A triathlete on the swimming leg of a triathlon is 150.0 m from the shore ( a ).

Question

A triathlete on the swimming leg of a triathlon is 150.0 m from the shore (a). The triathlete's bike is 80.0 m from the shore on the land (b). The component of her distance from the bicycle along the shore line, ( x + y in the diagram), is 207.6 m.

a) If the triathlete's running speed is 8.40 m/s and swimming speed is 1.680 m/s, calculate the value of x so that the triathlete reaches her bike in the least amount of time.

b) Calculate the minimum time required to reach the bicycle

- I set up the equation where the Total time = (sqrt(150^2 +x^2) / 1.680) + (sqrt(80^2 + (207.6-x^2)) / 8.4) and I know you are supposed to take the derivative of this equation and set it equal to zero and solve for x, but I got stuck trying to simplify the equation after taking the derivative. Could someone show me how they did it? Or a simplier way to do it?

Explanation / Answer

T =( (1502 + X2)1/2/1.68) +( (802 + (207.6-x)2)1/2 / 8.4 )
On differentiation
dT/dx = 0
dT/dx = (1502+x2)-1/2(2x/1.68) + (802 + (207.6-x)2)-1/2*2(207.6 -x)(-1) = 0
(1502+x2)-1/2(2x/1.68) = (802 + (207.6-x)2)-1/2*(2(207.6 -x) /8.4)
(2x/1.68) / (1502+x2)1/2 = (2(207.6 -x) /8.4) / (802 + (207.6-x)2)1/2
On squaring both side
x2 /(1.682*(1502+x2)) = (207.6 -x)2 / (8.42*(802 + (207.6-x)2))
Now on solving we get
x = 27.877 m

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