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hint ->
you want dy/dt
so:
dy/dt = (dy/dx) * (dx/dt)

which formula do i use?

you have dx/dt at this point and you have the point itself. that should help you

right.

got that.

dy/dt = (dy/dx) * (dx/dt)

dy/dx means derivative of y with respect to x
dx/dt is given at this point

i know.

so?

but what formula should i use, for example there are problems for volume, circle, triangle, etc.

distance formula?

what does s symbolize?

right.

s - distance

so that is it. now you have it all

thanks let me try

right the whole 5*sqrt(2x+2) is throwing me off though,

im not sure if i should find the derivative of that function with respect to time.

and the whole distance formula too.

cause
s=sqrt(x^2 + y^2) = sqrt(x^2+50x+50)
now
ds/dt = (ds/dx) * (dx/dt)

you just need to find ds/dx

i did this:
i found s=10.0499
then i found the derivative which i got 5/sqrt(2x+2) *dx/dt

i got ds/dt as 10 but i know it's wrong.

you dont need to find s.
you need to find ds/dx

ds/dx=10

what about dy?

since we express y in terms of x
we dont need to worry about it anymore.

ok

yeah im not sure what's happening.

plugged into the derivative x=1 and dx/dt = 4

understand what i did ?

sqrt(x^2+50x+50) ?

this is the distance..
s=sqrt(x^2 + y^2) = sqrt(x^2 + 50x + 50)

i know, how'd you come up with this sqrt(x^2+50x+50)

x^2+50x+50)

plugged y=5sqrt(2x+2)

10=10

50x+50?

i plugged in y and x

what ?

y=5sqrt(2x+2)
y^2 = 50x+50

plugged y=5sqrt(2x+2). i plugged in y and x.
10=10

so now
s = sqrt(x^2 + y^2)
but since y^2 = 50x+50
s = sqrt(x^2+50x+50)

not the value of y at the point. plug y as a function of x

okay thanks alot.

i got it

are you sure ?