contestada

A man 6.00 ft tall approaches a street light 15.0 ft above the ground at the rate of 4.00 ​ft/s. How fast is the end of the​ man's shadow moving when he is 14.0 ft from the base of the​ light?

Respuesta :

Answer:

[tex]\frac{dx}{dt} = 10 ft/s[/tex]

Explanation:

As per given figure let say the tip of the shadow is at distance "x" from the base of the lamp

so here we have

[tex]\frac{x}{15} = \frac{x - y}{6}[/tex]

so we have

[tex]6x = 15 x - 15 y[/tex]

[tex]15 y = 9 x[/tex]

now we have

[tex]5\frac{dy}{dt} = 2\frac{dx}{dt}[/tex]

[tex]5(4) = 2\frac{dx}{dt}[/tex]

[tex]\frac{dx}{dt} = 10 ft/s[/tex]

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