Question

5 mark for review let ( x ) and ( y ) be functions of time ( t ) such that the sum of ( x ) and ( y ) is constant. which of the following equations describes the relationship between the rate of change of ( x ) with respect to time and the rate of change of ( y ) with respect to time? a ( \frac{d x}{d t}=\frac{d y}{d t} ) b ( \frac{d x}{d t}=-\frac{d y}{d t} ) c ( \frac{d x}{d t}+\frac{d y}{d t}=\frac{d k}{d t} ), where ( k ) is a function of ( t ) d ( \frac{d x}{d t}+\frac{d y}{d t}=k ), where ( k ) is a function of ( t )

Explanation:

Step1: Set up the equation

Since \(x + y\) is constant, let \(x + y = C\) (where \(C\) is a constant).

Step2: Differentiate both sides with respect to \(t\)

Differentiate \(x + y = C\) with respect to \(t\). Using the sum rule of differentiation \(\frac{d}{dt}(u + v)=\frac{du}{dt}+\frac{dv}{dt}\), we get \(\frac{d}{dt}(x)+\frac{d}{dt}(y)=\frac{d}{dt}(C)\).

Step3: Evaluate the derivatives

We know that \(\frac{dC}{dt} = 0\) (the derivative of a constant is \(0\)), and \(\frac{d}{dt}(x)=\frac{dx}{dt}\), \(\frac{d}{dt}(y)=\frac{dy}{dt}\). So, \(\frac{dx}{dt}+\frac{dy}{dt}=0\).

Step4: Solve for \(\frac{dx}{dt}\)

Subtract \(\frac{dy}{dt}\) from both sides of the equation \(\frac{dx}{dt}+\frac{dy}{dt}=0\). We obtain \(\frac{dx}{dt}=-\frac{dy}{dt}\).

Answer:

B. \(\frac{dx}{dt}=-\frac{dy}{dt}\)