Answer:
Tip of the shadow is moving at the rate of [tex]\dfrac{20}{3}\:ft/s[/tex].
Step-by-step explanation:
Refer to the attachment.
Let AB be the height of the street light. So, AB = 15 ft.
Let DE be the height of the man. So, DE = 6 ft.
Let BE be the distance between the man and light pole. Assume BE = x
Let BC be the distance between tip of shadow and light pole. Assume BC = y
Now height of the shadow of the man is EC which can be calculated as follows,
BC = BE + EC
y = x + EC
Subtracting x,
y – x = EC.
So there are two triangles that is, triangle ABC and triangle DEC. Assume that both the triangles are similar.
So applying similar triangle property,
[tex]\dfrac{AB}{BC}=\dfrac{DE}{EC}[/tex]
Substituting the value,
[tex]\dfrac{15}{y}=\dfrac{6}{y-x}[/tex]
Cross multiplying,
[tex]15\times \left(y-x\right)=6\times y[/tex]
Using distributive property,
[tex]15y-15x=6y[/tex]
Subtracting 6y,
[tex]15y-6y-15x=0[/tex]
Adding 15x,
[tex]15y-6y=15x[/tex]
[tex]9y=15x[/tex]
Dividing by 9,
[tex]y=\dfrac{15}{9}x[/tex]
Multiply and divide by 3,
[tex]y=\dfrac{5}{3}x[/tex]
Differentiating above equation with respect to t,
[tex]\dfrac{dy}{dt}=\dfrac{d}{dt}\left(\dfrac{5}{3}x\right)[/tex]
Applying constant multiple rule,
[tex]\dfrac{dy}{dt}=\dfrac{5}{3}\dfrac{d}{dt}\left(x\right)[/tex]
[tex]\dfrac{dy}{dt}=\dfrac{5}{3}\dfrac{dx}{dt}[/tex]
Given that man walks away from the pole with a speed of 4 ft/s that is,
[tex]\dfrac{dx}{dt}=4[/tex]
Substituting the value,
[tex]\dfrac{dy}{dt}=\dfrac{5}{3}\left(4\right)[/tex]
[tex]\dfrac{dy}{dt}=\dfrac{20}{3}\:ft/s[/tex]
Therefore, tip of the shadow is moving at the rate of [tex]\dfrac{20}{3}\:ft/s[/tex].
