(Easy) Calculus problem dealing with a runner

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In summary, a runner was located 6 miles east and 5000 feet north of an intersection where a water main broke, causing a circular puddle with a radius of 5t feet (t being the time in seconds elapsed since the main broke). The runner is moving west at a rate of 17 feet per second. To find when the runner's feet will first touch the edge of the water, the equations x= 31680-17t, y= 5000, and x^2+y^2= 25t^2 are used to find the intersection of the runner's path and the water's boundary. The resulting quadratic equation yields a solution of 25.415 minutes.
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The_ArtofScience
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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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  • #2
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.
 
  • #3
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.
 
  • #4
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
 

1. What is the formula for finding the average speed of a runner?

The formula for average speed is distance divided by time. In calculus, this can be represented as average speed = ∆distance / ∆time.

2. How can I use calculus to determine the instantaneous speed of a runner?

To find the instantaneous speed of a runner, we can use the derivative of the position function with respect to time. This will give us the slope of the tangent line at a specific point, which represents the runner's instantaneous speed.

3. Can calculus help me predict the runner's position at a certain time?

Yes, calculus can be used to create a position function that models the runner's movement. By plugging in a specific time value, we can find the corresponding position of the runner at that time.

4. How does calculus help us understand the runner's acceleration?

In calculus, acceleration is represented as the derivative of velocity with respect to time. This means that by taking the derivative of the velocity function, we can find the runner's acceleration at any given time.

5. Is there a way to use calculus to optimize a runner's speed?

Yes, we can use optimization techniques in calculus to find the maximum or minimum value of a function. In the case of a runner, we can use these techniques to determine the optimal speed that will result in the fastest time for a given race distance.

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