(Easy) Calculus problem dealing with a runner

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Homework Help Overview

The problem involves a runner moving towards a circular puddle formed by a broken water main, with the radius of the puddle increasing over time. The runner's initial position and speed are given, and the objective is to determine when the runner's feet will get wet as the puddle expands.

Discussion Character

  • Exploratory, Problem interpretation, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss setting up equations for the positions of both the runner and the water, exploring the relationship between their movements and the geometry of the situation.
  • Some participants question the necessity of differentiation, suggesting that the problem may be solvable through algebraic manipulation of the equations derived from the positions.
  • There is mention of using the quadratic formula to find the time at which the runner reaches the edge of the puddle.

Discussion Status

The discussion is ongoing, with participants sharing different methods of approaching the problem. Some have provided equations and attempted to solve for time, while others express confusion about the problem's assignment in relation to the current chapter's content.

Contextual Notes

Participants note a potential mismatch between the problem and the chapter material, indicating that the problem may be challenging given the current learning context.

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Homework Statement



Water is flowing from a major broken water main at the intersection of two streets. The resulting puddle of water is circular and the radius r of the puddle is given by the equation r = 5t feet, where t represents time in seconds elapsed since the main broke.

When the main broke, A runner was 6 miles east and 5000 feet north of the intersection. The runner is due west at 17 feet per second. When will the runner's feet get wet?



The Attempt at a Solution



Not exactly sure how to solve this problem but here's all I got:

- Triangle, on the x-axis labelled it 31, 680 ft, on the y-axis labelled it 5,000 ft. The hypotenuse is then 32, 072.1437 ft.

- Diffierientiated x^2 + y^2 = r^2 and got 2x dx/dt + 2y dy/dt = 2 r dr/dt

- Looked back at the equation r = 5t, found r' to be 5, so dr/dt =5

- Just pluged the 5 back in like any algebra problem and got

5,000dx/dt + 31,680(-17) = 5(32072.1437)

dx/dt = 139. 784144

Not sure where I went wrong, I plugged the equation back into r = 5t and got 27.95 -- the book says 25.4154041 minutes so there's something wrong here
 
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Alright, well, I'm not sure if I did this the way you were supposed to, but...
I made equations for the locations of both the runner and the water.

For the runner, his x position: x = 31680-17t, and his y position: y = 5000.
Since the water forms a circle, the equation of which is x^2+y^2 = r&2, the water's location can be written as: x^2 + y^2 = (5t)^2 = 25t^2.

You want to find when the runner's position is the same as the edge of the water's position. So, you take the equations for x and y (from the runner) and put them into the equation for the water, giving you: (31680-17t)^2 + (5000)^2 = 25t^2.

With the help of a calculator and after moving all the terms to one side, you get:
264t^2-1077120t+1028622400=0.

Plug those numbers into the quadratic equation and you'll get two answers for t (in seconds). Take the smaller of those two answers (when the runner FIRST gets to the edge of the water) and divide by 60 to get 25.415 minutes.
 
Essentially, you are asking for the intersection of two graphs. The circle giving the boundary of the water is x2+ y2= 25t2 and the runners path is given by x= 6(5280)- 17t, y= 5000. Replacing x and y in the circle formula by those gives you (6(5280)- 17t)2+ (5000)2= 25t2, a quadratic equation for t. I see no reason to differentiate. This is just an algebra problem.
 
Thank you both for making that distinction. I am still scratching my head why this problem was assigned when the chapter is a total mismatch
 

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