Please help with Calc problem??

posted by .

The angle of elevation of the camera has to change at the correct rate in order to keep the rocket in sight. Also the mechanism for focusing the camera has to take into account the increasing distance from the camera to the rising rocket. Let's assume the rocket rises vertically and that its speed is 800 ft/s when it has risen 4000 ft.
How fast is the distance from the camera to the rocket changing in ft/s at that moment?


If the televison camera is always kept aimed at the rocket, how fast is the camera's angle of elevation changing in radians/s at that moment

  • Please help with Calc problem?? -

    It would help if we knew the distance of the camera to the launch pad, which would give us a constant.
    The way you have it involves 3 variables:
    the angle, the height, and the distance between rocket and camera.
    Are you sure there wasn't more information?

  • Please help with Calc problem?? -

    Sorry, I meant a television camera is positioned 2500 ft from the base of a rocket launching pad. The angle of elevation of the camera has to change at the correct rate in order to keep the rocket in sight. Also the mechanism for focusing the camera has to take into account the increasing distance from the camera to the rising rocket. Let's assume the rocket rises vertically and that its speed is 800 ft/s when it has risen 4000 ft.
    How fast is the distance from the camera to the rocket changing in ft/s at that moment?


    If the televison camera is always kept aimed at the rocket, how fast is the camera's angle of elevation changing in radians/s at that moment?

  • Please help with Calc problem?? -

    ok, that's better

    make a sketch of the situation, labeling the angle as Ø and the height as h, and the hypotenuse c.

    preliminary:
    when h = 4000
    c^2 = 2500^2 + 4000^2
    c = 4716.99
    tanØ = 4000/2500 = 1.6
    Ø = 57.995°
    at that moment: we have a similar triangle with base 5, height 8 and hypotenuse √89
    cosØ = 5/√89 ----> needed later
    sinØ = 8/√89 ----> needed later
    secØ = √89/5 ----> needed later

    tanØ = h/2500
    2500tanØ = h
    2500 sec^2Ø dØ/dt = dh/dt
    2500(89/25) dØ/dt = 800
    dØ/dt = 25(800)/(2500(89)) = .08988764

    in general:
    sinØ = h/c
    csinØ = h
    c cosØ dØ/dt + sinØ dc/dt = dh/dt

    for our given:
    4716.999(5/√89) (.08988764) + (8/√89) dc/dt = 800

    I will leave it up to you to do the "button-pushing" to find dc/dt

    also, please check my arithmetic

  • Please help with Calc problem?? -

    I haven't checked my answer against Reiny's, but I also took a stab at it in

    http://www.jiskha.com/display.cgi?id=1372720314

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Help! Urgent!

    You know tangentTheta=h/4 take the derivative sec^2 Theta * dTheta/dt= 1/4 dh/dt you know dtheta/dt solving for dh/dt You know sec Theta from the right triangle: sec theta = sqrt(h^2 +16) /h check me. Georgina: in the future, please …
  2. physics

    A hot air balloon is descending at a rate of 2.4 m/s when a passenger drops a camera. (a) If the camera is 50 m above the ground when it is dropped, how long does it take for the camera to reach the ground?
  3. pysics

    A hot-air balloon is descending at a rate of 1.5 when a passenger drops a camera.If the camera is 41 above the ground when it is dropped, how long does it take for the camera to reach the ground?
  4. Geometry One Question

    Three security cameras were mounted at the corners of a triangular parking lot. Camera 1 was 151 ft from camera 2, which was 122 ft from camera 3. Cameras 1 and 3 were 139 ft apart. Which camera had to cover the greatest angle?
  5. Calculus

    A recording camera is located 2500 feet from the launch pad for recording a rocket launch. As the rocket lifts off, the angle of elevation of the camera increases 4 degrees per second while recording the launch. Find the instantaneous …
  6. geometry

    Three security cameras were mounted at the corners of a triangular parking lot. Camera 1 was 151 ft from camera 2, which was 122 ft from camera 3. Cameras 1 and 3 were 139 ft apart. Which camera had to cover the greatest angle?
  7. Physics

    A hot-air balloon is descending at a rate of 1.7 m/s when a passenger drops a camera. If the camera is 44 m above the ground when it is dropped, how long does it take for the camera to reach the ground?
  8. Calculus

    A television camera is positioned 2500 ft from the base of a rocket launching pad. The angle of elevation of the camera has to change at the correct rate in order to keep the rocket in sight. Also the mechanism for focusing the camera …
  9. Calculus

    A television camera is positioned 2500 ft from the base of a rocket launching pad. The angle of elevation of the camera has to change at the correct rate in order to keep the rocket in sight. Also the mechanism for focusing the camera …
  10. Geometry

    three security cameras were mounted at the corners of a triangular parking lot. camera 1 was 110 ft. from camera 2, which was 137 ft from camera 3. cameras 1 and 3 were 158 ft. apart. which camera had to cover the greatest angle?

More Similar Questions