Related Rates - Finding dx/dt
I've attempted this problem 11 times now using veriest resources and I can't get it. Here's the problem I'm trying to solve:
Suppose that x = x(t) and y = y(t) are both functions of t. If y^2 + xy -3x = 5
and dy/dt = 1 when x = -1 and y = -1, what is dx/dt?I'm trying to start by converting the above into this:
2y + x (dx/dy) * y (dy/dt) - 3 = 0but it never comes out to the correct answer once I plug in the values so now i'm completely lost.
Thanks for your help!
$\endgroup$ 43 Answers
$\begingroup$The equation $y^2+xy-3x=5$ becomes, after differentiation using the chain rule and the product rule,
$$2y\frac{dy}{dt}+y\frac{dx}{dt}+x\frac{dy}{dt}-3\frac{dx}{dt}=0$$
Can you continue from there?
$\endgroup$ 2 $\begingroup$By taking the derivative of $y^2 + xy -3x = 5$ with respect to $t$, we get $$2y\frac{dy}{dt}+x\frac{dy}{dt}+\frac{dx}{dt}y-3\frac{dx}{dt}=0.$$ By substitution, $$-2-1-\frac{dx}{dt}-3\frac{dx}{dt}=0,$$ implying that $$\frac{dx}{dt}=-\frac34\cdot$$
$\endgroup$ $\begingroup$Doesn't $\frac{d}{dt} (y^2 +xy - 3x = 5) \Rightarrow 2y\frac{dy}{dt} + y\frac{dx}{dt} + x\frac{dy}{dt} - 3\frac{dx}{dt} $ ??
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