y^2+xy-3x=-5
dy/dt=-4
x=3
y=1
find dx/dt.
i tried doing 2y(dy/dt)+ydx/dt-3=0 but i am not getting the right answer. Please tell me where i am going wrong.
your derivative is wrong, the xy needs a product rule
2y dy/dt + x dy/dt + y dx/dt - 3dx/dt = 0
plugging in our stuff
2 (-4) + 3(-4) + 1(dx/dt) - 3 dx/dt = 0
-2 dx/dt = 20
dx/dt = -10
you need to use the product rule for xy
x dy/dt + y dx/dt
Thank you for the help! Completely forgot about that rule... woops
To find dx/dt, you can differentiate the given equation y^2+xy-3x=-5 with respect to t (time). Applying the chain rule, we have:
d/dt(y^2) + d/dt(xy) - d/dt(3x) = d/dt(-5)
Differentiating each term with respect to t:
2y(dy/dt) + x(dy/dt) + y(dx/dt) - 3(dx/dt) = 0
Now, plug in the given values:
2(1)(-4) + 3(-4) + (1)(dx/dt) - 3(dx/dt) = 0
-8 -12 + dx/dt - 3dx/dt = 0
Combine like terms:
-20 - 2dx/dt = 0
Rearrange the equation to solve for dx/dt:
-2dx/dt = 20
Divide by -2:
dx/dt = 20/-2
Simplifying further:
dx/dt = -10
Therefore, the value of dx/dt is -10.
It seems you made an error when differentiating the term "xy" with respect to t. The correct application of the chain rule for differentiating xy is:
d/dt(xy) = x(dy/dt) + y(dx/dt)
To avoid mistakes, it's important to carefully apply the chain rule when differentiating terms involving multiple variables.