# suppose..

• Mar 1st 2010, 12:21 PM
tbenne3
suppose..
• Mar 1st 2010, 12:32 PM
skeeter
$\displaystyle \frac{d}{dt}\left(xy = 3\right)$

$\displaystyle x \cdot \frac{dy}{dt} + y \cdot \frac{dx}{dt} = 0$

take it from here?
• Mar 1st 2010, 12:38 PM
tbenne3
Quote:

Originally Posted by skeeter
$\displaystyle \frac{d}{dt}\left(xy = 3\right)$

$\displaystyle x \cdot \frac{dy}{dt} + y \cdot \frac{dx}{dt} = 0$

take it from here?

is it -3?
• Mar 1st 2010, 12:53 PM
featherbox
$\displaystyle \frac{dx}{dt}=\frac{dx}{dy}\frac{dy}{dt}$
• Mar 1st 2010, 03:12 PM
tbenne3
• Mar 1st 2010, 03:50 PM
skeeter
Quote:

Originally Posted by tbenne3

this is not that difficult.

$\displaystyle xy = 3$

if $\displaystyle x = -3$ , then $\displaystyle -3y = 3$ ... $\displaystyle y = -1$

you were given $\displaystyle \frac{dy}{dt} = -3$ ... now use the derivative equation to solve for $\displaystyle \frac{dx}{dt}$
• Mar 1st 2010, 03:53 PM
tbenne3
Quote:

Originally Posted by skeeter
this is not that difficult.

$\displaystyle xy = 3$

if $\displaystyle x = -3$ , then $\displaystyle -3y = 3$ ... $\displaystyle y = -1$

you were given $\displaystyle \frac{dy}{dt} = -3$ ... now use the derivative equation to solve for $\displaystyle \frac{dx}{dt}$

sorry just not following.. what is the derivative of dx/dt
• Mar 1st 2010, 04:17 PM
skeeter
$\displaystyle x \cdot \frac{dy}{dt} + y \cdot \frac{dx}{dt} = 0$

$\displaystyle (-3) \cdot (-3) + (-1) \cdot \frac{dx}{dt} = 0$

solve for the value of $\displaystyle \frac{dx}{dt}$
• Mar 1st 2010, 04:41 PM
tbenne3
Quote:

Originally Posted by skeeter
$\displaystyle x \cdot \frac{dy}{dt} + y \cdot \frac{dx}{dt} = 0$

$\displaystyle (-3) \cdot (-3) + (-1) \cdot \frac{dx}{dt} = 0$

solve for the value of $\displaystyle \frac{dx}{dt}$

thanks