assume that (x = x(t)) and (y = y(t)). find (\frac{dx}{dt}), using the following information.(x^{2}+y^{2}=1.1…

Explanation:

Step1: Differentiate both sides with respect to t

Differentiate $x^{2}+y^{2}=1.13$ with respect to $t$ using the chain - rule. The derivative of $x^{2}$ with respect to $t$ is $2x\frac{dx}{dt}$, and the derivative of $y^{2}$ with respect to $t$ is $2y\frac{dy}{dt}$, and the derivative of the constant 1.13 with respect to $t$ is 0. So we get $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$.

Step2: Solve for $\frac{dx}{dt}$

First, factor out 2 from the left - hand side: $2(x\frac{dx}{dt}+y\frac{dy}{dt}) = 0$, then $x\frac{dx}{dt}+y\frac{dy}{dt}=0$. Rearrange the equation to solve for $\frac{dx}{dt}$: $\frac{dx}{dt}=-\frac{y}{x}\cdot\frac{dy}{dt}$.

Step3: Substitute the given values

We are given that $x = - 0.7$, $y = 0.8$, and $\frac{dy}{dt}=-2$. Substitute these values into the formula for $\frac{dx}{dt}$: $\frac{dx}{dt}=-\frac{0.8}{-0.7}\cdot(-2)$. Calculate $-\frac{0.8}{-0.7}\cdot(-2)=-\frac{1.6}{0.7}=-\frac{16}{7}$.

Answer:

$-\frac{16}{7}$