A rocket is launched from a hill above a lake. The rocket will fall into the lake after exploding at its maximum height. The roc
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Answer:
The shadow is decreasing at the rate of 3.55 inch/min
Step-by-step explanation:
The height of the building = 60ft
The shadow of the building on the level ground is 25ft long
Ѳ is increasing at the rate of 0.24°/min
Using SOHCAHTOA,
Tan Ѳ = opposite/ adjacent
= height of the building / length of the shadow
Tan Ѳ = h/x
X= h/tan Ѳ
Recall that tan Ѳ = sin Ѳ/cos Ѳ
X= h/x (sin Ѳ/cos Ѳ)
Differentiate with respect to t
dx/dt = (-h/sin²Ѳ)dѲ/dt
When x= 25ft
tanѲ = h/x
= 60/25
Ѳ= tan^-1(60/25)
= 67.38°
dѲ/dt= 0.24°/min
Convert the height in ft to inches
1 ft = 12 inches
Therefore, 60ft = 60*12
= 720 inches
Convert degree/min to radian/min
1°= 0.0175radian
Therefore, 0.24° = 0.24 * 0.0175
= 0.0042 radian/min
Recall that
dx/dt = (-h/sin²Ѳ)dѲ/dt
= (-720/sin²(67.38))*0.0042
= (-720/0.8521)*0.0042
-3.55 inch/min
Therefore, the rate at which the length of the shadow of the building decreases is 3.55 inches/min
