assume x and y are functions of t. evaluate dy/dt given that y² - 2x³ = 0; dx/dt = -5, x = 2, y = 4. answer…

Explanation:

Step1: Differentiate both sides with respect to t

Using the chain - rule, the derivative of $y^{2}$ with respect to $t$ is $2y\frac{dy}{dt}$, and the derivative of $2x^{3}$ with respect to $t$ is $6x^{2}\frac{dx}{dt}$. So, differentiating $y^{2}-2x^{3}=0$ with respect to $t$ gives $2y\frac{dy}{dt}-6x^{2}\frac{dx}{dt}=0$.

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

Rearrange the equation $2y\frac{dy}{dt}-6x^{2}\frac{dx}{dt}=0$ to isolate $\frac{dy}{dt}$. First, add $6x^{2}\frac{dx}{dt}$ to both sides: $2y\frac{dy}{dt}=6x^{2}\frac{dx}{dt}$. Then divide both sides by $2y$ to get $\frac{dy}{dt}=\frac{6x^{2}\frac{dx}{dt}}{2y}=\frac{3x^{2}\frac{dx}{dt}}{y}$.

Step3: Substitute the given values

We are given that $\frac{dx}{dt}=-5$, $x = 2$, and $y = 4$. Substitute these values into the formula for $\frac{dy}{dt}$: $\frac{dy}{dt}=\frac{3\times2^{2}\times(-5)}{4}$. First, calculate $2^{2}=4$. Then $3\times4\times(-5)=-60$. So, $\frac{dy}{dt}=\frac{-60}{4}=-15$.

Answer:

$-15$