Question

3. A person standing 3 feet away from the base of a lamp post casts a shadow 4 feet long. If the person is 6 feet tall and walks away from the lamp post at a speed of 400 feet per minute, at what rate will his shadow lengthen

Answer 1

Answer:

533.33 ft/min

Step-by-step explanation:

We are given that

x=3 feet

y=4 feet

Height of person,h=6 feet

$\frac{dx}{dt}=400 ft/min$

Triangle ABC and A'B'C are similar because all right triangles are similar

By using similarity property

$\frac{h'}{6}=\frac{7}{4}$

$h'=10.5 feet$

$x+y=3+4=7 feet$

$\frac{h'}{6}=\frac{x+y}{y}$

$\frac{10.5}{6}=\frac{x+y}{y}$

$10.5y=6x+6y$

$10.5y-6y=6x$

$4.5y=6x$

Differentiate w.r.t t

$4.5\frac{dy}{dt}=6\frac{dx}{dt}$

$\frac{dy}{dt}=\frac{6\times 400}{4.5}ft/min$

$\frac{dy}{dt}=533.33ft/min$

Hence, If the person is 6 feet tall and walks away from the lamp post at a speed of 400 feet per minute then his shadow moving at the rate 533.33 ft/min