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a spotlight on the ground shines on a wall 12 m away. if a man 2 m tall walks from the spotlight toward the building at a speed of 1.1 m/s, how fast (in m/s) is the length of his shadow on the building decreasing when he is 4 m from the building? (round your answer to one decimal place.)



Answer :

When he is 4 m away from the building, the speed at which his shadow lengthens is -0.41 m/s.

What do you mean by angles?

An angle is a shape that is made up of two rays or lines that have a shared terminal. The name "angle" comes from the Latin word "angulus," which means "corner." The common terminal of the two rays—known as the vertex—is referred to as an angle's sides. The angle in the plane does not always need to be in Euclidean space.

Given,

Distance of man from building = x

∴ initially, x = 12

dx/dt = -1.1 m/s

Let,

Shadow height = y

Now,

Solve for [tex]\frac{dy}{dt}[/tex] : when x=4

From the diagram it seen similar triangles with similar base/height ratios,

[tex]\frac{2}{12-x}[/tex] = [tex]\frac{y}{12}[/tex]

y = [tex]\frac{24}{12-x}[/tex]

[tex]\frac{dy}{dt}[/tex] = [tex]\frac{24}{(12-x)^{2} }[/tex] × [tex]\frac{dx}{dt}[/tex]

At x= 4,

dy/dt becomes:

[tex]\frac{24}{(12-4)^{2} }[/tex] × (-1.1)

∴ y = -0.41 m/s

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