Please help with a related rates baseball problem?

Question

A baseball diamond is a square 90 ft on a side. Chris is runnig from second base to third base at 20 ft/s. Sam is running from third base to home plate at 15 ft/s. When chris is 30ft from third base and Sam is 40 ft from third base, is the distance between them increasing or decreasing? At what rate?

Answer 1

1) The distance from Sam to third base is s = 40 + 15 t, the distance from Chris to third is c =30 - 20 t, where t is the time in seconds.

2) The distance x between the two runners is given by Pythagoras.
x^2 = s^2 + c^2 = 1600 + 1200 t + 225 t^2 + 900 - 1200 t + 400 t^2
x^2 = 2500 + 625 t^2
x = 25 * sqrt ( 4 + t^2)

3) The rate of change v = dx/dt = 25 t/ sqrt ( 4 + t^2)
This is positive for t>0, but equals 0 for t=0. So at that particular moment the rate of change is zero.

Answer 2

Place the diamond on a number plane so that home base is the origin (0, 0) and first base B1 ≡(90, 0).
Then second base B2 ≡(90, 90) and third base B3 ≡(0, 90)

Chris is at C ≡(30 - 20t, 90) and Sam is at S ≡(0, 40 - 15t)

|CS|² = ((30 - 20t) - 0)² + (90 - (40 - 15t))²
= (30 - 20t)² + (50 + 15t)²
= 25(6 - 4t)² + 25(10 + 3t)²
= 25(36 - 48t + 16t² + 100 + 60t + 9t²)
= 25(136 + 12t + 25t²)

So CS = 5√(136 + 12t + 25t²)
As t increases (136 + 12t + 25t²) increases So CS increases.
ie the distance between them is increasing

dCS/dt = 5/2 * (12 + 50t)/[√(136 + 12t + 25t²)]
= 5(12 + 50t)/[√(136 + 12t + 25t²)]
When t = 0 (at that given point)
dCS/dt = 60/√(136) ft/s

i e they are moving away from each other at a speed of 60/√(136) ft/s (~5.14 ft/s)

Answer 3

Imagine the situation 1 second before and 1 second after. Use triganometry to find the hypotenuse distance.

1 second before: Chris 50ft away, Sam 25ft away => 56ft apart
Now: Chris 30ft away, Sam 40ft away => 50 ft apart
1 second after: Chris 10ft away, Sam 55ft away => 56ft apart

So at that instance, they are 50ft apart and the rate is 0. They are not moving towards or away from each other.

Answer 4

first find the correlation between all values. this would be a triangle and pythagoreans (sp?) thm. would be the relation.

a^2 + b^2 = c^2

take derivative:

(2*a*a') + (2*b*b') = (2*c*c')

you have the values for everything except z', this would be the rate the distance between them is changing.

a = 40 ft
a' = 15 ft/s
b = 30 ft
b' = -20 ft/s
c = 50 = sqrt(30^2 + 40^2)

after solving this you will find that the rate the distance between the two players is not changing. and the distance between them is increasing.

Answer 5

x = third to second
y = third to home
z = sqrt(x^2 + y^2)

dz/dt = (2x*dx/dt + 2y*dy/dt)/sqrt(x^2 + y^2)

x = 30, dx/dt = -20, y = 40, dy/dt = 15

dz/dt = (-1200 + 1200)/50 = 0
So it is neither decreasing nor increasing.

Answer 6

It's increasing, tops 90FT!