Related Rates Problem

  • Thread starter cmajor47
  • Start date
  • #1
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Homework Statement


A balloon is rising at a constant speed of 5 ft/s. A boy is cycling along a straight road at a speed of 15 ft/s. When he passes under the balloon, it is 45 ft above him. How fast is the distance between the boy and the balloon increasing 3 seconds later.


Homework Equations


dx/dt=15 ft/s
dy/dt=5 ft/s
c2=x2+y2

The Attempt at a Solution


2c dc/dt = 2x dx/dt + 2y dy/dt
c(dc/dt) = x(dx/dt) + y(dy/dt)
c(dc/dt) = x(15) + y(5)
c(dc/dt) = 15x + 5y
My first problem is: Is the 45ft x or y?
 

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

c(dc/dt) = 15x + 5y
My first problem is: Is the 45ft x or y?

Hi cmajor47! Welcome to PF! :smile:

45 ft is the value of y at time 0 (and 0 ft is the value of x at that time).

In your equation, c(dc/dt) = 15x + 5y, c x and y must be the values at time 3. :smile:
 
  • #3
HallsofIvy
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There was NO "x" or "y" in the original problem! Since you were the one who wrote dx/dt= 5 ft/s and were told "A balloon is rising at a constant speed of 5 ft/s." I guess you were taking x to be the height of the ballon at any given time. What was the height of the ballon at the time in question?
 
  • #4
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Don't I have to assign a letter value to each piece of information? Did I assign them correctly?
 
  • #5
HallsofIvy
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No one can tell you what letters are "correct"- it is entirely your choice. You just have to be consistent.

What did YOU choose x and y to represent when you set up the equation? What are values of the quantities represented by x and y at the point in time in question?
 
  • #6
57
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I realized how to solve the problem. Thanks to everyone for their help.
 

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