Example: An airplane, flying at 475 km/h at a constant altitude h = 5 km, is approaching a camera on the ground. Let theta be the angle of elevation above the ground at which the camera is pointed. When theta = π /6 how fast does the camera have to rotate in order to keep the plane in view?
Solution: We suppose that the camera is at the point B and that the plane is above the point A. We denote the distance between A and B by x.
The question is to find the rate of change of theta with respect to (distance x/ altitude h/ elevation θ/ time t) ("... how fast ..."). Since the distance x between A and B is decreasing at 475 km/h we conclude that dt / dx=( ) km/h.
From the right triangle we see that
tan theta =5/x
It follows that
(1/(sin^2θ) or 1/(sec^2θ) or 1/(cos^2θ) or 1/(csc^2θ)) d(theta)/dt= 5 x^2 -5 x^(-2) 5 x^(-2) -5 x^2 dtdx.
To calculate d(theta)/td when = π /6 , first we calculate the distance between A and B from the right triangle as x = (5(tan π /6 ) or 5/(tan π /6 ) or (tan π /6 )/5 .
Therefore the camera must turn at
d(theta)/td=cos^2(pi/6)*(−5)*(5 / tan(pi/6)^−2*(−475)=
(2375 sin^2π /6 or (5 /475 )cos^2π or /6 (475 /5 )sin^2π /6
( ) radians/hour.
select the right answer or write down the answer in ().
Help me please
Solution: We suppose that the camera is at the point B and that the plane is above the point A. We denote the distance between A and B by x.
The question is to find the rate of change of theta with respect to (distance x/ altitude h/ elevation θ/ time t) ("... how fast ..."). Since the distance x between A and B is decreasing at 475 km/h we conclude that dt / dx=( ) km/h.
From the right triangle we see that
tan theta =5/x
It follows that
(1/(sin^2θ) or 1/(sec^2θ) or 1/(cos^2θ) or 1/(csc^2θ)) d(theta)/dt= 5 x^2 -5 x^(-2) 5 x^(-2) -5 x^2 dtdx.
To calculate d(theta)/td when = π /6 , first we calculate the distance between A and B from the right triangle as x = (5(tan π /6 ) or 5/(tan π /6 ) or (tan π /6 )/5 .
Therefore the camera must turn at
d(theta)/td=cos^2(pi/6)*(−5)*(5 / tan(pi/6)^−2*(−475)=
(2375 sin^2π /6 or (5 /475 )cos^2π or /6 (475 /5 )sin^2π /6
( ) radians/hour.
select the right answer or write down the answer in ().
Help me please
-
Plane's Distance = d
dd/dt = - 475 km./hr. (Negative because the plane is approaching the camera)
Altitude, h = 5 km.
Angle, Θ = π/6 rad.
Find dΘ/dt:
Tan Θ = h / d
d = h / Tan Θ
d = 5 / Tan (π/6)
d = 5 / 0.5774
d = 8.66
Since d is decreasing,
d = - 8.66
d Tan Θ = h
Differentiating Implicilty Over Time,
d (1 / Cos² Θ) (dΘ/dt) + (dd/dt) Tan Θ = dh/dt
Since the altitude does not change, h is constant and the derivative of a constant is zero, so dh/dt = 0.
- 8.66 [1 / Cos² (π/6)] (dΘ/dt) + (- 475) [Tan (π/6)] = 0
[- 8.66 / (0.866)²] (dΘ/dt) - 475 (0.5774) = 0
(- 8.66 / 0.75) (dΘ/dt) - 274.27 = 0
- 11.55 (dΘ/dt) = 274.27
dΘ/dt = 274.27 / - 11.55
dΘ/dt = - 23.75
The angle is decreasing at a rate of 23.75 rad./hr.
dd/dt = - 475 km./hr. (Negative because the plane is approaching the camera)
Altitude, h = 5 km.
Angle, Θ = π/6 rad.
Find dΘ/dt:
Tan Θ = h / d
d = h / Tan Θ
d = 5 / Tan (π/6)
d = 5 / 0.5774
d = 8.66
Since d is decreasing,
d = - 8.66
d Tan Θ = h
Differentiating Implicilty Over Time,
d (1 / Cos² Θ) (dΘ/dt) + (dd/dt) Tan Θ = dh/dt
Since the altitude does not change, h is constant and the derivative of a constant is zero, so dh/dt = 0.
- 8.66 [1 / Cos² (π/6)] (dΘ/dt) + (- 475) [Tan (π/6)] = 0
[- 8.66 / (0.866)²] (dΘ/dt) - 475 (0.5774) = 0
(- 8.66 / 0.75) (dΘ/dt) - 274.27 = 0
- 11.55 (dΘ/dt) = 274.27
dΘ/dt = 274.27 / - 11.55
dΘ/dt = - 23.75
The angle is decreasing at a rate of 23.75 rad./hr.