A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
A balloon released at point A rises vertically with a constant speed of 5m/s. Point B is level with and 100m distant from point A. How fast is the angle of elevation of the balloon at B changing when the balloon is 200m above A?
anonymous
 5 years ago
A balloon released at point A rises vertically with a constant speed of 5m/s. Point B is level with and 100m distant from point A. How fast is the angle of elevation of the balloon at B changing when the balloon is 200m above A?

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let at any time t height of balloon is x m. and angle of elevation is y...we have to find dy/dt. tan y=x/100 (sec^2y)dy/dx=1/100 dy/dx=1/(100sec^2y) (dy/dt)(dt/dx)=1/(100sec^2y) dy/dt=1/(100sec^2y) .(dx/dt) put the value from question dy/dt=1/(100sec^2y).(5)....................1 now sec^2y=1+tan^2y=1+(200/100)^2=5 put this value in equation 1 dy/dt=1/500.(5) dy/dt=1/100
Ask your own question
Sign UpFind more explanations on OpenStudy
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
 Engagement 19 Mad Hatter
 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.