Answer :
Answer:
At 28.93 km/min is the distance from the plane to the radar station increasing 5 minutes later
Step-by-step explanation:
Refer the attached figure
[tex]\angle A =45+90=135^{\circ}[/tex]
We are given A plane flying with a constant speed of 29 km/min passes over a ground radar station at an altitude of 15 km
[tex]\frac{dy}{dt}=29[/tex]
x=15 km
We will use cosine law
[tex]z^2=x^2+y^2-2xy Cos A\\z^2=15^2+y^2-30y Cos A\\z^2=225+y^2-30y Cos A[/tex]----1
Differentiating
[tex]2z\frac{dz}{dt}=2y\frac{dy}{dt}-30Cos(135^{\circ})\frac{dy}{dt}[/tex]----2
[tex]\frac{dy}{dt}=29\\y=5 \times 29 =145[/tex]
Substitute values in 1
[tex]z^2=225+(145)^2-30(145) Cos (135)\\z=\sqrt{225+(145)^2-30(145) Cos (135)}[/tex]
z=155.96 km
Substitute values in 2
So, [tex]2(155.96)\frac{dz}{dt}=2(145)(29)-30Cos(135^{\circ})(29)[/tex]
[tex]\frac{dz}{dt}=\frac{2(145)(29)-30Cos(135^{\circ})(29)}{2(155.96)}[/tex]
[tex]\frac{dz}{dt}=28.93[/tex]
Hence At 28.93 km/min is the distance from the plane to the radar station increasing 5 minutes later
The rate at which the distance from the plane to the radar station is increasing 5 minutes later is; 28.85 km/min
Since the train passes at an altitude of 15 km.
Let us say a = 15 km
Then it climbs at an angle of 45°
Plane is flying at a speed of 29 km/min
Thus, in t minutes, it will have flown a distance of 29t km
Now, the distance of the plane from the radar station t minutes later is gotten by Pythagoras theorem as;
x(t) = √[(29t)² + (15²)]
x(t) = √(841t² + 225)
x'(t) = dx/dt = 841t/(√(841t² + 225))
After t = 5 minutes, we have;
x'(5) = (841 × 5)/(√((841 * 5²) + 225))
x'(t) = 4205/145.7738
x'(t) = 28.85 km/min
Read more at; https://brainly.com/question/16813212