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anonymous

  • 5 years ago

PLEASE HELP!! --> A balloon rises at a rate of 4 meters per second from a point on the ground 40 meters from an observer. What is the rate of change of the angle of elevation (in rad/sec) when the balloon is 40 meters above the ground?

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  1. JamesJ
    • 5 years ago
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    Step 1: draw a diagram. If you done that, tell us what an expression for angle of elevation, call it A in terms of 40 meters and the balloon's height, h. And what is height h as a function of time t?

  2. anonymous
    • 5 years ago
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    I don't understand

  3. JamesJ
    • 5 years ago
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    When the balloon is h meters off the ground, what is the trigonometric expression for the angle of elevation. It involves something like sin or cos or tan or cot or sec or cosec.

  4. anonymous
    • 5 years ago
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    sin

  5. JamesJ
    • 5 years ago
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    The horizontal distance of 40 m and the vertical height of h, form two sides of right-angled triangle. The observer is sitting at the end of the horizontal 40 m looking up at the balloon. The angle at which she's looking is the angle of elevation, call it A. What is the trig relationship between that angle, h and 40?

  6. JamesJ
    • 5 years ago
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    That is, she's looking along the diagonal towards the balloon.

  7. anonymous
    • 5 years ago
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    opp/adj

  8. JamesJ
    • 5 years ago
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    I'm still waiting for an equation in which A appears.

  9. JamesJ
    • 5 years ago
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    tan A = h/40 That's the equation.

  10. JamesJ
    • 5 years ago
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    Now, the height h = 4t, because it's rising at a constant velocity. Hence tan A = 4t/40 = t/10 or A = arctan(t/10). The question now asks you for dA/dt when the balloon is 40 meters off the ground. So you need to a) calculate dA/dt in general b) find the value of t for which h = 40 m c) evaluate dA/dt for that value of t.

  11. anonymous
    • 5 years ago
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    ok i got .24

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