Question

SOMEONE HELP ASAP!!

The figure above shows a Ferris wheel with radius 5 meters as Jalen, whose eye level is at point (0,2), watches his friend, Ashanti, ride in one of the cars as the wheel turns. Let Z denote the distance from Jalen to Ashanti’s car.The diagram indicates the center of the Ferris wheel at the point (12,7) and the position of Ashanti’s car at the point (x,y). If x and y are functions of time t, in seconds, what is the rate of change of Z when x=15, y=11, and dxdt=1 ? (The equation of a circle with radius r and center (h,k) is (x−h)2+(y−k)2=r2.)

Answer 1

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Answer:

(c) dZ/dt = 11/√544, so moving away at about 0.47 m/s

Explanation:

The (x, y) coordinates of the car are related by the fact that they are on a circle centered at (12, 7) with a radius of 5. Then their rates of change are related by the derivative of the circle equation with respect to time.

(x -12)^2 + (y -7)^2 = 25

2(x -12)x' +2(y -7)y' = 0

y' = -(x -12)/(y -7)x'

At the time and point of interest, we have ...

y' = -(15 -12)/(11 -7)(1) = -3/4

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The distance (z) to the observer is given by the Pythagorean theorem:

z^2 = (x -0)^2 +(y -2)^2

and its rate of change with time is ...

2z·z' = 2(x-0)x' +2(y -2)y'

The distance d at the point of interest is ...

z = √((15 -0)^2 +(11 -2)^2) = √(225 +81) = √306 = 3√34

So, the rate of change of distance to the observer at the time and point of interest is ...

z' = (x(x') +(y -2)y')/z

z' = ((15)(1) +(11-2)(-3/4))/(3√34) = (33/4)/(3√34)

z' = 11/√544 ≈ 0.47 . . . m/s