22. For the camera and rocket shown in Figure 2.8.. as radians and increasing at

Question

22. For the camera and rocket shown in Figure 2.8.. as radians and increasing at a rate of 0.2 rad/s7 the Earth. Its Earth is a 23. A satellite is in an elliptical orbit around arth is given by distance r (in miles) from the center of the 4995 ing at the stant? Is 1 +0). 1 2 cos ? where ? is angle nearest the asured from the point on the s surface (see the accompanying figure on arting t up, the of the next page) the altitude of the satellite at perigee (the point t the surface of the Earth) and at apogee (the point nearestt farthest from the surface of the Earth). Use 3960 mi as the radius of the Earth (b) At the instant when ? is 120°, the angle ? is increas- ing at the rate of 2.7°/min. Find the altitude of the

Explanation / Answer

Question(24):

The height of Aircraft, y = 4000 ft.
Speed of Aircraft, dx/dt = - 300 mi./hr.
Distance Between Aircraft and Observation Point = z
Angle Between Aircraft and Observation Point, ? = 30°

Find d?/dt:

Tan ? = y / x
x = y / Tan ?
x = 4000 / Tan 30
x = 4000 / 0.5774
x = 6928

z² = x² + y²
z² = (6928)² + (4000)²
z² = 47997184 + 16000000
z² = 63991731.98
z = ?63991784
z ? 8000

Cos ? = x / z
Cos ? = x / 8000
Cos ?= 1/8000 x

Differentiate Implicitly:

- Sin ?(d?/dt) = (1/8000)(dx/dt)
- 0.5(d?/dt) = (1/8000)(- 300)
- 0.5(d?/dt) = - 300 / 8000
- 0.5(d?/dt) = - 0.0375
d?/dt = - 0.0375 / - 0.5
d?/dt = 0.075

The angle is changing at a rate of 0.075°/sec.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
z² = x² + y²
z² = x² + (4000)²

2z(dz/dt) = 2x(dx/dt) + 0
z(dz/dt) = x(dx/dt)
8000(dz/dt) = 6928(- 300)
8000(- 2078400 / 8000
dz/dt = - 259.8

The distance is changing at a rate of - 260 mi./hr.

I hope this answer helps.
Thanks.

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