Posted by Lucy on Thursday, September 22, 2011 at 11:09pm.
An object is moving along the graph of f (x)= x2 . When it reaches the point (2,4) the x coordinate of the object is decreasing at the rate of 3 units/sec. Give the rate of change of the distance between the object and the point (0,1) at the instant when the object is at (2,4).

calculus  Reiny, Friday, September 23, 2011 at 7:35am
Let P(x,y) be any point on f(x)
then the distance D between P and (0,1) is such that
D^2 = (x0)^ + (y1)^2
= x^2 + (x^2  1)^2
= x^2 + x^4  2x^2 + 1 = x^4  x^2 + 1
2D dD/dt = 4x^3 dx/dt  2x dx/dt
so at (2,4), x=2 and dx/dt = 3
and D^2 = 4 + 9 = 13
D = √13
√13 dD/dt = 4(8)(3)  2(2)(3)
dD/dt = 84/√13 = appr. 23.3 units/sec
check my arithmetic.

calculus  Steve, Friday, September 23, 2011 at 4:42pm
OK until you forgot that it was 2D dD/dt, and you just plugged in √13. So, the answer is half what you showed.
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