Answer:
[tex]\frac{dx}{dt}=0[/tex]
Step-by-step explanation:
We have:
[tex]y^2+xy-3x=8[/tex]
Where both x and y are functions of t.
To find our solution, let's first take the derivative of both sides with respect to t:
[tex]\frac{d}{dt}[y^2+xy-3x]=\frac{d}{dt}[8][/tex]
Expand:
[tex]\frac{d}{dt}[y^2]+\frac{d}{dt}[xy]+\frac{d}{dt}[-3x]=\frac{d}{dt}[8][/tex]
Differentiate. We must differentiate implicitly. Also, for the second term, we must use the product rule. So:
[tex](2y\frac{dy}{dt})+(\frac{dx}{dt}y+x\frac{dy}{dt})-(3\frac{dx}{dt})=0[/tex]
Simplify:
[tex]2y\frac{dy}{dt}+\frac{dx}{dt}y+x\frac{dy}{dt}-3\frac{dx}{dt}=0[/tex]
We know that dy/dt is 3 when x is -4 and y is 2.
So, to find dx/dt, substitute 3 for dy/dt, -4 for x, and 2 for y. This yields:
[tex]2(2)(3)+\frac{dx}{dt}(2)+(-4)(3)-3\frac{dx}{dt}=0[/tex]
Simplify:
[tex]12+2\frac{dx}{dt}-12-3\frac{dx}{dt}=0[/tex]
Simplify:
[tex]-\frac{dx}{dt}=0[/tex]
Divide both sides by -1:
[tex]\frac{dx}{dt}=0[/tex]
最终答案是零 :)
