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An airplane is flying overhead at a constant elevation of 3000ft. A man is viewing the plane from a position 4000ft from the base of a radio tower. The airplane is flying horizontally away from the man. If the plane is flying at a rate of 500ft/s, at what rate is the distance between the man and the plane increasing when the plane passes over the radio tower?

Answer :

Answer: The distance between the man and the plane increasing at a rate of 400ft/s

Explanation: Please see the attachments below

${teks-lihat-gambar} Abdulazeez10
${teks-lihat-gambar} Abdulazeez10
${teks-lihat-gambar} Abdulazeez10
Cricetus

The rate of distance will be "300 ft/s".

Let,

  • The horizontal distance between the man and the base will be "x".
  • The distance between man and plane will be "D".

We know,

The rate,

  • 500 ft/s

then,

  • [tex]\frac{dx}{dt} = 500[/tex]...(1)

By using Pythagoras theorem, we get

→ [tex]x^2+3000^2 = D^2[/tex]

                [tex]D = \sqrt{x^2+3000^2}[/tex]

Differentiating both sides with respect to time, we get

→ [tex]\frac{dD}{dt} = \frac{x}{\sqrt{x^2+3000^2} }\times \frac{dx}{dt}[/tex]

By substituting the "x=4000" and using (1), we get

→ [tex]\frac{dD}{dt} = \frac{4000}{\sqrt{4000^2+3000^2} }\times 500[/tex]

        [tex]= \frac{3000}{1000\sqrt{16+9} }\times 500[/tex]

        [tex]=\frac{1500}{5}[/tex]

        [tex]= 300 \ ft/s[/tex]

Thus the above approach is correct.  

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