Question

related rate problem: help please. (attached)

Answer 1

Recall the distance formula:
\[d^2=(x-x_0)^2+(y-y_0)^2\]
In this particular problem, you have \((x_0,y_0)=(0,0)\) (the origin), since you want to find the distance between any given point on the curve \((x,y)\) and the origin.
\[d^2=x^2+y^2\]
Substitute the given equation of the curve:
\[d^2=x^2+\left(4\sqrt{2x+2}\right)^2\\
d^2=x^2+16(2x+2)\\
d^2=x^2+32x+32\]
Now differentiate both sides implicitly with respect to some dummy variable \(t\):
\[2d\frac{dd}{dt}=(2x+32)\frac{dx}{dt}\]
You're told that at the point \((1,8)\), \(x\) is increasing at a rate of 3 units/s, which translates to \(\dfrac{dx}{dt}=3\).
\[2d\frac{dd}{dt}=(2\cdot1+32)(3)\\
2d\frac{dd}{dt}=102\]
You're asked to find \(\dfrac{dd}{dt}\), so first you need to find \(d\). To do this, find the distance between \((1,8)\) and \((0,0)\), then plug it into the above equation. Lastly, solve for \(\dfrac{dd}{dt}\).

Answer 2

@SithsAndGiggles omg thank you!! :)

Answer 3

Yer welkum

Answer 4

@SithsAndGiggles hey, i found the distance which is 8 right? then i plugged in 8 into the equation and got 51/8... however, something is wrong?

Answer 5

i get \(\frac{19}{8}\)

Answer 6

oh nvm i got it.

Answer 7

it's 6.32577

Answer 8

to find the distance is this:

Answer 9

i got it... thank you! @satellite73

Answer 10

\[d^2=x^2+32x+32\]
\[2dd'=2xx'+32x'\]
\[dd'=xx'+32x'\]
\[8d'=3+96\]

Answer 11

oh right doh

Answer 12

distance isn't 8 is it?
distance is \(\sqrt{65}\)

Answer 13

which is pretty close to 8...

Answer 14

yes

Answer 15

is the final answer \(\frac{99}{\sqrt{65}}\) ?

Answer 16

close yes... i thought i had to do this: x2-x1/y2-y1

Answer 17

it's 51/sqrt(65)/65

Answer 18

oh damn i forgot to divide 32 by 2
must be past my bedtime

Answer 19

lol