An ant travels from the point A(0,-63) to the point B(0,74) as follows. It first crawls straight to (x,0) with x>=0, moving at a constant speed of sqrt(2) units per second. It is then instantly teleported to the point (x,x). Finally, it heads directly to B at 2 units per second. What value of x should the ant choose to minimize the time it takes to travel from A to B?
I was able to get to \(Total time = \frac{\sqrt{x^2+\left(74-x\right)^2}}{2}+\frac{\sqrt{x^2+63^2}}{\sqrt{2}}\)
How do I find x such that it take "t" least amount of time
You have the correct Total Time function
We could use Calculus to solve this but the derivative is a little messy....I solved it here with a graph :
https://web2.0calc.com/questions/interesting-geometry
However...if you want me to, I think I can solve it with pure Calculus.....
Yes I did see your solution as 23.31 is the value of x. I assume that it would have some rounding because of sqrt. I am trying to get to the value of x in terms of sqrt root so I do not make any approximations. Would calculus help with that?
With calculus would it be d/dt(above function) -> 0? Could you please help with calculus
T = ( x^2 + (74 - x)^2 )^(1/2) / 2 + ( x^2 + 63^2) / sqrt (2)
T = ( x^2 + x^2 - 148 + 5476 )^(1/2) / 2 + ( x^2 + 3969)^(1/2) / sqrt (2)
T' = (1/4) ( 4x - 148)/ ( 2x^2 - 148x + 5476)^(1/2) + (2x) /[sqrt (2)(x^2 + 3969)^(1/2)] = 0
( x - 37) x
_____________________ + ______________________ = 0
(2x^2 - 148x + 5476)^(1/2) sqrt (2) (x^2 + 3969)^(1/2)
Note that we can write the denomnator of the first fraction as
2 ( x^2 - 74x + 2738)
So we can write
(x - 37) x
__________________________ + _____________________ = 0
sqrt (2) ( x^2 - 74x + 2738)^(1/2) sqrt (2) ( x^2 + 3969)^(1/2)
(x - 37) -x
____________________ = _________________ sqaure both sides
(x^2 - 74x + 2738)^(1/2) (x^2 + 3969)^(1/2)
(x - 37)^2 x^2
_______________ = ___________ cross-multiply
x^2 - 74x + 2738 x^2 + 3969
(x - 37)^2 ( x^2 + 3969) = x^2 ( x^2 - 74x + 2738) simplify
x^4 - 74 x^3 + 5338 x^2 - 293706 x + 5433561 = x^4 - 74 x^3 + 2738 x^2
After a little manipulation we get
2600 x^2 - 293706 x + 5433561 = 0
Putting this into the quadratic formula and evaluating we get that
x = [ 293706 - sqrt ( 293706^2 - 4*2600*5433561)] / (5200) =
[ 293706 - sqrt ( 29754180036) ] / 5200 =
[ 293706 - 172494 ] /5200 = 23.31 / 100 = 23.31 = x
(There is another value of x but it does not give us a minimum.......this is probably due to the fact that we squared the derivative )