Calculus question involving a dog.?

Question

Rover, the dog who loves calculus, can run twice as fast as he can swim. On a sunny day, you take him to the beach, which is very straight and long, and throw a ball into the water for him to retrieve. The ball lands at point B, which is 5 meters offshore and 20 meters down the beach from Rover R. After calculating with his paws in the sand for a few seconds, Rover chooses the path that gets him to the ball as quickly as possible. What path does he choose?

Is there pythagorean theorem in this? Is dr/dt = 2s (i.e. two times swimming time)?

Answer 1

Let Rover run to E 20 - x metres from R and then swim to P which is the hypotenuse of a right angled triangle with sides x and 5

So Rover runs (20 - x) m and swims √(x² + 25) m

Now running rate = twice swimming rate.
So r = 2s

So time for running = (20 - x) /r and time for swimming = √(x² + 25)/s

Total time T = (20 - x) /2s + √(x² + 25)
= 1/(2s)*(20 - x + 2√(x² + 25)

dT/dx = 1/(2s)*( -1 + 2x/√(x² + 25)) = 0 for stationary points
ie -1 + 2x/√(x² + 25) = 0
ie √(x² + 25) = 2x
Square both sides
x² + 25 = 4x²
3x² = 25
x = ±5/√3
= ±(5√3)/3 Since x > 0 the negative result is ignored.
T = 1/2s (20 - (5√3)/3 + 2 √(100/3))
= 1/2s (20 - (5√3)/3 + 20/√3)
= 1/2s(20 + 5√3)

If he swam all the way T = √(425)/s = 5√(17)/s = 10√(17)/2s
If he walked to P and then swam straight in T = 20/2s + 5/s
= 30/2s

So the quickest time is when he runs [20 - (5√3)/3] m down the beach and swims the rest of the way

Answer 2

Wow that's a sweet math problem!!!!

ok, look at this diagram

________North
Water----Water---Water---BALL
Water----Water---Water---Water
ROVER-Beach---Beach---Beach
________South

Lets assume he runs 10 meters a minute and swims 5 meters a minute.

#1 If he runs 20 meters East (2 minutes) and then swims 5 meters
North (1 minutes) it takes him 3 minutes.

#2 If he swims straight for it diagonally, he goes a distance of the square root of 20 squared plus 5 squared, √(20^2 + 5^2), which is the square root of 425 and about 20.6, then it takes him about 4.12 minutes.

The quickest path is not #1.
There is a quicker path.

For example, if he runs 15 meters East (1.5 minutes) then he swims diagonally to the ball, going a distance of the square root of 5 squared plus 5 squared, √(5^2 + 5^2), which is the square root of 50 and about 7.1 (1.4 minutes), then it takes him about 2.9 minutes.

You need to find out how far horizantally from the ball that Rover should be when he starts to swim diagonally instead of run.

So, we have a triangle in which A is length East-West, B is height North South, and C is the diagonal. B is always 5 meters long.

We know that C^2 = A^2 + 5^2. So when will it be quicker for Rover to swim diagonally instead of continuing to run along the beach?
For this triangle, it takes him C/5 minutes to go diagonally.

So, in total, he is going to run 20-A meters, then swim C meters.
Remember, C^2 = A^2 + 25.
For now, sqrt(Q) = Square root of Q
Remember, sqrt(x)= x^0.5

TIme = (20-A)/10 + C/5 = 2 - (A/10) + (√(A^2 + 25))/5
Time = 2 - (A/10) + ( (A^2 + 25)^0.5 )/5

Now, we want Time to be as small as possible, right?

Now with derivative, whenever the derivative of a function equals zero, then there is a minimum, or a maximum there. Let's hope that we find a maximum.

Time = 2 - (A/10) + ( (A^2 + 25)^(1/2) )/5
Derivative of Time =
0 - 1/10 + (1/5)(1/2) ((A^2 + 25)^(-1/2)) (2A) =
-1/10 + (1/10)(2A) ((A^2 + 25)^(-1/2)) =
(Remember that F * (X^(-Z)) = F/(X^Z) )
-1/10 + (2A/10) / ((A^2 + 25)^(1/2)) =
Common divisors

-((A^2 + 25)^(1/2)) +2A
___________________ =
10 ((A^2 + 25)^(1/2))


2A - √(A^2 + 25)
___________________ =
10 √(A^2 + 25)


Conjugate? (A - B) (A + B) = A^2 - B^2

(2A - √(A^2 + 25) ) * (2A + √(A^2 + 25) )
________________________________________ =
10 √(A^2 + 25) * (2A + √(A^2 + 25) )



4A^2 - (A^2 + 25)
__________________________________ =
(20A √(A^2 + 25) ) + 10A^2 + 250


We are almost there...

4A^2 - A^2 - 25
_______________________________ =
10 (2A √(A^2 + 25) + A^2 + 25)



3A^2 - 25
_______________________________ =
10 (2A √(A^2 + 25) + A^2 + 25)


You can ignore the bottom, because what you want, is for the entire thing to equal 0, hence, 3A^2 - 25 must equal zero and all of the calculations to get the bottom were pointless.
3A^2 - 25 = 0
3A^2 = 25

A^2 = 25/3
A = 5/sqrt(3) = about 2.887

Okay?
So Rover ran 20- 5/sqrt(3) meters (1.71 seconds) then he swam diagonally
sqrt( (5/sqrt(3))^2 + 5^2 ) = sqrt ( 25/3 + 25) = sqrt (100/3) = 10/sqrt(3) meters (1.15 seconds)

Totaling about 2.86 seconds

I suggest that you give your answer in exacts (using the sqrt in your answer.)

Good luck.

Answer 3

Since I have no idea about the math involved, I will instead prove to you that a piece of paper is a stupid dog.

OK - a piece of paper is a plane, right?
And a printed piece of paper is an ink-lined plane...
An inclined plane is a slope up.
And a slow pup is a stupid dog!

[try reading it aloud if it don't work for ya]