SOLUTION: A point is moving along the circle x2+y2= 25 in the first quadrant in such a way that its x coordinate changes at the rate of 2cm/sec. how fast is its y coordinate changing as the

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Question 650829: A point is moving along the circle x2+y2= 25 in the first quadrant in such a way that its x coordinate changes at the rate of 2cm/sec. how fast is its y coordinate changing as the point passes through (3,4)?
Answer by Edwin McCravy(20086) About Me  (Show Source):
You can put this solution on YOUR website!
A point is moving along the circle x² + y² = 25

>>...x coordinate changes at the rate of 2cm/sec...<<
The vector pointing right is %28dx%29%2F%28dt%29 = 2 cm/sec (right because it's
positive. The vector pointing down is %28dy%29%2F%28dt%29 = ? cm/sec.
So the point is moving clockwise, and since the point is moving downard
toward the x-axis, %28dy%29%2F%28dt%29 vector is pointing downward, so we can
expect its speed to be to be negative.  


>>how fast is its y=coordinate changing as the point passes through (3,4)?<<
We want to know: %28dy%29%2F%28dt%29 = ? when x=3 and y=4

CAUTION:
x and y are varying as the point is moving before it reaches (3,4),
so DO NOT substitute for x or y until after we have taken the derivative.
Only then do we freeze the motion at the point (3,4)

                  x² + y² = 25

            2x%28dx%29%2F%28dt%29 + 2y%28dy%29%2F%28dt%29 = 0

Divide through by 2

              x%28dx%29%2F%28dt%29 + y%28dy%29%2F%28dt%29 = 0

                     y%28dy%29%2F%28dt%29 =  -x%28dx%29%2F%28dt%29

                      %28dy%29%2F%28dt%29 =  -x%2Fy·%28dx%29%2F%28dt%29

Now finally we can freeze the motion by substituting x=3, y=4, %28dx%29%2F%28dt%29 = 2

                      %28dy%29%2F%28dt%29 =  -3%2F4·(2)

                      %28dy%29%2F%28dt%29 =  -3%2F2

                      %28dy%29%2F%28dt%29 =  -1.5 cm/sec

That means that at the instant when the point passes through (3,4) 
it is falling toward the x-axis at a rate of 1.5 cm/sec.

Edwin