A plane flying with a constant speed of 5 km/min passes over a ground radar station at an altitude of 4 km and climbs at an angle of 35 degrees. At what rate, in km/min is the distance from the plane to the radar station increasing 10 minutes later?

I got it down to [250-20cos(125)]/[sqrt(40000-400cos(125))] but it's wrong.

User Icon for Reiny Reiny answered
15 years ago

I would call the time after passing over the station as t minutes

then by cosine law
D^2 = 4^2 + (5t)^2 - (2)(4)(5t)cos125
= 25t^2 * (40cos125)t + 16

I usually don't take the square root, but rather do it implicitly, so
2D(dD/dt) = 50t + 40cos125
dD/dt = (25t + 20cos125)/D

when t=10
D^2 = 2500 - 400cos125 + 16
D = 47.818
and dD/dt = (250 + 20cos125)/47.818
= 4.988 km/min

User Icon for Z32 Z32 answered
15 years ago

Not right for some reason.

User Icon for Reiny Reiny answered
15 years ago

re-checked my calculations, found a few typing and calculator errors. The old brain just doesn't work properly after midnight

NEW VERSION:

I would call the time after passing over the station as t minutes
then by cosine law
D^2 = 4^2 + (5t)^2 - (2)(4)(5t)cos125
= 25t^2 - (40cos125)t + 16

I usually don't take the square root, but rather do it implicitly, so
2D(dD/dt) = 50t - 40cos125
dD/dt = (25t - 20cos125)/D

when t=10
D^2 = 2500 - 400cos125 + 16
D = 47.818
and dD/dt = (250 - 20cos125)/52.3969
= 4.99 km/min

check my calculations

User Icon for Z32 Z32 answered
15 years ago

Ahh, there we go. Can't believe such a little number could cause it to be wrong. Thanks.

User Icon for Explain Bot Explain Bot answered
11 months ago

To solve this problem, we can use trigonometry and the concept of rates. Let's break it down step by step:

1. Label the relevant quantities:
- Distance from the plane to the radar station: Let's call it d (in km).
- Altitude of the plane: 4 km.
- Angle of climb: 35 degrees.
- Speed of the plane: 5 km/min.
- Time: 10 minutes later.

2. Find the horizontal and vertical components of the plane's motion:
- The horizontal component is given by cos(35) * speed = cos(35) * 5.
- The vertical component is given by sin(35) * speed = sin(35) * 5.

3. Use the distance formula to relate the variables:
- In a right triangle, the hypotenuse (d) can be calculated using the Pythagorean theorem:
d^2 = (horizontal component)^2 + (vertical component + altitude)^2.

4. Differentiate the equation with respect to time (t):
- Take the derivative of both sides of the equation implicitly with respect to t:
2d * dd/dt = 2(horizontal component) * (d(horizontal component)/dt) + 2(vertical component + altitude) * (d(vertical component + altitude)/dt).

5. Simplify the equation and find the rate of change of distance, dd/dt:
- Substitute the given values: horizontal component = cos(35) * 5, vertical component = sin(35) * 5, altitude = 4.
- Substitute the rate of change of altitude (d(vertical component + altitude)/dt) = 0 (since it is not changing).
- Solve for dd/dt by substituting t = 10 minutes (remember to convert it to km/min).

By following these steps, you should be able to calculate the correct answer.