Homework Help: DIFF. CALCULUS

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 x-coordinate 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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