A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 5 years ago

I'm drowing in calculus over here. An observer stands 200 meters from the launch site of a hot-air balloon. Balloon rises 4m/sec at a constant rate. How fast is the angle of elevation of the balloon increasing 30sec after the launch? Help mehh!

  • This Question is Closed
  1. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hey, how's it going?

  2. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Dont drown!! haha, hold on.

  3. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok, you helping michael?

  4. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i got 1/6 radians/sec as an answer. i don't even know if i was suppose to get radians

  5. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Okay, so first we need to draw a picture. You can turn radians into degrees by multiplying 180/π and degrees into radians by multiplying by π/180. So, here's the picture O - [] | | | | 4m/s | \O/ | | | | |____________________Θ/\ |----200m--------| Just imagine there is a diagonal between the balloon and the person and you have the triangle. Notice that the balloon is traveling up at a constant rate of 4m/sec, so we can use this to calculate the height of the balloon at 30sec using this equation speed = distance/time, speed * time = distance, 4m/sec * 30sec = 120m We will need this in order to solve for the rate of change in the angle. Now, we can use the tan(Θ) = Opposite/Adjacent to solve for what the angle is at 30sec as well. \[\tan (\Theta)=O/A, \tan(\Theta)=120m/200m, \tan^{-1}(120m/200m) = \Theta \] So, Θ=30.9638 degrees @ t=30sec Now, will make a little change to the diagram by making the opposite side of the angle a variable, like so: O - [] | | | | y | \O/ | | | | |____________________Θ/\ |----200m--------| because we know as the balloon rises its height relative to the ground is changing as well as the angle. So, now we have this equation: \[\tan(\Theta)=y/200m\]Take the derivative \[\sec ^{2}(Θ)Θ\prime=y\prime/200m\]We know y'=4m/sec and theta = 30.9638 degrees at t=30sec so all we do is plug those in and solve for the rate of change in theta, i.e. Θ' \[Θ\prime=(4m/s) / (200m*\sec^{2}(30.9638))=(4m/s * \cos^{2}(30.9638))/200m\] because sec^2(Θ) = 1 / cos^2(Θ) So,\[Θ'=.0147059\]degrees/sec or \[Θ'=.0147059*π/180=0.257\]radians/sec

  6. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    whoops i meant 1/68 radians per sec not 1/6 so i guess i got the same answer just thought it was radians not degrees which confuses me a little

  7. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    tan(a) = y/200 (d/dt)(tan(a)) = (d/dt)(y/200) (da/dt) sec^2(a) = (dy/dt)(1/200) da/dt = a' and dy/dt = y' a' sec^2(a) = y'/200 solve for a' a' = y'/200(sec^2(a)) y' = 4 a' = 4/200(sec^2(a)) a' = 1/50(sec^2(a)) sec(a) = "c"/200 where c=sqrt(a^2 + b^2) c = sqrt(200^2 + 120^2) ...30sec(4m/sec) = 120 c = sqrt(40000 + 14400) = sqrt(54400) c = 233.24 sec(a) = 233.24/200 sec(a)^2 = 1.08 a' = 1/[50(1.08)] a' = 0.0185 radians per sec if I did it right :) might need to double check meself tho

  8. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    You can do it that way, but that's the long way. You made an error at sec^2(a) = 1.08 (233.24/200)^2 = 1.36002 a' = 1/(50 * 1.36002) = .0147057 radians/sec I made an error to saying that .0147057 was degrees/sec when it should have been radians/sec

  9. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thanx for the check :)

  10. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    After getting some sleep, yeah....i went for a long walkabout on that one :) tan(a) = 120/200 = 12/20 = 6/10 = 3/5 which makes the triangle 3^2 + 5^2 = c^2 and c=sqrt(34). sec(a) = sqrt(34)/5 sec^2(a) = 34/25 a' = 4/(200(34)/25) = 4/(8(34)) = 1/(2(34)) = 1/68 a' = (1/68) radians per sec; or if we want the french version: a' = 0.0147058823529412 radians per sec :)

  11. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.