Question

how do you find -8/9 from this?

Find dy/dt when x=-2, dx/dt= 2, x^3 + y^3 =19.

So far I have got 3(-2)^2 (2) + 2( )^2 dy/dt= 0
x^3 + y^3 =19
(-2)^3 + y^3= 19= -8 + y^3 = 19

what do i do next

Answer 1

This is how I did it, though I'm not sure if it's the easiest way:\[x^{3}+y^{3}=19\] Use implicit differentiation with respect to t:\[3x^{2}\frac{dx}{dt}+3y^{2}\frac{dy}{dt}=0\]Rearrange to get dy/dt:\[\frac{dy}{dt} = \frac{-x^{2}}{y^{2}}\frac{dx}{dt}\]Using the original equation, solve for y:\[y = \sqrt[3]{19-x^{3}}\]Then substitute y in the derivative:\[\frac{dy}{dt} = \frac{-x^{2}}{(\sqrt[3]{19-x^{3}})^{2}}\frac{dx}{dt}\] You are given both x and dx/dt, so plug in those values and solve.\[\frac{dy}{dt} = \frac{-(-2)^{2}}{(\sqrt[3]{19-(-2)^{3}})^{2}}(2)\]\[\frac{dy}{dt} = \frac{-8}{(\sqrt[3]{19+8})^{2}}\]\[\frac{dy}{dt} = \frac{-8}{(\sqrt[3]{27})^{2}}\]\[\frac{dy}{dt} = \frac{-8}{9}\]