A man standing 9 feet from the base of a lamppost casts a shadow 6 feet long. if the man is 6 feet tall and walks away from the lamppost at a speed of 30 feet per minute, at what rate, in feet per minute, will his shadow lengthen?

let x be the distance of shadow and y the distance of the man from the lampost

Height of lampost can be found using similar traingles and it comes to 15 feet.

x /(x+ y) = 6/15 = 2/5

5x = 2x + 2y
3x = 2y

so y = (3/2)x and x = (2/3)y dy/dx = 3/2 and dx/dy = 2/3

speed at which y is increasing = dy/dt = 30

speed of shadow lengthening = dx / dt = dx/dy * dy/dt = 2/3 * 30 = 20 ft/min

Answer is 20 ft / minute.

boffeemadrid boffeemadrid · Answer 2 · 2021-10-14T12:43:41+02:00

The rate at which the shadow is lengthening is required.

The shadow will lengthen at a rate of 20 feet per minute.

Let [tex]s[/tex] be shadow length = 6 ft

[tex]x[/tex] be distance from pole = 9 ft

[tex]m[/tex] = Man's height = 6 ft

[tex]h[/tex] = Pole height

Since the triangles ABD and ECD are similar by Angle Angle similarity.

[tex]\dfrac{s+x}{h}=\dfrac{s}{m}\\\Rightarrow h=\dfrac{(s+x)s}{m}\\\Rightarrow h=\dfrac{(6+9)6}{6}\\\Rightarrow h=15\ \text{feet}[/tex]

Again using the similarity property we have

[tex]\dfrac{s+x}{h}=\dfrac{s}{m}\\\Rightarrow \dfrac{s+x}{15}=\dfrac{s}{6}\\\Rightarrow 6s+6x=15s\\\Rightarrow (15-6)s=6x\\\Rightarrow 9s=6x\\\Rightarrow s=\dfrac{2}{3}x[/tex]

Differentiating with respect to time we have

[tex]\dfrac{ds}{dt}=\dfrac{2}{3}\dfrac{dx}{dt}\\\Rightarrow \dfrac{ds}{dt}=\dfrac{2}{3}\times 30\\\Rightarrow \dfrac{ds}{dt}=20\ \text{feet/minute}[/tex]

The shadow will lengthen at a rate of 20 feet per minute.

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