Question

please help :)

A particle is moving along the curve y=5sqrt(4x+4). As the particle passes through the point (3,20) its x-coordinate increases at a rate of 3 units per second. Find the rate of change of the distance from the particle to the origin at this instant.

Answer 1

I have...
dy/dt=(3)(0.3)[(4x+4)^(-1/2)](5dx/dt)

Answer 2

dy/dt = 4(0.5)[(4x+9)^(-1/2)](4dx/dt)
so when x=4
dy/dt = (8/5)(dx/dt)=8 units/sec
D = (x^2+y^2)^0.5
dD/dt = 0.5(x^2+y^2)^(-0.5) (2dx/dt+2ydy/dt)
= (5 + 20*8) / sqrt(16+400) = 165/20.396=8.089 units/sec

Answer 3

that's not right.. that's what I had the first time and it said it was wrong.. I don't know what I'm doing wrong.

Answer 4

@Chlorophyll can you help me? I have done this problem over and over and can't seem to get the right answer?

Answer 5

guess you need the distance formula right?

Answer 6

I don't know what I need.. I obviously can't seem to get it right... It's bothering me :/

Answer 7

or at least the square of the distance
square of the distance from \((0,0)\) so \((x,5\sqrt{4x+4})\) is
\[d^2=x^2+25(4x+4)\]

Answer 8

sqrt(409) right?

Answer 9

we can simplify this a little
\[d^2=x^2+100x+100\] and you are looking for \(d'\)
taking derivatives wrt \(t\) using the chain rule we get
\[2dd'=2xx'+100x'\]

Answer 10

maybe easier to start with
\[dd'=xx'+100x'\] and you are told that \(x'=3\) and also that \(x=3\) so you can solve for \(d'\)

Answer 11

oh damn i mean
\[dd'=xx'+50x\] sorry

Answer 12

dd'=159?

Answer 13

now \(x=3,x'=3\) so we get
\[dd'=9+150=159\] right

Answer 14

you want \(d'\) so all that is left is to find \(d\) which you also know since it is the distance between \((0,0)\) and \((3,20)\)

Answer 15

i get \(d=\sqrt{409}\) making \(d'=\frac{159}{\sqrt{409}}\) but my arithmetic sucks, so i wouldn't bet money on it

Answer 16

that is the idea however

Answer 17

I got sqrt(409) as well.

Answer 18

ok then i am happy with that answer

Answer 19

you?

Answer 20

my gosh, thank so much...

Answer 21

yeah, it's correct!

Answer 22

yw, just hope it is right
notice, however, that when working with distance problems it is much easier to work with the square of the distance
makes all the computations a lot nicer