Related Rates Help: Solve Distance Change in 1 Sec

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Discussion Overview

The discussion revolves around a related rates problem involving a balloon rising vertically and an automobile traveling horizontally beneath it. Participants explore how to determine the rate at which the distance between the two is changing after one second, incorporating aspects of geometry and calculus.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses uncertainty about how to begin solving the problem, despite familiarity with related rates.
  • Another participant suggests that external factors could affect the rates of change, implying that the situation may not be straightforward.
  • One participant proposes a method involving a right triangle, defining the vertical side as 200 + 15t and the horizontal side as 66t, and calculates dx/dt to be 2.9 ft/s at t=1 second.
  • A different participant describes a similar approach, using the relationship d^2 = x^2 + y^2, and derives dd/dt to be 33.7 ft/s at t=1 second after substituting the known values.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct rate of change, as two different calculations yield different results (2.9 ft/s vs. 33.7 ft/s). The discussion remains unresolved regarding the correct approach and final answer.

Contextual Notes

There are potential limitations in the assumptions made by participants regarding the conditions of the problem, such as the neglect of external influences on the balloon's path and the specific definitions of the variables involved.

ashleyk
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A balloon is 200 ft. off the ground and rising vertically at the constant rate of 15 ft/sec. An automobile passes beneath it traveling along a straight road at the constant rate of 45 mph or 66 ft/sec. How fast is the distance between them changing 1 sec. later?

i know how to do related rates problems...i just don't know where to start with this one...please help!
 
Last edited:
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Do you know what the answer is?

Personally I would just say: "There is no telling what the rate of change of the balloon and of the car will be 1 second later"... maybe a big wind will blow the balloon sideway and maybe in 0.5 seconds later the car will crash into a tree.
 
here is my suggestion, have to wait for some one else to check to see if its right.

i formed a right angled triangle
the vertical side has length 200+15t, the horizontal side has length 66t and let the hypotenuse be x.
we want to find dx/dt
using pythagorus we get x^2=291t^2+600t+40000
from this calculate dx/dt
pop in t=1 second to get dx/dt=2.9ft/s
 
I did sort of the same thing, creating a right triangle with sides x, y, and hyp of d. the taking d^2=x^2+y^2, then taking the derivative, getting d(dd/dt)=x(dx/dt)+y(dy/dt), then subbing in the given numbers, and finding d after one second, I get (224.9ft)(dd/dt)=(66ft)(66ft/sec)+(215ft)(15ft/sec), and when solved, I get (dd/dt)=33.7ft/sec at t=1
 

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