4/5)² × 3

d∅/dt = 12/15 = 0.48 rad/sec

the least value out of the data shown is 5, maybe if you post a picture with it i could a little more.

step-by-step explanation:

question 1 is 8, question 2 is 9, question 3 is 24

step-by-step explanation:

now, the numbers they give you at the top are the key. from there, you just have to plug and chug! let's try 1, 2, and 3!

key: a=3, b=5, c=6

1. a+5

remember, a=3, so we plug that in to get 3+5=8

2. 15-c

c=6, so we plug that in to get 15-6=9

3. 4b

b=6, so we plug it in to get 4*6=24

time to try the rest on your own! if you put the answers below in a comment, i'll check them for you!