Valishati
contestada

A man walks in a straight path away from a streetlight at a rate of 4 feet per second. The man is 6 feet tall and the streetlight is 15 feet tall.

At what rate is the length of his shadow changing?
At what rate is the distance from the top of the streetlight to the tip of his shadow changing when the man is 12 feet from the streetlight?

Respuesta :

amna04352

Answer:

a) 8/3 feet/second

b) 16/3 feet/second

Step-by-step explanation:

Let the length of the shadow be X

And initial distance from the streetlight be D

15/6 = (D + X)/X

15X = 6D + 6X

9X = 6D

X = ⅔D

dX/dD = ⅔

⅔ = dX/dt × dt/dD

dX/dt = ⅔ × 4 = 8/3

Distance from the tip is the hypotenuse 'H'

H² = 15² + (D + X)²

H² = 225 + (D + X)²

Differentiate both sides wrt time

2H × dH/dt = 2(D + X) × (dD/dt + dX/dt)

When D = 12, X = ⅔(12) = 8

D + X = 20

H² = 20² + 15²

H² = 625

H = 25

2(25) × dH/dt = 2(20) × (4 + 8/3)

dH/dt = 16/3

kapoorprachi783

The answer is:

  1. 8/3 feet/second
  2. 16/3 feet/second

How to find the length of his shadow changing?

Let the length of the shadow be x and the initial distance from the streetlight be D

⇒ 15/6 = (D + X)/X

⇒ 15X = 6D + 6X

⇒ 9X = 6D

⇒ X = ⅔D

⇒ dX/dD = ⅔

⇒ ⅔ = dX/dt × dt/dD

⇒ dX/dt = ⅔ × 4 = 8/3

Distance from the tip is the hypotenuse 'H'

⇒ H² = 15² + (D + X)²

⇒ H² = 225 + (D + X)²

Differentiate both sides wrt time

⇒ 2H × dH/dt = 2(D + X) × (dD/dt + dX/dt)

When D = 12, X = ⅔(12) = 8

⇒ D + X = 20

⇒ H² = 20² + 15²

⇒ H² = 625

⇒ H = 25

⇒ 2(25) × dH/dt = 2(20) × (4 + 8/3)

⇒ dH/dt = 16/3

To differentiate is defined as separating out two or more things, or looking at and understanding what makes things different or distinctive.

Learn more about differentiability here: brainly.com/question/25081524

#SPJ2

Otras preguntas