I got this question in my advanced functions class for the trig unit and don't understand it. Please help!

A circular dining room at the top of a skyscraper rotates in a counterclockwise direction so diners can see the entire city. A woman sits next to the window ledge and places her purse on the ledge. Eighteen minutes later she realizes that her table has moved but her purse is on the ledge where she left it. The coordinates of her position are (x,y) = (20cos(7.5t)°, 20 sin(7.5t)°), where t is the time in minutes and x and y are in metres. What is the shortest distance she has to walk to retrieve her purse?

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The woman travels in a circle given by the equation x(t) = 20cos(7.5t), y(t) = 20sin(7.5t). Thus the circle's radius is sqrt(x^2 + y^2) = 20 metres.

Take the purse as θ1 = 0°. Then after 18 min, the woman sits at (-14.1, 14.1) by ( x(t),y(t) ); thus θ2 = 135°.

The shortest distance to the purse is the chord from θ1 to θ2 (length d). Draw a line from the center to the chord bisecting the angle to form two identical right triangles. Now,

d/2 = 20Sin(135/2) = 18.48 m

**d = 36.96 m**

We assume three possiblities. (i) the window is attached to the rotating room itself. (ii) window is outside the dyning room and not a rotating part and the rotating room could be crossed in any direction and (iii) the window is out side and you can't cross the rotating room. You have to go out and reach the window by the nearest path along and outside of the circumference of the room.

The given details indicate that the when t = 0, window coordinates =(20, 0) suggests the radius of the rotating room is 20 meter.

When t = 18 minutes, the coordinates ( x,y)= {20 cos (7.5*18),20 sin(7.5*18) }. This indicates that the room has rotated by 7.5*18 dgerees = 135 degrees or 135/360 = 0.375 of a rotation.

Case(i): She need not walk any distance.

case (ii): The coordinates of the window ledge (20 sin 7.5*0 , 20 cos 7*5*0) = (20sin0, 20 cos0) = (20*0, 20*1) = (0 , 20). The the coordinates of the woman after 18 minutes = { 20cos(7.5*18), 20sin(7.5*18) } = (20cos 135, 20 sin135) = {(-20/(2^0.5) , 20/(2^0.5)}

Therefore the chord distance between the points of widow ledge and the lady = { square of the diifference of x coordintes of the window ledge and that of the lady+square of the diiference of the y cordinates of the window ledge and that of the lady}^(0.5)

= sqrt{[20+20/(2^0.5)]^2+ [0-20/(2^05]^2 }

=20*(2+sqrt2) = 36.9552 m

Case (iii):

You can reach the window ledge coming out of the rotating room and walk along the circumferece of 135 degree clock wise or another way (360 -135)= 225 degree anti clockwise.

Going about 135 degree clockwise the possible nearest.That is (135/360)(2*pi*r), where r = 20 meter, the radius of the dining room.So, (135/360)(2*pi*r) = (135/360)(2*pi*20) = 47.1234 meter but not less.

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