Posted by KAREEN on Thursday, November 18, 2010 at 8:43am.
A particle is moving along the parabola 4y = (x + 2)^2 in such a way that
its xcoordinate is increasing at a constant rate of 2 units per second. How
fast is the particle's distance to the point (2, 0) changing at the moment
that the particle is at the point (2, 4)?

DIFF. CALCULUS  Reiny, Thursday, November 18, 2010 at 9:15am
Let the point be at the position P(x,y)
The parabola has vertex at (2,0), so the question is how fast is P moving away from the vertex.
Let the distance be D
D^2 = (x+2)^2 + y^2
= (x+2)^2 + (x+2)^4/16
2D dD/dt = 2(x+2) dx/dt + (1/4)(x+2)^3 dx/dt
dD/dt = [2(x+2) dx/dt + (1/4)(x+2)^3 dx/dt]/(2D)
when x = 2
D^2 = 16 + 256/16
= 32
D = √32
dD/dt = [2(4)(2) + (1/4(64)(2)]/(2√32)
= 24/√32 = 6/√2 = appr. 4.243 units per second
check my arithmetic
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