Question

21:11 1076 - 78% h 0 5 5 2 Question 4 It is known that pigeons fly faster over land than over water. Assume a pigeon flies 10 metres per second over land but only 8 metres per second over water. If a pigeon is released at the edge of a straight river 500 metres wide and must fly to its home 1300 metres away diagonally across the river. What path would minimise its flying time? Pigeon Home (b) if (distance travelled) = (rate) x (time) show that the time taken by the pigeon is given by Vx2 + 5002 1200 - X T'(x) = = 8 10 - +- (c) Show that the value of x which minimises the time taken for the pigeon to return to its home is 2000/3 metres. (d) Use the First Derivative test to check the answer in part c) is a minimum. (You do not need to check the end points of x.)

Answer 1

From the given information we get,

From triangle ABC we get,

500 m 1300 m B xm C 1200 m

AB + BC = ACC

i.e.,

AC? = 500+24

i.e.,

AC = V5002 +22

From triangle ABD we get,

BD = ADP - AB

i.e.,

BD= 1300 - 500

i.e., $BD^2=1690000-250000$

i.e.,

BD2 = 1440000

i.e., $BD=\sqrt{1440000}$

i.e., $BD=1200$

and, $CD=BD-BC$

i.e., $CD=1200-x$

Therefore, the distance covered by pigeon over water is
V5002 + 2
meters and the distance covered by pigeon over land is meters.

Now, the time taken by pigeon to cover
V5002 + 2
meters over water is $\frac{\sqrt{500^2+x^2}}{8}$ seconds and the time taken by pigeon to cover meters over land is $\frac{1200-x}{10}$ seconds.

If T(x) is the total time taken by pigeon, then we have,

$T(x)=\frac{\sqrt{500^2+x^2}}{8}+\frac{1200-x}{10}$

(c) Graphing the above function we get,

500 (2000/3,157.5) 200 400 600 800 1000 1200

This shows that the required value of x is : x = 2000/3.

(d) Here, $\frac{d}{dx}T(x)=0$

i.e., $\frac{x}{8\sqrt{500^2+x^2}}-\frac{1}{10}=0$

i.e., $\frac{x}{8\sqrt{500^2+x^2}}=\frac{1}{10}$

i.e., $10x=8\sqrt{500^2+x^2}$

i.e., $100x^2=64(500^2+x^2)$

i.e., $100x^2=64*500^2+64x^2$

i.e.,

100.r? - 64x2 = 64 * 500

i.e., $36x^2=64*500^2$

i.e., $x^2=\frac{64*500^2}{36}$

i.e., $x^2=\frac{8^2*500^2}{6^2}$

i.e., $x=\frac{8*500}{6}$

i.e., $x=\frac{4000}{6}$

i.e., $x=\frac{2000}{3}$

Therefore, the required value is : $x=\frac{2000}{3}$ .