A plane flying horizontally at an altitude of 2 miles and a speed of 460 mi/h passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it is 5 miles away from the station. (Round your answer to the nearest whole number.)

Respuesta :

Answer:

The rate at which the distance from the plane to the station is increasing when it is 5 miles away from the station is 421 mi/hr

Step-by-step explanation:

Assume a plane flying horizontally at an altitude of 2miles

h= 2

Considering speed 460mi/hr

Let [tex]\frac{db}{dt}= 460mi/hr[/tex]

Let the distance from the plane to the station be 'c'

c= 5miles

[tex]\frac{dc}{dt}[/tex] = ?

h is constant , so [tex]\frac{dh}{dt} =0[/tex]

The plane diagram is attached below ,

now,

Using pythagorean theorem ,

[tex]h^2+b^2 =c^2[/tex]

[tex]2^2+b^2 =5^2[/tex]

[tex]4+ b^2= 25[/tex]

[tex]b^2=21[/tex]

[tex]b=\sqrt{21}[/tex]

Now, differentiate with respect to 't'

[tex]2h\frac{dh}{dt} +2b\frac{db}{dt} =c\frac{dc}{dt}[/tex]

divide by  2,

[tex]h\frac{dh}{dt} +b\frac{db}{dt} =c\frac{dc}{dt}[/tex]

[tex]2(0)+ (\sqrt{21}(460) =5\frac{dc}{dt}[/tex]

       [tex]\frac{dc}{dt} =\frac{460(\sqrt{21})}{5}[/tex]

        =     421.36 mi/hr

        = 421 mi/hr

Hence , the answer is 421 mi/hr

The rate at which the distance from the plane towards the station would be  increasing when it's 5 miles away from the station will be:

"421 mi/hr".

Speed and Distance

According to the question,

Altitude, h = 2 miles

Speed, [tex]\frac{db}{dt}[/tex] = 460 mi/h

Distance from plane, c = 5 miles

Let,

Constant, h = 0

By using Pythagoras theorem,

→ h² + b² = c²

By substituting the values,

(2)² + b² = (5)²

    4 + b² = 25

          b² = 25 - 4

               = √21

By differentiating with respect to "t", we get

→        2h [tex]\frac{dh}{dt}[/tex] + 2b [tex]\frac{db}{dt}[/tex] = c [tex]\frac{dc}{dt}[/tex]

By substituting the values,

   2(0) + √21 × 460 = 5 [tex]\frac{dc}{dt}[/tex]

                            [tex]\frac{dc}{dt}[/tex] = [tex]\frac{460(\sqrt{21} )}{5}[/tex]

                                = 421 mi/hr

Thus the above answer is appropriate.  

Find out more information about distance here:

https://brainly.com/question/4931057

ACCESS MORE
EDU ACCESS