Question

Please help! (: Been stuck on this problem for 2 hrs now.

A particle is moving along the curve y=3(sqrt(4x+4)). As the particle passes through the point (3, 12), its x-coordinate increases at a rate of 4 units per second. Find the rate of change of the distance from the particle to the origin at this instant.

Answer 1

looks like you need to set up parametric equations, because this deals with 3 variables:
x, y, and t (time)

Answer 2

How do I go about doing that.
I started with \[d ^{2}=x^{2}+9(4x+4)\]
and got
\[d^{2}=x^{2}+36x+36\]

then I used the points (0,0) and (3,12) to find the distance. but I don't know where to go from there.

Answer 3

I got 2dd'=2xx'+6x'
=2(4)(4)+6(4)
=56

Answer 4

^what's that?that ain't integration

Answer 5

oh my bad..i had to say derivative..GOSH why did i say that

Answer 6

first of all differentiate this one \[d ^{2}=x ^{2}+36x+36\] @pdd21

Answer 7

@pdd21 That answer doesn't look plausible. At t = 0, the particle is at (3,12). 1 second later, it's at \((7,3\sqrt{4*7+4})\) or \((7,12\sqrt{2})\). If you take the distance between those points, \[d = \sqrt{(7-3)^2+(12\sqrt{2}-12)^2} \approx 6.4\]

Answer 8

wouldn't it be 2x+36?
@stgreen

Answer 9

2dd'=2x+36
where d'=rate of change of distance w.r.t x-coordinate=4

Answer 10

and d would be \[\sqrt{153}\] right?
@stgreen

Answer 11

what?why?

Answer 12

yes, \(3\sqrt{17}\) is distance from origin to (3,12)

Answer 13

oh yeah right

Answer 14

I used the distance formula to find d?
sorry, I suck at math.
@stgreen

Answer 15

So now that I have
2dd'=2x+36 and the distance where do I go from there?

Answer 16

if you have distance d..you should know x too

Answer 17

\[d = \sqrt{(x-0)^2+(3\sqrt{4x+4} - 0)^2} = \sqrt{x^2+9(4x+4)} = \sqrt{x^2+36x+36}\]\[\frac{d}{dx}[(x^2+36x+36)^{1/2} = \frac{1}{2}*(2x+36)(x^2+36x+36)^{-1/2}\]Evaluate that at \(x=3\), multiply by \(dx/dt = 4\) and you've got your answer....

Answer 18

^yeah right

Answer 19

The way you layed out the problem made so much more sense. I see where I went wrong.. I derived wrong. Thank you so much! I got 6.79 and it was correct! @whpalmer4

Answer 20

You're welcome. I knew there was something wrong with what you were doing, but I hadn't figured out just what!