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
Answer is b ,-4x^2 - 13y^2
Answer: You need to combine like terms the answer is 2y+2x
slope =(y2-y1)/(x2-x1)
m=(-1-5)/(1--1)
m=-6/2
m=-3
y-y1=m(x-x1) point slope form
y-5 = -3 (x--1)
y-5 = -3 (x+1)
distribute
y-5 = -3x -3
add 5 to each side
y = -3x+2
Choice D