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).

Answer 1

Let P(x,y) be any point on f(x)
then the distance D between P and (0,1) is such that
D^2 = (x-0)^ + (y-1)^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.

Answer 2

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