Assume that x and y are both differentiable functions of t and the required values of dy/dt and dx/dt
xy=6
a) Find dy/dt, given x=8 and dx/dt=12
b) Find dx/dt, given x=1 and dy/dt=-8
I started with a and got dy/dt=dx/dt(-y/x)
Plugging with the given values leaves me with dy/dt=12(-y/8)
There's a y that is not given, how would I find out y in this case???
Finally figured out, simply plug the given x value into the original equation!
This leaves with
dy/dt=12(-.75/8)
dy/dt=-1.125
dy/dt=-9/8
a) y = 6 x^-1
... dy/dt = -6 x^-2 dx/dt
... dy/dt = (-6 / 64) * 12
b) missing y a typo?
To find the value of y in this case, you can use the equation xy = 6, and substitute the given value of x into it. In part a), you are given x = 8.
Substituting x = 8 into xy = 6 gives you 8y = 6.
Next, solve the equation for y:
y = 6/8 = 3/4.
So, in part a), when x = 8 and xy = 6, the value of y is 3/4.
Now, you can substitute the values of x, y, and dx/dt into the equation dy/dt = dx/dt(-y/x) to find dy/dt.
dy/dt = 12(-3/4 / 8)
Simplifying this will give you the final value of dy/dt.