Question

SOMEONE HELP ASAP!!

Answer 1

9514 1404 393

Answer:

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

Step-by-step 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