Answer:
(a) [tex]\frac{dy}{dt}=-3\frac{3}{4}[/tex]
(b) [tex]\frac{dx}{dt}=3\frac{3}{4}[/tex]
Step-by-step explanation:
[tex]x^{2} +y^{2}=25[/tex]
Take [tex]\frac{d}{dt}[/tex] of of each term.
[tex]\frac{d}{dt}(x^{2})+\frac{d}{dt}(y^{2})=\frac{d}{dt}(25)\\\\(\frac{d}{dx}(x^{2})*\frac{dx}{dt}) +(\frac{d}{dy}(y^{2})*\frac{dy}{dt})=\frac{d}{dt}(25)\\\\2x\frac{dx}{dt} +2y\frac{dy}{dt} = 0\\\\[/tex]
For Question a
[tex]2y\frac{dy}{dt}=-2x\frac{dx}{dt}\\\\\frac{dy}{dt}=\frac{-2x\frac{dx}{dt}}{2y} \\\\\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}[/tex]
Given that x = 3, y = 4, and dx/dt = 5.
[tex]\frac{dy}{dt}=-\frac{3}{4}*5=-\frac{15}{4}\\ \\\frac{dy}{dt}=-3\frac{3}{4}[/tex]
For Question b
[tex]2x\frac{dx}{dt}=-2y\frac{dy}{dt}\\\\\frac{dx}{dt}=\frac{-2y\frac{dy}{dt}}{2x} \\\\\frac{dx}{dt}=-\frac{y}{x}\frac{dy}{dt}[/tex]
Given that x = 4, y = 3, and dx/dt = -5.
[tex]\frac{dx}{dt}=-\frac{3}{4}*-5=\frac{15}{4}\\ \\\frac{dx}{dt}=3\frac{3}{4}[/tex]
