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.
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