calculus
posted by Lucy on .
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).

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. 
OK until you forgot that it was 2D dD/dt, and you just plugged in √13. So, the answer is half what you showed.