Question

A particle is moving along the curve y=3sqrt(2x+5). As the particle passes through the point (2,9) , its x-coordinate increases at a rate of units per second. Find the rate of change of the distance from the particle to the origin at this instant.

Answer 1

ur missing dx/dt in your problem....

Answer 2

ohh its increasing at a rate of 2 units per second

Answer 3

let d = distance from the point to the origin...

so: \(\large d=\sqrt{x^2+y^2} \) or \(\large d^2=x^2+y^2 \)

since \(\large y=3\sqrt{2x+5} \), the equation becomes: \(\large d^2=x^2+9(2x+5)^2 \)

can you differentiate wrt time here?

Answer 4

im not sure a little confused

Answer 5

2d=2x+9(4x^2+10x+25) ?

Answer 6

2d=2x+9(8x+10) ?

Answer 7

maybe simplifying it a bit will help.....
\(\large d^2=x^2+9(2x+5)^2 \)
\(\large d^2=x^2+9(4x^2+20x+25) \)
\(\large d^2=x^2+36x^2+180x+225 \)
\(\large d^2=37x^2+180x+225 \)

so differentiating wrt time:
\(\large [d^2]'=[37x^2+180x+225]' \)
\(\large 2dd'=74xx'+180x' \)

is this ok ??? lemme know if you need explanation of any step...

Answer 8

i understand

Answer 9

since you're given x', and x, you'll need to find d....
do you know how to do that?

Answer 10

sub in for x and x prime?

Answer 11

not yet... find d first....

Answer 12

when x=2, y=9 (the point is given) find the distance from this point to the origin.

Answer 13

how do we do that?

Answer 14

use the formula above... \(\large d=\sqrt{x^2+y^2} \)

Answer 15

\(\large d=\sqrt{(\color {red}2)^2+(\color {red}9)^2} \) = ???

Answer 16

once you have d, you can substitute that value in along with x' into the equation:

\(\large 2dd'=74xx'+180x' \rightarrow d'=\frac{74xx'+180x'}{2d} \)

Answer 17

so now you substitute?

Answer 18

d=sqrt 85

Answer 19

ok... good...
so now just use the formula for d' (the rate at which the distance to the origin is changing) by replacing x=2, x'=2, d=sqrt(85)...

Answer 20

= 74(2)(2)+180*2/2sqrt)85) ?

Answer 21

= [74(2)(2)+180*2]/[2sqrt(85)]
if that's what you meant, then yes....