# circular problem

• Jan 6th 2009, 04:40 PM
jenny woodland
circular problem
I need help with this.... Can anyone help me??? Many thanks

The figure shows a circular field with centre O, and radius 2 km. AC is the diameter of the circle, and B lies on the circumference. Jack cycles along the straight line from A to B at a speed of 12 km/hr. He then walks along the straight line from Bto C at a speed of 5 km/hr. Given that the distance from B to C is x km, and the total time taken is T hours, find the value of x so that Jack can complete his journey in the shortest time possible.
• Jan 6th 2009, 05:03 PM
skeeter
a triangle inscribed in a semicircle is a right triangle.

$\displaystyle AC = 4$ km

$\displaystyle BC = x$ km

$\displaystyle AB = \sqrt{16 - x^2}$ km

distance/speed = time

$\displaystyle T = \frac{\sqrt{16 - x^2}}{12} + \frac{x}{5}$

find $\displaystyle \frac{dT}{dx}$ and minimize ... don't forget about endpoint extrema.
• Jan 6th 2009, 05:16 PM
jenny woodland
what is endpoint extrema?
Hi skeeter, thanks for your answer. But erm can you teach me what is endpoint extrema? My teacher didn't teach me about it and this is the first time I come across this term. I know that I'm a bit of a noob but I hope you can teach me as I want to know more. And thanks again!!!!
• Jan 6th 2009, 06:05 PM
skeeter
what is the domain of the function $\displaystyle T(x)$ ?

have you found $\displaystyle \frac{dT}{dx}$ and completed an analysis of the function $\displaystyle T(x)$ ?
• Jan 6th 2009, 06:08 PM
jenny woodland
Re: Circular problem
I work out the problem. Can the answer be 3.69 km?
• Jan 6th 2009, 06:13 PM
skeeter
is x = 3.69 the location of a minimum or a maximum?

how do you tell?
• Jan 6th 2009, 06:48 PM
jenny woodland
Domain of T= -1.538 to 1.538

dT/dx = -12x / [144 (√16-x)] + 1/5

and 3.69 is a maximum.

Am I correct? If not, can you correct my mistakes? Thanks!
• Jan 7th 2009, 02:18 PM
skeeter
the domain for x is 0 $\displaystyle \leq x \leq 4$

T(3.69) is a maximum ... the function T(x) increases and then decreases because T'(x) changes sign from (+) to (-).

This means that the minimum for T(x) is either T(0) or T(4) ... endpoint minimums. the absolute min is the lesser of the two.