Answer :
[tex]\frac{dx}{dy} = \frac{5}{3}. 4\frac{ft}{s} = \frac{20}{3}\frac{ft}{s}[/tex] is moving at the speed of the shadow
x is the distance from the man to the pole, and y is the distance from the tip of the man's shadow to the pole. I assume the man and pole are standing straight up, which means the two cases are similar.
[tex]\frac{y-x}{y} = \frac{6}{15}[/tex]
15(y-x) = 6y
9y = 15x
[tex]\frac{5}{3}x\\[/tex] = y
differentiate both sides with respect to t or time.
[tex]\frac{dx}d{y} = \frac{5}{3}\frac{dx}{dt}[/tex]
you know [tex]\frac{dx}{dy} = 4\frac{ft}{s}[/tex] because the man is walking that speed away from the pole. you want to find [tex]\frac{dx}{dy}[/tex] , how fast the tip of the shadow is moving.
that means
[tex]\frac{dx}{dy} = \frac{5}{3}. 4\frac{ft}{s} = \frac{20}{3}\frac{ft}{s}[/tex]
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