Question

How do I find the time when a line intersects a circle with a constantly changing radius?

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=7t feet where t represents time in seconds elapsed since the main broke.

(a) When the main broke, a runner was located 6 miles from the intersection. The runner continues toward the intersection at the constant speed of 15 feet per second. When will the runner's feet get wet?

(b) Suppose, instead, that when the main broke, the runner was 6 miles east, and 5000 feet north of the intersection. The runner runs due west at 15 feet per second. When will the runner's feet get wet? (Round your answer to one decimal place.)

Answer 1

24 minutes, about 24 minutes 58 seconds

Explanation:

For part a, we can solve this in one dimension. The runner will run right into the puddle, toward the center. Therefore, we consider the puddle expanding at 7 t and the runner getting closer at 15 t. Therefore, the distance between them is 6 miles - 7t - 15t = 6 * 5280 - 22 t

We can solve when this equals 0:
`d = 5280 - 22 t = 0 implies t = 5280 / 22 = 1440 s = 24 min`

For part b, we have to consider the absolute distance. We know that the North/South distance of the runner is a constant 5000 and the East/West distance is shrinking as 6 miles - 15t, as before. However, now, we just need to consider when the absolute distance is 7t. We can apply the distance formula:
`7t = sqrt(5000^2 + (6 * 5280 - 15t)^2)`
`49t = 225t^2 - 30 * 6 * 5280 t + (36 * 5280^2+5000^2)`

Doing some factoring to simplify just a little bit,
`0 = 2^4*11 t^2- 2^7 3^3 5^2 11 t + 2^6 5^2 (2^6 3^4 11^2 + 5^6)`
`0 = 11 t^2- 2^3 3^3 5^2 11 t + 2^2 5^2 (2^6 3^4 11^2 + 5^6)`
`0 = 11t^2 - 59400 t + 64288900`

You could apply the quadratic formula here, or just plug it into a calculator and get two values: `t = 1497.7, 3902.3`
These represent when the runner enters the puddle and when the runner exits the puddle, something we don't care about.

Therefore, the solution to b is `t = 1497.7 s approx 24 min 58 s`