O lvl A Maths: Applications of Differentiation (Rates of Change)

A man 1.5m tall is walking at a speed of 2 ms^-1 away from a lamppost which has a lamp 5m above the ground, as shown in the figure.

Find
(a) the rate at which his shadow is increasing,
(b) the speed of the top of his shadow.

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Answer:

Let the length of shadow = x
Let the distance from the lamppost to the man = y

Since the man is 1.5 m tall,
using similar triangles,

(x+y)/5 = x/1.5
y/5 = 7x/15
y = 7x/3

(a) Given: dy/dt = 2
To find: dx/dt

From y=7x/3, dy/dx = 7/3

dy/dt = dy/dx * dx/dt
2 = 7/3 * dx/dt
dx/dt = 6/7 ms^-1

Thus, the shadow is increasing at 6/7 ms^-1


(b) Let z = distance of top of shadow from lamppost
z = x + y

Thus,
speed of the top of his shadow
= dz/dt
= d(x + y) / dt
= dx/dt + dy/dt
= 20/7 ms^-1

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