Answer:
The rate of change of the height is - 4 ft/s
Solution:
As per the question:
Height of the person, y = 5 ft
The rate at which the person walks away, [tex]\frac{dx}{dt} = 4\ ft/s[/tex]
Distance of the spotlight from the wall, x = 40 ft
Now,
To calculate the rate of change in the height, [tex]\frac{dy}{dt}[/tex] of the person when, x = 10 m:
From fig 1.
[tex]\Delta ABC[\tex] ≈ [tex]\Delta PQC[\tex]
Thus
[tex]\frac{BC}{AB} = \frac{PQ}{QC}[/tex]
[tex]\frac{y}{40} = \frac{5}{x}[/tex]
xy = 200 (1)
Differentiating the above eqn w.r.t time t:
[tex]x\frac{dy}{dt} + y\frac{dx}{dt} = 0[/tex]
Thus
[tex]\frac{dy}{dt} = - \frac{y}{x}\frac{dx}{dt}[/tex] (2)
From eqn (1):
When x = 10 ft
10y = 200
y = 20 ft
Using eqn (2):
[tex]\frac{dy}{dt} = - \frac{20}{10}\times 2 = - 4\ ft[/tex]
